First, consider this question from 2001:

Derivatives and Antiderivatives When I run the formulaint(sqrt(cos(x^2)),x=0..1)in Maple it just returns the formula. I know it's because there is no antiderivative, butwhy isn't there one?

Robert is trying to calculate a **definite integral**, $$\int_0^1\sqrt{\cos(x^2)}dx$$ using the mathematics software Maple. With the numeric option, that would produce a number, as Wolfram Alpha does today:

But his request didn’t work, because it was trying to produce a “closed-form expression”, not just a numeric value, and it couldn’t. For comparison, here is what WA gives when I ask it for the **indefinite integral**:

The question is, why doesn’t that work?

Doctor Rob answered:

Thanks for writing to Ask Dr. Math, Robert. There are many, many functions that behave similarly. For any of them, its antiderivative exists, but there isno expression for it in closed form in terms of familiar functions and constants. Examples are x^x, e^(x^2), e^(e^x)), e^(-x)/x, 1/ln(x), ln(1-x)/x, sin(x)/x, sin(x^2), sin(sin(x)), sqrt(sin(x)), e^sin(x), ln(sin(x)), sqrt(x^3+a*x^2+b*x+c), and so on.

Let’s see what Wolfram Alpha does for each of these; the differences are illuminating:

$$\int x^x dx$$

$$\int e^{x^2}dx$$

$$\int e^{e^x}dx$$

$$\int\frac{e^{-x}}{x}dx$$

$$\int\frac{1}{\ln(x)}dx$$

$$\int\frac{\ln(1-x)}{x}dx$$

$$\int\frac{\sin(x)}{x}dx$$

$$\int\sin(x^2)dx$$

$$\int\sin(\sin(x))dx$$

$$\int\sqrt{\sin(x)}dx$$

$$\int e^{\sin(x)}dx$$

$$\int\ln(\sin(x))dx$$

$$\int\sqrt{x^3+ax^2+bx+c}dx$$

SomeĀ of these, it just gives up on, and provides only an infinite **series** or **approximation**; others define a **new function** of their own; and others are expressed *in terms of* those new functions: \(\mathrm{erf}\), \(\mathrm{erfi}\), \(\mathrm{Ei}\), \(\mathrm{li}\), \(\mathrm{Li}_n\), \(\mathrm{Si}\), \(S\), \(\mathrm{H}_n\), \(E(x|m)\). We’ll be discussing these options as we proceed.

(Incidentally, when I am helping someone with an integral that looks difficult, I will sometimes give it to Wolfram Alpha just to see whether it’s worth trying; if the answer looks like any of these above, I ask the student to check whether they’ve copied the problem correctly!)

Now, back to the question:

The nonexistence of such an expression is a consequence of the fact that we only have afairly small restricted set of functionsto use to try to express antiderivatives: constant, rational, irrational algebraic, exponential, logarithmic, trigonometric, inverse trigonometric, hyperbolic, and inverse hyperbolic functions. This may seem like a big list to you, but it isnot big enoughfor those purposes.

These basic functions arise from algebra or trigonometry; we can find the derivative of all of them in terms of others, but their antiderivatives open out into a bigger universe, so to speak.

One way to think of this is that the antiderivative is the solution of a certaindifferential equation: dy/dx = sqrt(cos(x^2)). You probably are aware that not all such solutions are expressible that way.Bessel functionsand many other so-called "special" functions that crop up in engineering and physics arise as such solutions.

A course in differential equations necessarily includes a study of the use of infinite series, because many important equations can’t be solved any other way; the special functions are invented to provide an easier way to talk about the solutions.

A similar situation exists with trying to solve thegeneral quintic equation. The rootscannot be expressedin terms of the equation's coefficients using the above functions. The rootsexist, and are continuous functions of the coefficients, but those functionsjust don't have simple expressions. Sorry!

Note the important distinction here: The solutions *exist*; we just can’t find a *formula* for them. We have a quadratic formula that gives the solutions of a general quadratic equation \(ax^2+bx+c=0\), and there are (much more elaborate) “formulas” for cubic and quartic equations; but there is no such formula to solve *every* quintic, \(ax^5+bx^4+cx^3+dx^2+ex+f=0\), even though every such equation has at least one real solution (and 5 complex solutions). The solution of *some* can be known exactly (such as \(2x^5-13x^4+6x^3+48x^2-56x+16=0\), whose solutions are \(-2,\frac{1}{2},2,3-\sqrt{5},3+\sqrt{5}\).

(I made up the equation starting with the solutions – otherwise there would be little chance of solving it.)

If I make a little change, say to \(2x^{5}-12x^{4}+6x^{3}+48x^{2}-56x+16=0\), we can no longer find exact solutions, but we can still see that they are there:

I often tell students that this is the dark secret of math teachers: We teach you only the problems for which we have methods; if you gave us a random problem, even in algebra, we probably couldn’t solve it. Math is a wild forest, but we give tours mostly of the cultivated orchards.

Anyway, the same is true of integrals: Many can’t be found exactly, and yet the function is still *integrable*: there is an antiderivative, but we just can’t write it out.

A 1996 question takes us deeper:

Integrating X^x, Closed Form (1) How would you go aboutintegrating x^x? I don't have a clue where to start. (2) Also, how do you express the equationy = xcosx in terms of y? For both these problems, I have asked other people who have told me thatthere is no simple solution. If this is the case, please send the solution anyway, or at least give me some idea of the theory on which it is based.

Only the first question is directly related to our topic, but the second has an interesting indirect relationship. In both cases, it’s not so much that there is no simple *way* of solving it, but that there is no way to *write* the solution.

$$\int x^xdx$$

Doctor Jerry answered, going directly to the bigger question of how to tell whether there is a solution:

Dear Simon, (1) There is an algorithm due to a contemporary mathematician namedRischwhich can decide whether the anti-derivative of a continuous function f can be expressed as afinite combination of elementary functions. The elementary functions include polynomials, the trig functions, the inverse trig functions, the exponential function and its inverse, etc.

I know nothing about this algorithm, and with good reason: Wikipedia says,

Risch called it a decision procedure, because it is a method for deciding whether a function has an elementary function as an indefinite integral, and if it does, for determining that indefinite integral. However, the algorithm does not always succeed in identifying whether or not the antiderivative of a given function in fact can be expressed in terms of elementary functions.

The complete description of the Risch algorithm takes over 100 pages.

There are many functions - called special functions - which fail to have an anti-derivative expressible as a finite combination of elementary functions. The so-calledelliptic functions, theerror function, and thegamma functionare a few examples. The error function, which is extremely useful in both physics and statistics, is defined as:erf(x)= (2/sqrt(pi))integral from 0 to x of e^(-t^2)dt Extensive tables of the error function would not exist if the anti-derivative of e^(-t^2) were expressible as a finite combination of elementary functions.

$$\mathrm{erf}(x)=\frac{2}{\sqrt{\pi}}\int_0^x e^{-t^2}dt$$

This is one of the special functions we saw above; we’ll look at it a little below. Then, there were tables of values for this (and other special functions) in which you could look up a numerical value; today, we let computers either use the same sorts of numerical method that were used to make the tables, or look them up in a table, or both. If you’ve ever looked up a probability for a “z-score” in a “standard normal distribution table”, that’s essentially what this is!

Theanti-derivative of x^xis not expressible as a finite combination of elementary functions. I'm not sufficiently familiar with the Risch algorithm to even hint at a proof of it.

As I’ve hinted above, the only evidence I could point to (besides articles about this integral!) would be that Wolfram Alpha, or its relatives, give the sort of answer we saw.

(2) As to solving the equationy = xcosxfor x in terms of y, thiscan't be done in "closed form"or"by classical methods."Both of these phrases are used to mean that this equation can't be solved by algebraic methods or with pencil and paper methods. Of course, given a value of y, one can determinenumerically(using Newton's method, for example) the values of x for which xcosx = y.

The goal here was to find the inverse of the function \(f(x)=x\cos(x)\), which is equivalent to solving the equation \(x\cos(x)=a\) for any given value of *a*.

Even more than the quintic equation, a “transcendental equation” like this can’t be solved exactly, even though solutions exist, as we can see, again, by graphing:

Notice that not only can’t we find an exact value of *x* for a given *y*, there are in fact infinitely many solutions, so this doesn’t have an inverse *function*.

But this raises an interesting question, even for a simpler problem of the same type:

I'd like to give a parallel question to y = xcos x.Can you solve the equation y = sinx for x in terms of y?If you say yes and writex = arcsin y, then you have only recognized that the function inverse to sine is a well known function and you know its name. You haven't actually found a finite expression for x. The fact is,neither sine nor arcsin are knownin the sense that y = x^2 or x = sqrt(y) is known. Except for a few special cases, specific values of sine and arcsin must be looked up in a table or calculated numerically.

In other words, when we find a function that can’t be expressed in terms of known functions, we **give it a name**, and now it is a known function! This is the same thing we do when we need a function like erf or si. **The arcsin function is just as special**, and just as “made up”. We’re just more familiar with it, and don’t need calculus to understand what it does.

We’ll close with another question from 2001 that takes us to the next step:

Trying to Integrate f(x) = exp(-ax^2) How would you go about integrating a function of the formf(x) = exp(-ax^2)?

This is a very important integral, closely related to the erf function mentioned above: $$\int e^{-ax^2}dx$$

Doctor Jubal answered:

Hi Craig, Thanks for writing to Dr. Math. The function f(x) = exp(-ax^2) doesn't have an antiderivative that can be expressed in terms of other commonly used functions. What this means isyou can't analytically integrate it.

As we’ll see, the antiderivative has a name, but that name doesn’t tell how to evaluate a definite integral, just as the name arcsin doesn’t provide a way to evaluate it.

Yet, the function pops up a lot in many places where it would be nice to be able to integrate it. So mathematicians have come up with severalworkarounds. Your options are:(i)Hope your limits of integration are zero and infinity. For these limits of integration, the integral does have an analytical solution.The integral of exp(-ax^2) from 0 to infinity is (1/2)sqrt(pi/a). If you'd like to see a proof of this, feel free to write back.

This one (improper) definite integral is important, because it’s used in normalizing the normal distribution:

$$\int_0^{\infty}e^{-ax^2}dx=\frac{1}{2}\sqrt{\frac{\pi}{a}}$$

We’ll be seeing that soon, because he did write back!

(ii)Solve it numerically. So your limits of integration aren't zero and infinity. Use Simpson's Rule or thenumerical integration methodof your choice to approximate the integral.

All sorts of software can approximate such an integral. For instance, we can use Wolfram Alpha:

or we can use Desmos:

This presumably uses some numerical approximation method.

(iii)Use theerror function. Mathematicians like this integral a lot. They like it so much, in fact, that they defined a function just so this integral would have an "analytical" solution. The error function is defined 2 /xerf(x)= --------- | exp[-t^2] dt sqrt(pi) /0 and /x sqrt(pi/a) | exp[-at^2] dt = ---------- erf(x) /0 2 The error function has been evaluated numerically at many, many points, and the results tabulated in mathematical handbooks. Essentially this is just doing (ii), but taking advantage of the fact thatsomeone else has probably already done the work for you.

We don’t use tables much any more; but software knows how to handle it.

Craig did respond by asking for the proof:

Hi, Could I see aproofof the integral of f(x) = exp(-ax^2) between 0 and infinity, please? Thanks.

Doctor Jubal replied:

Hi Craig, thanks for writing back. We want to evaluate /inf | exp(-ax^2) dx /0 However, using any of our usual arsenal of integration tricks to produce an antiderivative of this will fail. When I was an undergraduate, I spent five hours one evening attempting various ways to integrate this by parts and trying various substitutions.Little did I know there was no antiderivative.The solution came to me over breakfast, within half an hour of my waking up, thus illustrating that in mathematics, there is no substitute for a good night's sleep. But I digress.

His solution is a classic, probably one that has been reinvented many times!

Let'scall the value of the definite integral value I. For reasons that will become apparent shortly, it's a lot easier to evaluate I^2. /inf /infI^2= | exp(-ax^2) dx | exp(-ax^2) dx /0 /0

$$I=\int_0^{\infty}e^{-ax^2}dx\\I^2=\left[\int_0^{\infty}e^{-ax^2}dx\right]^2\\=\left[\int_0^{\infty}e^{-ax^2}dx\right]\left[\int_0^{\infty}e^{-ax^2}dx\right]$$

It's rather irrelevant what symbols we use for the variables of integration, so I'm going touse x in the first integral and y in the second integral. This is okay because the two integrals are completely independent of each other. /inf /inf I^2 = | exp(-ax^2) dx | exp(-ay^2) dy /0 /0

This is because the variable is just a placeholder: $$I^2=\left[\int_0^{\infty}e^{-ax^2}dx\right]\left[\int_0^{\infty}e^{-ay^2}dy\right]$$

And because the two integrals are independent of each other, we can combine them into a single double integral. /inf /inf I^2 = | | exp(-ax^2) exp(-ay^2) dx dy /0 /0 /inf /inf = | | exp[-a(x^2 + y^2)] dx dy /0 /0

$$I^2=\int_0^{\infty}\int_0^{\infty}e^{-ax^2}e^{-ay^2}dxdy\\=\int_0^{\infty}\int_0^{\infty}e^{-a(x^2+y^2)}dxdy$$

We’re integrating one entire quadrant of this function:

Now so far, it looks as if all we've managed to do is make a mess of things. This integral doesn't look any easier to deal with than what we started with, and we have an extra variable to boot. But we have managed to create an x^2 + y^2 term, which meansthis double integral may be more manageable in polar coordinates.

This is the key step. And we couldn’t do it for a finite integral, because a square region doesn’t fit polar coordinates.

The region of integration is the first quadrant, so we're integrating over r from 0 to infinity and theta from 0 to pi/2. I'll use Q for theta. Remember that in polar coordinates,r^2 = x^2 + y^2, anddx dy = r dr dQ. /(pi/2) /inf I^2 = | | exp(-ar^2) r dr dQ /Q=0 /r=0 /(pi/2) /inf = | dQ | exp(-ar^2) r dr /Q=0 /r=0 This integral can be solved analytically, but I'll leave this as something for you to work out on your own. (Hint: try the substitution u^2 = ar^2).

He’s using Q as a stand-in for theta: $$I^2=\int_0^{\infty}\int_0^{\infty}e^{-a(x^2+y^2)}dxdy\\=\int_{\theta=0}^{\pi/2}\int_{r=0}^{\infty}e^{-ar^2}rdrd\theta\\=\int_{\theta=0}^{\pi/2}d\theta\cdot\int_{r=0}^{\infty}e^{-ar^2}rdr$$

This splits the double integral back into a product of two simple integrals.

Let’s finish:

The first integral is easy: $$\int_{0}^{\pi/2}d\theta=\left[\theta\right]_{0}^{\pi/2}=\frac{\pi}{2}$$

The second is a simple substitution: Let \(u=-ar^2\), so \(du=-2ardr\). Then $$\int e^{-ar^2}rdr=\frac{1}{-2a}\int e^{-ar^2}\cdot(-2ardr)=\frac{1}{-2a}\int e^udu=\frac{1}{-2a}e^u=\frac{1}{-2a}e^{-ar^2}$$

So the definite integral is $$\int_{0}^{\infty}e^{-ar^2}dr=\lim_{b\to\infty}\int_{0}^{b}e^{-ar^2}dr\\=\lim_{b\to\infty}\frac{1}{-2a}\left[e^{-ar^2}\right]_{0}^{b}=\lim_{b\to\infty}\frac{1}{-2a}\left[e^{-ab^2}-1\right]=\frac{1}{2a}$$

Putting it all together,

$$I^2=\int_{0}^{\pi/2}d\theta\cdot\int_{0}^{\infty}e^{-ar^2}dr\\=\frac{\pi}{2}\cdot\frac{1}{2a}=\frac{\pi}{4a}$$

$$I=\sqrt{\frac{\pi}{4a}}=\frac{1}{2}\sqrt{\frac{\pi}{a}}$$

as we were told.

Now, why is the error function, erf, defined with that strange constant multiplier? So that this will happen (taking \(a=1\)):

$$\lim_{x\to\infty}\mathrm{erf}(x)=\frac{2}{\sqrt{\pi}}\int_0^{\infty}e^{-t^2}dt=\frac{2}{\sqrt{\pi}}\cdot\frac{1}{2}\sqrt{\pi}=1$$

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This 2008 question ends up with a little twist:

Integration by Parts I'm trying to integrate e^x(x+1)/(x+2)^2 and am stuck. I tried letting x+2 = t with 1 = dt/dx but didn't get anywhere. Can you help me?

It’s a reasonable substitution; if we try it, we get

$$\int\frac{e^x(x+1)}{(x+2)^2}dx=\int\frac{e^{t-2}(t-1)}{t^2}dt$$

which might help a little, by making the denominator simpler, so we can, for example, split the fraction into a sum (in fact, it’s quite a nice start), but doesn’t get us very far by itself.

To my eye, since we have algebraic and exponential factors, integration by parts seems likely; but the algebraic part doesn’t invite us to either differentiate or antidifferentiate it! What can we do?

Doctor Ali took this one, just getting it started:

Hi Arpita! Thanks for writing to Dr. Math. Here's how I'd approach it: - | x+1 x | --------- * e dx | (x+2)^2 - Write it this way in order to expand the numerator: - | x+2-1 x | --------- * e dx | (x+2)^2 - Break the fraction into two separate fractions: - - | e^x | e^x | ------- dx - | --------- dx | (x+2) | (x+2)^2 - -

This is the technique I called “partial cancellation” a couple weeks ago; splitting into a sum this way is just what I would have done next after the substitution above. So far, we have

$$\int\frac{e^x(x+1)}{(x+2)^2}dx=\int\frac{(x+2)-1}{(x+2)^2}e^xdx\\=\int\left(\frac{x+2}{(x+2)^2}e^x-\frac{1}{(x+2)^2}e^x\right)dx=\int\frac{e^x}{x+2}dx-\int\frac{e^x}{(x+2)^2}dx$$

Let the first integral be as it is.

I’m not [initially!] sure what he means here, as there is work to do for this integral; we’ll do it below, and see why he sets it aside.

Useintegration by partsto integrate the second one. We take dv = dx/(x+2)^2 We do this way because we know the integral of dv: 1 v = - ----- x+2 Use the main formula of integration by parts: - - | udv = uv - | vdu - - Can you continue the process yourself?

Now, 16 years later, we may as well try to finish.

Taking \(u=e^x\) and \(dv=\frac{1}{(x+2)^2}dx\), we have \(du=e^xdx\) and \(v=\frac{-1}{x+2}\), and

$$\int \underset{u}{\underbrace{e^x\strut}}\underset{dv}{\underbrace{\frac{1}{(x+2)^2}dx}}=\underset{u}{\underbrace{e^x\strut}}\underset{v}{\underbrace{\frac{-1}{x+2}}}-\int\underset{v}{\underbrace{\frac{-1}{x+2}}}\underset{du}{\underbrace{e^xdx\strut}}\\=\frac{-e^x}{x+2}+\int\frac{e^x}{x+2}dx$$

Here we took the algebraic part (the fraction) as \(dv\) because integrating it would make it simpler; it doesn’t matter where we put \(e^x\). And now we are left with another copy of that first integral we set aside!

So we repeat the process, and use parts for this integral; but here we run into a problem (quite possibly what Arpita found). If we take \(u=e^x\) and \(dv=\frac{1}{x+2}dx\), then \(du=e^xdx\) and \(v=\ln(x+2)\), so we have

$$\int \underset{u}{\underbrace{e^x\strut}}\underset{dv}{\underbrace{\frac{1}{x+2}dx}}=\underset{u}{\underbrace{e^x\strut}}\underset{v}{\underbrace{\ln(x+2)}}-\int\underset{v}{\underbrace{\ln(x+2)}}\underset{du}{\underbrace{e^xdx\strut}}=e^x\ln(x+2)-\int e^x\ln(x+2)dx$$ That doesn’t look like an improvement.

But if we take \(u=\frac{1}{x+2}\) and \(dv=e^xdx\), then \(du=\frac{-1}{(x+2)^2}dx\) and \(v=e^x\), so we have

$$\int \underset{u}{\underbrace{\frac{1}{x+2}}}\underset{dv}{\underbrace{e^xdx\strut}}=\underset{u}{\underbrace{\frac{1}{x+2}}}\underset{v}{\underbrace{e^x\strut}}-\int\underset{v}{\underbrace{e^x\strut}}\underset{du}{\underbrace{\frac{-1}{(x+2)^2}dx}}=\frac{e^x}{x+2}+\int\frac{e^x}{(x+2)^2}dx$$ bringing us back where we started.

But wait!

Putting the pieces together without doing that integral, we have

$$\int\frac{e^x(x+1)}{(x+2)^2}dx=\int\frac{e^x}{x+2}dx-\int\frac{e^x}{(x+2)^2}dx\\=\int\frac{e^x}{x+2}dx-\left(\frac{-e^x}{x+2}+\int\frac{e^x}{x+2}dx\right)\\=\int\frac{e^x}{x+2}dx+\frac{e^x}{x+2}-\int\frac{e^x}{x+2}dx=\frac{e^x}{x+2}$$

Interesting: We don’t really have to evaluate that first integral after all, since it cancels out! Now we know why Doctor Ali just set it aside. (In fact, if I ask Wolfram Alpha to integrate the first integral, it gives an answer in terms of something called “the exponential integral Ei”, which means this can’t be integrated in terms of elementary functions.)

Two days later, Ali answered another one:

Challenging Trig Expression Integration For the life of me, I can't figure out how to integrate this function: (sin x)^3 ----------------------- [(sin x)^3]+[(cos x)^3] There is no convenient form that this can take so thatsubstitution,integration by parts, or both can be performed. I've looked in several textbooks and online sites and haven't seen anything remotely close to this, but the answer is exact and symbolic (no Taylor series needed). answer:(x/2) - (1/6)ln(cos x + sin x) + (1/3)ln(2 - sin(2x)) I've tried a huge number of differenttrig identitiesto force this into a form which is integrable by substitution or by parts (probably need both). I also completed the square on several of them to make two pieces that are perhaps then able to be converted to integrable forms. Nothing works. I got the answer from Wolfram Integrator. I've commonly found that it can be misleading because a more human approach will produce anequivalent formthat isn't necessarily obvious that it matches Wolfram's answer. The first part, (x/2), bothers me because every integration I've performed with trig functions produces a trig function or an exponential. SoI used that as my starting pointto try and figure out how that can even come about from a trig function, but so far I've gotten nowhere with that. Maybe this needs to be performed in complex or hypergeometric?

Wolfram Alpha gives essentially the same answer today.

There are some good ideas here (including the recognition that the answer he gets may look very different from what software produces). But also, it’s dangerous to “reverse-engineer” a solution! That linear term is not central to the answer, but incidental, as we’ll see.

Unfortunately, Ryan has missed one method: **partial fractions**! We’ll start, though, with some trig identities.

Doctor Ali said:

Hi Ryan! Thanks for writing to Dr. Math. It sounds like you've done a lot of good thinking so far. I think the trick to this is doing some algebraic manipulation. I'd start it this way: - | sin(x)^3 | --------------------- dx | sin(x)^3 + cos(x)^3 - Divide the top and bottom by cos(x)^3: - | tan(x)^3 | -------------- dx | tan(x)^3 + 1 -

$$\int\frac{\sin^3(x)}{\sin^3(x)+\cos^3(x)}dx=\int\frac{\frac{\sin^3(x)}{\cos^3(x)}}{\frac{\sin^3(x)}{\cos^3(x)}+\frac{\cos^3(x)}{\cos^3(x)}}dx=\int\frac{\tan^3(x)}{\tan^3(x)+1}dx$$

I often prefer to write an expression in terms of sine and cosine, rather than the reverse; but sometimes you just have to try everything to find something that works. At least here we have only one trig function!

Now we can try a substitution using the obvious “inner” function, the tangent, and work out the differential using an identity:

Assume TAN(x) = u, so (1 + tan(x)^2) dx = du - | u^3 | ------------------- du | (u^3 + 1)(u^2 + 1) - Does that make sense?

Here are the details: Letting \(u=\tan(x)\), \(du=\sec^2(x)dx=(1+\tan^2(x))dx\), and therefore $$dx=\frac{du}{1+\tan^2(x)}=\frac{du}{1+u^2}$$ This gives

$$\int\frac{\tan^3(x)}{\tan^3(x)+1}dx=\int\frac{u^3}{u^3+1}\frac{dx}{u^2+1}$$

Now we have a rational function, which we can always integrate by partial fractions.

Let's try to break that fraction into an equivalent sum of fractions. This is the key algebraic step I referred to above: - | / 2u - 1 u - 1 1 \ | |--------------- - ---------- - -------- | du | \ 3(u^2 - u + 1) 2(u^2 + 1) 6(u + 1) / - Can you do this yourself? This way, we change the difficult integral to three not so difficult integrals.

Here’s one version of the work to obtain this. We first factor the sum of cubes:

$$\frac{u^3}{(u^3+1)(u^2+1)}=\frac{u^3}{(u+1)(u^2-u+1)(u^2+1)}$$

Now we want

$$\frac{u^3}{(u^2-u+1)(u^2+1)(u+1)}=\frac{Au+B}{u^2-u+1}+\frac{Cu+D}{u^2+1}+\frac{E}{u+1}$$

Solving this, I get $$A=\frac{2}{3},B=-\frac{1}{3},C=-\frac{1}{2},D=\frac{1}{2},E=-\frac{1}{6}$$

So we have $$\frac{u^3}{(u^2-u+1)(u^2+1)(u+1)}=\frac{2u-1}{3(u^2-u+1)}+\frac{-u+1}{2(u^2+1)}+\frac{-1}{6(u+1)}$$

Now we have to integrate the sum, using these hints:

Hints: 1) The first integral can be solved using LN, because the top is a multiple of the bottom's derivative. 2) The second integral splits to give one LN and one ARCTAN. 3) The third integral is a simple LN. Can you continue the solution yourself? By the way, don't forget to substitute TAN(x) back for u at the end.

In his order, we have $$\int\left[\frac{2u-1}{3(u^2-u+1)}-\frac{u-1}{2(u^2+1)}-\frac{1}{6(u+1)}\right]du$$

I’ll take one term at a time.

**Part 1**: The numerator is just the derivative we need, so let \(v=u^2-u+1\) and \(dv=(2u-1)du\):

$$\frac{1}{3}\int\frac{2u-1}{u^2-u+1}=\frac{1}{3}\int\frac{dv}{v}=\frac{1}{3}\ln|v|=\frac{1}{3}\ln\left|u^2-u+1\right|=\frac{1}{3}\ln\left|\tan^2(x)-\tan(x)+1\right|$$

**Part 2**: If \(v=u^2+1\), then \(dv=2udu\), so we split into two integrals:

$$-\frac{1}{2}\int\frac{u-1}{u^2+1}du=-\frac{1}{2}\int\frac{udu}{u^2+1}+\frac{1}{2}\int\frac{du}{u^2+1}$$

**Part 2a**: Let \(v=u^2+1\), and \(dv=2udu\): $$-\frac{1}{4}\int\frac{2udu}{u^2+1}=-\frac{1}{4}\int\frac{dv}{v}=-\frac{1}{4}\ln|v|=-\frac{1}{4}\ln\left|u^2+1\right|=-\frac{1}{4}\ln\left|\tan^2(x)+1\right|$$

**Part 2b**: $$\frac{1}{2}\int\frac{du}{u^2+1}=\frac{1}{2}\arctan(u)=\frac{1}{2}x$$

Aha! We found the source of the linear term that baffled Ryan.

**Part 3**: Here we can use a substitution so simple I would not explicitly define the new variable, but would do it all in my head. Let \(v=u+1\), so \(dv=du\):

$$-\frac{1}{6}\int\frac{du}{u+1}=-\frac{1}{6}\int\frac{dv}{v}=-\frac{1}{6}\ln|v|=-\frac{1}{6}\ln|u+1|=-\frac{1}{6}\ln|\tan(x)+1|$$

Putting it together, our solution is $$\frac{1}{3}\ln\left|\tan^2(x)-\tan(x)+1\right|-\frac{1}{4}\ln\left|\tan^2(x)+1\right|+\frac{1}{2}x-\frac{1}{6}\ln|\tan(x)+1|+C$$

But we expected to get $$\frac{1}{3}\ln\left|2-\sin(2x)\right|-\frac{1}{6}\ln|\cos(x)+\sin(x)|+\frac{1}{2}x+C$$

Our answer looks entirely different; but when I graph these two solutions together, I find that they differ by a constant, about 0.23.

So our answer appears to be correct.

But when I graph the three non-linear terms of our solution individually (broken), none of them agrees with either of the two non-linear terms of the computer solution (solid):

The first thing I think of to rewrite the computer answer is to express it in terms of trig functions of *x* only, by changing \(\frac{1}{3}\ln\left|2-\sin(2x)\right|\) to $$\frac{1}{3}\ln\left|2(1-\sin(x)\cos(x))\right|=\frac{1}{3}\ln2+\frac{1}{3}\ln\left|1-\sin(x)\cos(x)\right|.$$ And lo and behold, \(\frac{1}{3}\ln2\approx0.231\)! That explains the constant difference (which is absorbed into the arbitrary constant). So we’d like to show that our solution is equal to this: $$\frac{1}{3}\ln\left|1-\sin(x)\cos(x)\right|-\frac{1}{6}\ln|\cos(x)+\sin(x)|+\frac{1}{2}x+C$$

Let’s take our three terms one by one, and try to express them in terms of the arguments in the desired solution:

**Part 1**: $$\frac{1}{3}\ln\left|\tan^2(x)-\tan(x)+1\right|\\=\frac{1}{3}\ln\left|\sec^2(x)-\tan(x)\right|$$

$$=\frac{1}{3}\ln\left|\frac{1-\sin(x)\cos(x)}{\cos^2(x)}\right|\\=\frac{1}{3}\ln\left|1-\sin(x)\cos(x)\right|-\frac{1}{3}\ln\left|\cos^2(x)\right|$$

$$=\frac{1}{3}\ln\left|1-\sin(x)\cos(x)\right|-\frac{2}{3}\ln\left|\cos(x)\right|$$

**Part 2a**: $$-\frac{1}{4}\ln\left|\tan^2(x)+1\right|=-\frac{1}{4}\ln\left|\sec^2(x)\right|$$

$$=-\frac{1}{4}\ln\left|\frac{1}{\sec^2(x)}\right|=\frac{1}{4}\ln\left|\cos^2(x)\right|$$

$$=\frac{1}{2}\ln\left|\cos(x)\right|$$

**Part 3**: $$-\frac{1}{6}\ln|\tan(x)+1|\\=-\frac{1}{6}\ln|\frac{\cos(x)(\tan(x)+1)}{\cos(x)}|\\=-\frac{1}{6}\ln|\cos(x)(\tan(x)+1)|+\frac{1}{6}\ln|\cos(x)|\\=-\frac{1}{6}\ln|\sin(x)+\cos(x))|+\frac{1}{6}\ln|\cos(x)|$$

Adding these up, together with part 2b, a piece of each cancels out, and we get $$\frac{1}{3}\ln\left|1-\sin(x)\cos(x)\right|{\color{Red}{-\frac{2}{3}\ln\left|\cos(x)\right|}}\\{\color{Red}{+\frac{1}{2}\ln\left|\cos(x)\right|}}+\frac{1}{2}x\\-\frac{1}{6}\ln|\sin(x)+\cos(x))|{\color{Red}{+\frac{1}{6}\ln|\cos(x)|}}\\=\frac{1}{3}\ln\left|1-\sin(x)\cos(x)\right|+\frac{1}{2}x-\frac{1}{6}\ln|\sin(x)+\cos(x))|+C$$

And this is what we wanted!

Again, we didn’t have to do all this, but it’s fun to see it all come together. Computer systems do things differently than humans, and often come up with oddly different results.

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Let’s start with a general question, from 1998:

Tips for Integrating Functions Hi! I have been looking all over the Net for the past few weeks for a good tutorial on finding anti-derivatives, and haven't had much luck. There is an abundance of excellent sources on Derivative Calculus but none on Integral, so I am turning to you for help. Unfortunately it seemsthere is not an easy formulafor finding an Anti-Derivative, as there is for finding a derivative (f'(x)=(f(x+h)-f(x))/h). So if a good explanation would be too lengthy, even a pointer to a tutorial would be great. Thanks for maintaining such a wonderful site, as it has helped me with numerous issues, not to mention it's just cool going "Ahhh, so that's why." =) Brian Cowan

Doctor Rob answered:

It is an unfortunate fact that while every familiar function of calculus has a derivative,not all of them have antiderivatives expressible in terms of those familiar functions. A classic example is the so-called elliptic integrals, which first appear when trying to find the arc-length of an ellipse. The arc-length s from x = 0 to x = a * sin(T) is given by: s = a * Integral sqrt(1 - k^2 * sin^2[t]) dt from t = 0 to T where k is the eccentricity of the ellipse. For ellipses that are not circles, k is not 1, andthis integral is not possible in closed form. Other examples are the integrals of sqrt(sin(x)), e^(-x^2), and so on.

Not only is integration *difficult*, it is sometimes literally *impossible* (in the sense of finding an expression for the answer in terms of the basic functions, which is what “closed form” means).

You are correct that there isno standard methodfor doing integration. Some classes of functions have such methods, however. Forrational functions, the idea is to decompose the rational function into a polynomial plus a rational function whose numerator has smaller degree than the denominator. Then this new rational function can be split into simpler ones by factoring the denominator, and splitting it up via the method ofpartial fractions. These can be integrated individually.

So if the integrand is a rational function, you know it can be integrated, because there is a standard method (as long as the denominator can be factored, and you have enough patience to carry out the process).

Forrational functions of trigonometric functions, the method is to use multiple angle formulas to express the function in terms of sines and cosines of a single angle t. Then the substitutionz = tan(t/2)will reduce the function to a rational function in z. This reduces it to the previous case.

This is a special method that *always* works for such integrands; beginners are generally taught easier methods that *often* work, similar to what we’ll do below (but for non-rational functions).

For more elaborate functions, and those of other types, a high degree ofpattern-recognitionandexperiencein choosing substitutions that will simplify the problem at hand are involved. Often when there is a common subexpression among two or more parts of the integrand, it is useful tosubstitutesome new variable for it. When an expression of the form sqrt(a^2-t^2) is encountered, the substitutiont = a*sin(z)will often help, ort = q*coth(z). When it is sqrt(a^2+t^2), the substitution t =a*tan(z)will often help, or elset = a*cosh(z).

I often say that integration is an art; it takes a lot of experience to see quickly what to try, and even then you may not be able to see a way. The methods of “*u*-substitution” (as we’ll be seeing below) and “trig substitution” or “hyperbolic substitution” (like those he just mentioned) often help simplify a problem.

Another very useful technique isintegration by parts. When to use it is not, however, very clear. When the integrand is aproduct, one of whose parts is integrable, and the other of which has a relatively simple derivative, then integration by partsmaybe beneficial. In summary, there are just a few useful techniques that are useful in integration. They have to beapplied imaginatively. Even so, some functions will turn out not to be integrable at all.

The key to integration is to have a full toolbox, and to learn to recognize when each tool might be needed. And sometimes you have to try one you didn’t think would help – just because nothing else did!

As a first example, I want to start very small, with this question from 2001:

Integrate Cos 2a Dear Dr. Math, I was doing some integration and one of the problems was tointegrate cos2a. The answer given in the book is sin 2a/2. The textbooks that I have tell me that integrating cos a gives me sin a, and tell me in detail how that was arrived at. But I have searched everywhere for information on cos2a and how the result sin 2a/2 is arrived at. Does this follow a pattern, i.e. cos 3a = sin 3a/3 cos 4a = sin 4a/4 or cos na = sin na/n ?

Katherine is using *a* (which sounds like a constant) where I would expect to see *x*; to avoid confusion, we may want to think of this as integrating $$\int\cos(2x)dx$$ and more generally $$\int\cos(nx)dx$$ Books commonly give this integral as $$\int\cos(ax)dx=\frac{\sin(ax)}{a}+C$$

I answered, using the variable *x*:

Hi, Katherine. In general, theformulais [INT] cos(ax)dx = sin(ax) / a You can prove this by taking the derivative of the right side: d/dx[1/a sin(ax)] = 1/a cos(ax) d/dx(ax) = 1/a cos(ax) * a = cos(ax) This required only a simple application of the chain rule.

Here I’ve taken the formula as given, and checked this antiderivative by differentiating it. The formula is worth memorizing. But what if you haven’t done so?

Not knowing this more general formula, you canobtain it by substitution. If we let u = ax, then du = a dx and dx = du/a, so [INT]cos(ax)dx = [INT]cos(u) du/a = sin(u)/a = sin(ax)/a Here I firstreplaced ax with uanddx with du/a, then integrated, and finallyreplaced u with axagain. The same method is useful everywhere, so you should learn it well and even be able to do it in your head.

For details on the method, see Integration by Substitution. We defined a new variable *u* to be equal to the quantity found inside the function, *ax*; found how to express *dx* in terms of *du*; and then replaced each in the integral. The resulting integral was one we’ve memorized, so we just wrote the answer. Finally, we back-substituted to put the answer in terms of the original variable, *x*.

Now, what I often do for problems like this, if I’m not quite sure of the formula, is to start with a knowingly wrong “guess”, and correct it by checking:

I want the integral of \(\cos(ax)\); I know that the antiderivative of \(\cos(x)\) is \(\sin(x)\) (that is, that the derivative of the latter is the former), so I expect the answer to be something like \(\sin(ax)\).

Then I take the derivative of that using the chain rule, finding that it is \(a\cos(ax)\), which is *a* times as large as I want. To fix this, I just need to divide by *a*, so my answer is \(\frac{1}{a}\sin(ax)\).

I can do all this in my head, and it frees my memory from having to be certain what to multiply or divide by. (Each function has its own behavior, which I don’t have to remember this way.)

This can be described as a version of the method of false position. We’ll be using it again below.

For a harder example, here’s a question from 1999:

Techniques of Integration - Change of Variables Hello. I would like to know how to solve this question.(integral sign) sin 2x / sqrt(9-cos^4 x) dxHow should I begin? Jen

We want to integrate $$\int\frac{\sin(2x)}{\sqrt{9-\cos^4(x)}}dx$$ It is not obvious that this is a function of a function, which is what you look for to make a substitution (giving a name to the inner function); but rewriting it will change things.

Doctor Luis answered:

You can approach this question more easily if you use the identity sin(2x) = 2sin(x)cos(x), and if you rewrite your integral in a more suggestive way, like this: / | 2 sin(x) cos(x) | ----------------------- dx | sqrt(9 - (cos^2(x))^2) /

Getting everything in terms of the same angle (*x* itself) is a good start; moreover, we see that the numerator is the derivative of \(\cos^2(x)\)! Also, writing the radicand as a difference of squares suggests something …

At this point it is clear that achange of variableswill do the trick. Letcos^2(x) = 3u. Then, differentiating implicitly, 2cos(x)*(-sin(x))dx = 3du This means that 2cos(x)sin(x) = -3du, and so, rewriting our integral in terms of u, we have / / | -3 du | du | ---------------- = - | ------------- | sqrt(9 - 9 u^2) | sqrt(1-u^2) / /

The way he substituted may be unfamiliar to you.

We often define a new variable directly as something like \(u=\cos(x)\) (a “*u*-substitution”), or, in reverse, something like \(x=\cos(\theta)\) (a “trig substitution”). In those cases, we can directly differentiate to find \(du\), or to solve for \(d\theta\).

Here, we are replacing one expression by another (\(\cos^2(x)\) by \(3u\)), and implicit differentiation shows that we can replace all of \(2\cos(x)\sin(x)dx\) with \(-3du\).

But this last integral is already known. It's nothing more than the inverse cosine function. So, / | du - | ------------- = arccos(u) = arccos((1/3)cos^2(x)) | sqrt(1-u^2) / Notice that we have expressed our final answer in terms of the original variable of integration, using the fact that u = (1/3)cos^2(x) . Obviously, I have neglected the arbitraryconstant of integration, but you can add that at any time.

This integral is commonly just **memorized**. If you hadn’t, you could use a **trig substitution** to derive it; and if you had just half-memorized it (like me) you could **check** that you had all the signs right, as Doctor Luis now shows:

Now, to show that theintegral of -1/sqrt(1-u^2)is thearccos(u)function, what you can do is check that the derivative of the arccos(u) function is -1/sqrt(1-u^2). We can do that as follows: Let y = arccos(x); then x = cos(y). By implicit differentiation on this last equation you can obtain, 1 = -sin(y) * dy/dx (notice we used the chain rule here) Solving for dy/dx, you get -1 dy/dx = --------- sin(y) Expressing sin(y) in terms of cos(y) (use the identity sin^2(y) + cos^2(y) = 1 for this step) we get: -1 dy/dx = ----------------- sqrt(1-cos^2(y)) But, by definition, x = cos(y). Therefore, -1 dy/dx = ------------- sqrt(1-x^2) Now, y was just y = arccos(x). Therefore, we have proven that the derivative of the arccos(x) function is -1/sqrt(1-x^2) .

Alternatively, you might know that \(\int\frac{1}{\sqrt{1-x^2}}=\arcsin(x)\), and write the answer as $$-\int\frac{du}{\sqrt{1-u^2}}=-\arcsin(u)=-\arcsin\left(\frac{1}{3}\cos^2(x)\right)+C$$ Believe it or not, this is equivalent to his answer (because arcsin and arccos add up to \(\frac{\pi}{2}\)).

Now consider this, from 2003:

Substituting to Simplify the Integral What is the integral of tan^3 x * sec x dx?

We want $$\int\tan^3(x)\sec(x)dx$$ Where can we start?

Doctor Barrus answered:

Hi, Trista. This is a good question. Finding the indefinite integral of a bunch of trigonometric functions is often challenging, andsometimes it takes a lot of tries to get something that works. One strategy that you'll want to adapt is tolook for a good substitutionthat will make the integral simpler. I'm going to assume that you're familiar with substitutions; let us know if you need more information on this.

We need to consider possible inner functions, and what their derivatives are:

What I'll do is look at different choices for ournew variable u, based on what you have already in the integral, and figure out what thedifferential duwould be with such a choice... If then u = tan x du = sec^2 x dx u = tan^2 x du = 2(tan x)(sec^2 x) dx u = tan^3 x du = 3(tan^2 x)(sec^2 x)dx u = sec x du = sec x tan x dx The first three didn't look promising. All of them required a sec^2 x in order to substitute du into the integral. The last one looks a lot better, though, because we DO have a sec x tan x dx in the indefinite integral.

Since we have only one secant, \(u=\sec(x)\) seems like the only choice in order to have its derivative present. Note the key idea here: We want *u* to be something whose **derivative** is present in the integral, even if we don’t (yet) see a function of that *u*!

What we'll want to do, then, is to pull this quantity apart from the rest of the indefinite integral. In other words, we'll rewrite (tan^3 x)(sec x)dx as (tan^2 x)(sec x tan x dx) Remember that it doesn't matter in which order we multiply things. Then we know that if we make the substitution u = sec x, we'll be able to change the sec x tan x dx into du.

Now we look at the rest of the integrand.

The question now is whether we can write that remaining tan^2 x in terms of sec x, so we can replace it by something involving u. What relations (identities, etc.) exist between the functions tan x and sec x? I hope you're familiar with one of the Pythagorean identities that relates these two: 1 + tan^2 x = sec^2 x We want to write tan^2 x in terms of sec x, so we'll solve for tan^2 x in this equation: tan^2 x = sec^2 x - 1 Or, putting it in terms of u = sec x, we have tan^2 x = u^2 - 1

So now our integral becomes $$\int\tan^3(x)\sec(x)dx=\int\underset{f(u)}{\underbrace{\tan^2(x)}}\cdot\underset{du}{\underbrace{\sec(x)\tan(x)dx}}\\=\int\underset{u^2-1}{\underbrace{(\sec^2(x)-1)}}\cdot\underset{du}{\underbrace{\sec(x)\tan(x)dx}}=\int(u^2-1)du$$

We can substitute this into the indefinite integral to arrive at the following indefinite integral: INT (u^2 - 1) du and we hope you know how to finish the problem from there. Does this make sense? Substitution is a good strategy to consider in problems like this, particularly in instances where one of the trig functions is raised to an odd power, as in this case.

Here is the rest of the work: First we integrate, $$\int(u^2-1)du=\frac{1}{3}u^3-u+C$$ and then we back-substitute, replacing *u* with \(\sec(x)\), $$\frac{1}{3}\sec^3(x)-\sec(x)+C$$

Finally, if we are unsure, we can check by differentiating: $$\frac{d}{dx}\left(\frac{1}{3}\sec^3(x)-\sec(x)+C\right)\\=\frac{1}{3}\cdot3\sec^2(x)\cdot\sec(x)\tan(x)-\sec(x)\tan(x)\\=\left(\sec^2(x)-1\right)\sec(x)\tan(x)\\=\tan^2(x)\sec(x)\tan(x)=\tan^3(x)\sec(x)$$

Sometimes it takes more than one substitution, as in this final question from 1998:

A Trigonometry Integral Requiring Two Substitutions What is the integral of sqrt(1 + sin(x)), where sqrt stands for "square root of"?

We want $$\int\sqrt{1+\sin(x)}dx$$ With nothing outside of the radical, what can we use for *du*?

Doctor Sam answered:

Paul, This is a tricky problem. It will take (I think) two different substitutions. We want to find: INT sqrt(1 + sin x) dx

(We’ll see that it could be done with only one substitution, but what he’s doing is just what I would do, trying something simple first. We just try whatever seems like it might help.)

I am going totry substituting u = sin(x)to try to remove the trig function. When you make a substitution, you must also substitute for dx. So: u = sin(x) and du = cos(x) dx This gives dx = du/cos(x), and changes the integral to sqrt(1 + u) INT ----------- du cos(x)

Since this time there is no differential sitting there waiting for us, we’re taking a step in the dark, trying an “inner function” and hoping its derivative will turn up. If this didn’t work, we might instead try \(u=1+\sin(x)\), or even \(u=\sqrt{1+\sin(x)}\) to see what happened.

I don’t like to write an integral with two different variables in it as he does here, but it’s okay as long as we see it as a fleeting intermediate step. We immediately want to express that denominator in terms of the new variable *u*:

This is no good.We need to get an integral in terms of the u variable alone.Here's where a little right-triangle trigonometry can help. We made the substitution u = sin(x), so we can visualize a triangle with an acute angle x whose sine is u. Here is one such triangle: /| / | / | / | 1 / | u / | /x | -------- Now we can use the Pythagorean Theorem to find the third side, and then the cosine of x. The third side is sqrt(1 - u^2), and so: cos(x) = sqrt(1 - u^2)

Alternatively, we could just recall the Pythagorean identity without drawing the picture; We want to express \(\cos(x)\) in terms of \(\sin(x)\), so we observe that \(\sin^2(x)+\cos^2(x)=1\); therefore, \(\cos(x)=\sqrt{1-\sin^2(x)}=\sqrt{1-u^2}\).

But we’ve skipped over an important question: There are *two* square roots; shall we take the positive or negative root? This depends on what quadrant angle *x* is in! If \(-\frac{\pi}{2}\le x\le \frac{\pi}{2}\), then the cosine will be positive; otherwise it would be negative, and the actual integral would have a different sign. Let’s just restrict *x* to that interval for simplicity; it turns out that making a single formula for the integral over all real numbers is tricky (even Wolfram Alpha doesn’t quite manage it).

Our integral is now: sqrt(1 + u) INT ------------- du sqrt(1 - u^2) Now I can't help but notice that 1 - u^2 = (1 - u)(1 + u) so this fraction simplifies to: 1 INT ----------- du sqrt(1 - u)

So we have a new, and simpler, integral; but it still is not one we’ve memorized.

We are almost done. We have now transformed our trig integral into an algebraic integral. Now asecond substitution:w = 1 - ushould finish the job. If w = 1 - u, then dw = -du, so du = -dw. This gives: 1 - INT ------- dw sqrt(w)

Now we just have to rewrite this, and it will be one of our basic forms:

Interpret this as w^(-1/2), and we can use the formula for antidifferentiating u^n: - INT w^(-1/2) dw = -2w^(1/2) + C Now change back from w to u using w = 1 - u: -2w^(1/2) = -2 sqrt(1 - u) + C And now change back from u to x using u = sin(x): -2w^(1/2) = -2 sqrt(1 - u) + C =-2 sqrt(1 - sin(x)) + C

We can check our answer by differentiating: $$\frac{d}{dx}\left(-2\sqrt{1-\sin(x)}+C\right)=\frac{d}{dx}\left(-2\left(1-\sin(x)\right)^{1/2}\right)\\=-2\cdot\frac{1}{2}\left(1-\sin(x)\right)^{-1/2}\cdot\left(-\cos(x)\right)=\frac{\cos(x)}{\sqrt{1-\sin(x)}}\\=\frac{\sqrt{1-\sin^2(x)}}{\sqrt{1-\sin(x)}}=\frac{\sqrt{1+\sin(x)}\sqrt{1-\sin(x)}}{\sqrt{1-\sin(x)}}=\sqrt{1+\sin(x)}$$

Do you see how this suggests a trick method we could have used instead? Work backward:

$$\int\sqrt{1+\sin(x)}dx=\int\frac{\sqrt{1+\sin(x)}\sqrt{1-\sin(x)}}{\sqrt{1-\sin(x)}}dx\\=\int\frac{\sqrt{1-\sin^2(x)}}{\sqrt{1-\sin(x)}}dx=\int\frac{\cos(x)dx}{\sqrt{1-\sin(x)}}$$ Letting \(u=1-\sin(x)\), we have

$$\int\frac{-du}{\sqrt{u}}=-\int u^{-1/2}du=-2u^{1/2}\\=-2\sqrt{1-\sin(x)}+C$$

And we could have made this same substitution from the start; it combines the two we did above into one. We can let \(w=1-\sin(x)\) all at once, so that \(\sin(x)=1-w\) and $$dw=-\cos(x)dx=-\sqrt{1-\sin^2(x)}dx\\=-\sqrt{1-(1-w)^2}dx=-\sqrt{2w-w^2}dx$$

Then $$\int\sqrt{1+\sin(x)}dx=\int\sqrt{1+(1-w)}\cdot\frac{dw}{-\sqrt{2w-w^2}}\\

=-\int\frac{\sqrt{2-w}\,dw}{\sqrt{2w-w^2}}=-\int\frac{\sqrt{2-w}\,dw}{\sqrt{w}\sqrt{2-w}}\\

=-\int\frac{dw}{\sqrt{w}}=-\int w^{-1/2}dw\\

=-2w^{1/2}=-2(1-\sin(x))^{1/2}=-2\sqrt{1-\sin(x)}+C$$

But we’d never have thought of that. Step by step is the way to go.

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Our first question is from 2001:

Proof of Integration by Parts Can you please tell me the integration formula for u(x)v(x)?

Kamrul is asking how to integrate the **product of two functions** of the independent variable; there is no “product rule” for integrals, as there is for derivatives, but there is something very similar.

Doctor Jordi answered:

Hello, Kamrul - thanks for writing to Dr. Math. I imagine you are talking about a procedure calledintegration by parts. It really is nothing more than aby-product of the product rule(no pun intended) for differentiation. Check it out: d(f(x)*g(x))/dx = f(x)*g'(x) + f'(x)*g(x) We solve for f(x)*g'(x): f(x)*g'(x) = d(f(x)*g(x))/dx - f'(x)*g(x) Integrate both sides and recall that an integral is the antiderivative: INT[f(x) g'(x)dx] = f(x)*g(x) - INT[ f'(x) g(x) dx]

$$\int f(x)g'(x)dx=f(x)g(x)-\int f'(x)g(x)dx$$

We are turning the product rule for derivatives “inside-out”.

This is the function-notation version; we often memorize it in terms of variables and differentials instead:

Then change to the common notation: u = f(x) v = g(x) du = f'(x)dx dv = g'(x)dx This gives us the formula for integration by parts: INT[u dv] = uv - INT[v du]

$$\int u dv=uv-\int v du$$

What this does is to swap the roles of *u* and *v*, in the hope that the new integral will be something we can work out. Like all methods for integration, this is an art, not a mechanical rule that will always give us an answer.

The hard part is often to decide what part of the integrand to call “*u*“. We need to be able to find *v*, first, and then to be able to carry out the integral.

Consider this question from 1996:

Integral of Ln x Dr. Math, Can you please tell me what theintegral of the natural log of xis (ln x)?

It isn’t immediately obvious that this could benefit from integration by parts, since there is only one “part”; I’ll do this directly in a moment. Doctor Sebastien took a different way, which made it more obvious to use parts:

Here's the question: Integral (ln x) dx Let'ssubstituteln x by y: x = e^y dx/dy = e^y So Integral (ln x) dx = Integral (ln x) dx/dy dy = Integral (y e^y) dy

A more traditional way to do this substitution would be to replace \(x\) with \(e^y\), and \(dx\) with \(e^ydy\): $$\int\ln(x)dx=\int\ln(e^y)(e^ydy)=\int ye^ydy$$

So this substitution changed an integral of a single function into a product.

Usingintegration by parts, Integral (y e^y) dy = uv - Integral (v du/dy) dy Let u = y. du/dy = 1 Integral dv = Integral (e^y) dy v = e^y

How did he decide whether to take \(y\) or \(e^y\) as \(u\) in the formula? The exponential function can be either differentiated or integrated without even changing; differentiating \(y\) makes it simpler (reducing the degree), while integrating it *increases* the degree, making it harder. So we take the latter as \(u\), so the new integral will be simpler:

Therefore, Integral (y e^y) dy = y e^y - Integral (e^y * 1) dy = y e^y - (e^y) + c, where c is a constant. Therefore, Integral (ln x) dx = y e^y - (e^y) + c = (ln x)e^(ln x) - e^(ln x) + c = x(ln x) - x + c

Restating this for clarity, $$\int ye^ydy=\int \underset{u}{\underbrace{y}}\cdot\underset{dv}{\underbrace{e^ydy}}=\underset{u}{\underbrace{y}}\cdot\underset{v}{\underbrace{e^y\strut}}-\int\underset{v}{\underbrace{e^y\strut}}\underset{du}{\underbrace{dy}}=ye^y-e^y$$

And back-substituting \(y=\ln(x)\), he got $$\int\ln(x)dx=ye^y-e^y=\ln(x)e^{\ln(x)}-e^{\ln(x)}=x\ln(x)-x+C$$

What he didn’t notice is that **we can integrate directly by parts**, taking \(u=\ln(x)\) and \(dv=dx\). At first, this looks silly, but it allows us to differentiate the log, which we know how to do, rather than integrate it, which we don’t (until we’ve solved this problem!):

$$\int \ln(x)dx=\int \underset{u}{\underbrace{\ln(x)}}\cdot\underset{dv}{\underbrace{dx\strut}}=\underset{u}{\underbrace{\ln(x)}}\cdot\underset{v}{\underbrace{x\strut}}-\int\underset{v}{\underbrace{x\strut}}\underset{du}{\underbrace{\frac{1}{x}dx}}\\=x\ln(x)-\int dx=x\ln(x)-x+C$$

Comparing the two methods, we see that they are really doing the same thing.

Note that a key to the work in each case is the fact that a **logarithm** can be differentiated easily, but not integrated (the latter, indeed being this problem!), and differentiating a **power** helps more than integrating it, and that an **exponential** function is happy no matter what we do to it. So we preferred to put \(\ln(x)\) in *u* over anything else, and we preferred to put *x* in *u* rather than the exponential.

A mnemonic I have seen in at least one textbook is LIATE (or, sometimes, ILATE or LIPET):

- Logarithm
- Inverse trigonometric function
- Algebraic function (polynomial, power, root)
- Trigonometric function
- Exponential function

This is the order in which you might consider functions to play the role of *u*; they are listed in order of the value or ease of differentiation as opposed to integrating. (If there’s a log, definitely take it as *u*; only use an exponential if nothing else works!)

I prefer just to think about what I can differentiate, and what I can integrate; LIATE isn’t a rigid rule, and doesn’t cover everything that can happen, so I wouldn’t want to be dependent on it.

Next, take a more complicated question, from 2003:

Choosing Factors When Integrating by Parts I use integration by parts to find 1 / | (r^3)/[(4+r^2)^(1/2)] dr / 0 What I find most difficult isfinding u, v, du, and dv. I tried separating the integral by makingone simple integralto take the anti-derivative of and then using thedifficult integralto do integration by parts. I just don't know how to start it.

This is a definite integral; we’ll be focusing on the indefinite integral in the answer; at the end of the work I’ll bring the limits of integration back in. (This is how I often prefer to handle definite integrals anyway.) So our problem for now is $$\int\frac{r^3}{\sqrt{4+r^2}}dr$$

Consuelo describes, a little awkwardly, the right basic approach: Looking for a part to call *dv* that can be easily integrated to get *v*, and letting the rest be *u*. But it can be harder than it sounds.

Doctor Fenton answered:

Hi Consuelo, Thanks for writing to Dr. Math. I like to write the integration by parts formula in the form / / | u v' dx = uv - | u' v dx , / / because it arises from the Product Rule for derivatives: [u v]' = u' v + u v' . The idea is to look at the integrand as a product of two functions, andone of them must be an exact derivative(or maybe a constant times an exact derivative). The trick is to choose the factors correctly.

This version replaces \(dv\), as it is often written, with \(v’dx\), to keep the focus on the fact that *u* and *v* are both functions of *x*; use whichever makes more sense to you.

The first requirement will be that you can find an antiderivative of \(v’\) reasonably easily; another will be that the result is nicer than what you started with:

Formally, it looks as if you are "moving" the derivative from one factor to the other. This will help onlyif the new integrand is simplerin some sense than the original. In your example, the integrand is (r^3)/[(4+r^2)^(1/2)] The square root is a unit which you can't break up, so the only question is how to distribute the powers of r. Essentially, the only choices are to take the factors to be one of the following: r^3 and (4+r^2)^(-1/2) r^2 and r*(4+r^2)^(-1/2) r and r^2*(4+r^2)^(-1/2) Of course, we also have tochoose one to be u(the factor to be differentiated) andone to be v'(the factor to be integrated).

We don’t want to differentiate the more complicated part (on the right):

You can differentiate (4+r^2)^(-1/2) but if you do, you will obtain a factor of (4+r^2)^(-3/2) plus an additional power of r from applying the Chain Rule. You would also have to integrate r^3, which would produce r^4, so the integrand in the new integral after using this approach would be a constant times r^5*(4+r^2)^(-3/2) Thatlooks worse than the original integral: it has more powers of r, and a larger power of (4+r^2) in the denominator.

Sometimes what looks bad at first turns out to be the best choice, so you have to go with it; but this approach is at least worth setting aside for now!

On the other hand, integrating such a complicated piece seems scary … unless we can arrange things just right:

This observation strongly suggests that you want tointegrate(4+r^2)^(-1/2). However, if you try to use the substitution u = 4+r^2 then du = 2rdr, so you need a power of r to integrate this term. That means that you should write the original integrand as the product of r^2 and r*(4+r^2)^(-1/2) takingu = r^2andv' = r*(4+r^2)^(-1/2).

So we have chosen a pair \(u\) and \(v’\) such that differentiating \(u\) decreases the degree, and integrating \(v’\) is at least possible.

After integrating, you will have an integrand which is a constant times r*(4+r^2)^(1/2) and that can be evaluated with another integration by substitution. If you have any questions or need more help, please write back and show me what you have been able to do, and I will try to offer further suggestions.

Let’s carry this out:

First, making the substitution \(t=4+r^2,dt=2rdr\), we find$$v=\int (4+r^2)^{-1/2}rdr=\int t^{-1/2}\frac{1}{2}dt=t^{1/2}=(4+r^2)^{1/2}$$ so that

$$\int\frac{r^3}{\sqrt{4+r^2}}dr=\int \underset{u}{\underbrace{r^2\strut}}\cdot\underset{dv}{\underbrace{r(4+r^2)^{-1/2}dr}}\\=\underset{u}{\underbrace{r^2\strut}}\cdot\underset{v}{\underbrace{(4+r^2)^{1/2}}}-\int\underset{v}{\underbrace{(4+r^2)^{1/2}}}\underset{du}{\underbrace{2rdr\strut}}\\=r^2(4+r^2)^{1/2}-2\int r(4+r^2)^{1/2}dr$$

(Note that in my substitution I used *t*, not the usual *u*, as the variable, to avoid confusion with *u* in the parts. This is a reason I am troubled when students are taught to call the method “*u*-substitution”, as if the name mattered.)

As Doctor Fenton said, the remaining integral is much like what we solved to find *v*, and can be solved by again letting \(t=4+r^2,dt=2rdr\): $$\int(4+r^2)^{1/2}rdr=\int t^{1/2}\cdot\frac{1}{2}dt=\frac{1}{2}\cdot\frac{2}{3}t^{3/2}=\frac{1}{3}(4+r^2)^{3/2}$$

Therefore, the entire integral is $$\int\frac{r^3}{\sqrt{4+r^2}}dr=r^2(4+r^2)^{1/2}-\frac{2}{3}(4+r^2)^{3/2}+C$$

This can be simplified a little, to $$(4+r^2)^{1/2}\left(r^2-\frac{2}{3}(4+r^2)\right)\\

=\frac{1}{3}(4+r^2)^{1/2}\left(3r^2-2(4+r^2)\right)\\

=\frac{1}{3}(4+r^2)^{1/2}\left(r^2-8\right)+C$$

Finally, we can find the definite integral: $$\int_0^1\frac{r^3}{\sqrt{4+r^2}}dr=\left[\frac{1}{3}\sqrt{4+r^2}\left(r^2-8\right)\right]_0^1\\=\left[\frac{1}{3}\sqrt{5}\left(-7)\right)\right]-\left[\frac{1}{3}\sqrt{4}\left(-8\right)\right]\\=-\frac{7}{3}\sqrt{5}+\frac{16}{3}$$

Another challenge comes from 1999:

Integration Can you help me withINTEGRAL x tan^2 x dx? I think you have to use some kind of substitution or maybe trigonometry identities Thank you.

Seeing a product of two different kinds of functions, integration by parts seems likely; but what are the parts? We can guess that \(x\) is easier to differentiate (and LIATE suggests that that’s more worth doing than the tangent); but then we’ll have to integrate \(\tan^2(x)\) …

Doctor Luis answered:

The first thing that comes to mind isintegration by parts. My primary motivation for this method is the factor of x that appears on the integrand, because I know that differentiating that factor will give me a constant. Quick review: integration by parts is essentially based on the following formula / / | | | u dv = u*v - | v du | | / / In this case the substitutions should beu = x----->du = dxdv = tan^2(x) dx-----> v = ?

Often the antiderivative of \(dv\) is obvious, but not here. We have to go off to the side and work out \(\int\tan^2(x)dx\), which requires a trig identity:

Note that there is a small problem here. We need to integrate the function tan^2(x) in order to obtain v from dv. But this does not really present that much of a problem, if you remember that tan(x) = sin(x)/cos(x) and that cos^2(x)+sin^2(x) = 1. / / | sin^2(x) | 1 - cos^2(x) v = | --------- dx = | ------------- dx | cos^2(x) | cos^2(x) / / / | = | (sec^2(x) - 1)dx = tan(x) - x | / Therefore,v = tan(x) - x.

By the way, it’s worth noting that *v* can be *any* antiderivative of \(\tan^2(x)\), so we don’t need “+ C” here.

Now we can come back to the integration by parts and continue:

Now, substituting into our "integration by parts formula," you get / / | | | (x)*(tan^2(x)dx) = (x)*(tan(x)-x) - | (tan(x)-x)*(dx) | | / / where I have enclosed the substitutions we made in parentheses.

The hope in integration by parts is that the new integral will be easier; and it is, when you take time. Doctor Luis does it a different way than I would:

Now, the second term on the integral on the righthand side of the equation presents no problem. It's just (1/2)x^2, but the first term seems a bit more difficult. However, if yourewrite tan(x) as sin(x)cos(x)/cos^2(x), you can see that we almost have a complete differential; the derivative of cos^2(x) is 2*cos(x)*(-sin(x)). So, concentrating on the integral of tan(x), we have: / / | | sin(x)cos(x) | tan(x)dx = | ------------ dx | | cos^2(x) / / If you make thesubstitution u = cos^2(x), which implies that du = 2cos(x)*(-sin(x))dx or cos(x)sin(x)dx = -du/2), then this last integral becomes / / | sin(x)cos(x) 1 | du | ------------ = - --- | -- = -(1/2)ln(u) = -(1/2)ln(cos^2(x)) | cos^2(x) 2 | u / /

I would have made the substitution \(t=\cos(x),dt=-\sin(x)dx\): $$\int\tan(x)dx=\int\frac{\sin(x)}{\cos(x)}dx=\int\frac{-dt}{t}=-\ln|t|=-\ln|\cos(x)|$$ This is equivalent. (Do you see why?)

Therefore the integral of tan(x) is just -(1/2)ln(cos^2(x)) and so our original integral becomes: / / | | | (x)*(tan^2(x)dx) = (x)*(tan(x)-x) - | (tan(x)-x)*(dx) | | / / = x*tan(x)-x^2 - (-(1/2)ln(cos^2(x))-(1/2)x^2) = x*tan(x) - (1/2)x^2 + (1/2)ln(cos^2(x)) I hope this explanation helps, although it would have been better if you had included an explanation of how you attempted this problem and where you got stuck. (I did not know how or why you got stuck, so I tried to explain everything.) Feel free to reply if you have any questions about this answer.

Stepping up the difficulty, consider this from 1995:

Integration by Parts Since most people did calculus in high school, I assume this question you folks will be able to answer. I've been working all day on it, and no luck. I need theintegral from 0 to 1 of x^4 * exp(-x/2). Can you help? Stuck in Seattle

This was very early in the history of *Ask Dr. Math*, when it was assumed we were entirely for pre-college math. Again, the question is about a definite integral; this time we’ll be working that way from the start, finding $$\int_0^1x^4e^{-x/2}dx$$

Doctor Ken answered, starting with a substitution, which is not necessary, but helpful:

Hello there! This is a classic example of a problem that can be solved by using the technique ofIntegration by Parts. But first, let's make the problem a little bit nicer by getting rid of the -x/2. Let's lety = -x/2(i.e. x = -2y), and then we have dx = -2dy. So we have theintegral from 0 to -1/2 of 16y^4 * e^y (-2)dy. We can pull the -32 outside of the integral. So now we've got -32 times the integral from 0 to -1/2 of y^4 * e^y dy.

So far we have $$\int_0^1x^4e^{-x/2}dx=\int_0^{-1/2}\left(-2y\right)^4e^{y}(-2)dy=-32\int_0^{-1/2}y^4e^ydy$$ (I’ve corrected a numerical error in the original, which wasn’t noticed because the problem wasn’t finished.)

We’ll let *u* be the algebraic part (so that differentiating it reduces the degree), and let *dv* be the exponential part (which is essentially unchanged by integrating):

Using Integration by Parts, letu = y^4and letdv = e^y dy. So thendu = 4y^3 dy, andv = e^y. Remember the formula for integration by parts? It's Integral(u dv) = uv - Integral(v du). So we get -32 Integral(y^4 e^y dy) = -32*y^4 e^y + 32*Integral(4y^3 e^y dy) = -32/(16Sqrt{e}) + 128*Integral(y^3 e^y dy)) = -2/Sqrt{e} + 128*Integral(y^3 e^y dy). ***note: all integrals are between 0 and -1/2*** Notice that we've knocked the power down on the y^4 so that it became y^3. You canapply this method againand knock the power down another notch,and another, until you get the integral of e^y. Which you can do, because it's e^y. I hope this makes sense to you.

Writing it as an indefinite integral, the entire process looks like this:

$$-32\int (y^4)(e^ydy)=-32(y^4)(e^y)+32\int(e^y)(4y^3dy)\\

=-32y^4e^y+128\int(y^3)(e^ydy)\\

=-32y^4e^y+128\left((y^3)(e^y)-\int(e^y)(3y^2dy)\right)\\

=-32y^4e^y+128y^3e^y-384\int(y^2)(e^ydy)\\

=-32y^4e^y+128y^3e^y-384\left((y^2)(e^y)-\int(e^y)(2ydy)\right)\\

=-32y^4e^y+128y^3e^y-384y^2e^y+768\int(y)(e^ydy)\\

=-32y^4e^y+128y^3e^y-384y^2e^y+768\left((y)(e^y)-\int e^ydy)\right)\\

=-32y^4e^y+128y^3e^y-384y^2e^y+768ye^y-768e^y\\

=(-32y^4+128y^3-384y^2+768y-768)e^y$$

and substituting for *y*, $$=\left(-32\left(-\frac{x}{2}\right)^4+128\left(-\frac{x}{2}\right)^3-384\left(-\frac{x}{2}\right)^2+768\left(-\frac{x}{2}\right)-768\right)e^{-x/2}\\

=\left(-2x^4-16x^3-96x^2-384x-768\right)e^{-x/2}\\

=-2\left(x^4+8x^3+48x^2+192x+384\right)e^{-x/2}$$

Evaluating this from 0 to 1, we have $$\left[-2\left(x^4+8x^3+48x^2+192x+384\right)e^{-x/2}\right]_0^1\\

=\left[-2\left(1+8+48+192+384\right)e^{-1/2}\right]-\left[-2\left(384\right)e^{0}\right]\\

=768-1266e^{-1/2}$$

If, instead, we do all the work as a definite integral, continuing Doctor Ken’s work, we get the same result:

$$-32\int_0^{-1/2}y^4e^ydy\\

=-32y^4e^y|_0^{-1/2}+128\int_0^{-1/2}y^3e^ydy\\

=-2e^{-1/2}+128\int_0^{-1/2}y^3e^ydy\\

=-2e^{-1/2}+128\left(y^3e^y|_0^{-1/2}-\int_0^{-1/2}3y^2e^ydy\right)\\

=-2e^{-1/2}-16e^{-1/2}-384\int_0^{-1/2}y^2e^ydy\\

=-18e^{-1/2}-384\left(y^2e^y|_0^{-1/2}-\int_0^{-1/2}2ye^ydy\right)\\

=-18e^{-1/2}-96e^{-1/2}+768\int_0^{-1/2}ye^ydy\\

=-114e^{-1/2}+768\left(ye^y|_0^{-1/2}-\int_0^{-1/2}e^ydy\right)\\

=-114e^{-1/2}-384e^{-1/2}-768e^y|_0^{-1/2}\\

=-498e^{-1/2}-768e^{-1/2}+768\\

=768-1266e^{-1/2}$$

We’ll close with this from 2001:

Integral of e^xsinx I have tried to integrate e^x sinx by parts, but I always get e^x sinx or e^x cosx to integrate, andit just keeps going. Could you show me the steps to this indefinite integral?

We did this same integral in Two Integration Puzzlers.

Doctor Rob answered:

Thanks for writing to Ask Dr. Math, George. There is a pretty clever trick to this one. Integrate by parts twice as follows: INTEGRAL u dv = u*v - INTEGRAL v du, u = sin(x), dv = e^x dx, du = cos(x) dx, v = e^x: INTEGRAL e^x*sin(x) dx = e^x*sin(x) - INTEGRAL e^x*cos(x) dx. INTEGRAL u dv = u*v - INTEGRAL v du, u = cos(x), dv = e^x dx, du = -sin(x) dx, v = e^x: INTEGRAL e^x*sin(x) dx = e^x*sin(x) - [e^x*cos(x) - INTEGRAL -e^x*sin(x) dx], = e^x*[sin(x)-cos(x)] - INTEGRAL e^x*sin(x) dx.

That is, $$\int e^x\sin(x)dx=e^x\sin(x)-\int e^x\cos(x)dx\\=e^x\sin(x)-\left[e^x\cos(x) – \int -e^x\sin(x) dx\right]\\= e^x\left(\sin(x)-\cos(x)\right)-\int e^x\sin(x)dx.$$

Now it may seem that you didn't get anywhere by doing this, but notice that the integral on the left and the one on the right are the same. Now just combine like terms by adding that term to both sides, and finallysolve for that integralto get your answer. You can easily verify the answer by differentiation.

If we define \(I=\int e^x\sin(x)dx\), then we have $$I=e^x\left(\sin(x)-\cos(x)\right)-I$$

so that $$2I=e^x\left(\sin(x)-\cos(x)\right)$$

$$I=\frac{1}{2}e^x\left(\sin(x)-\cos(x)\right)+C$$

If we differentiate this as a check, we get $$\frac{dI}{dx}=\frac{1}{2}e^x\left(\sin(x)-\cos(x)\right)+\frac{1}{2}e^x\left(\cos(x)+\sin(x)\right)=e^x\sin(x)$$

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First, a question from Amia in June:

Hi Dr math,

I want your opinion for the solution attached:

Amia started by multiplying the numerator and denominator by a negative power of *x*; that amounts to dividing the numerator and denominator by the greatest power in the denominator, which is a common technique for dealing with limits at infinity; perhaps that is the motivation.

Then he splits the numerator into a multiple of the denominator, plus a remainder; this amounts to long division, as we would do if the numerator had a higher degree than the denominator. The second term can then be integrated by substitution (\(u=1+x^{-2}\)).

We can rewrite the answer without negative exponents, to make it look more like we expect: $$\frac{x^{-1}}{-1}+\frac{1}{2}\ln\left|1+x^{-2}\right|=\frac{1}{2}\ln\left|\frac{x^2+1}{x^2}\right|-\frac{1}{x}+C$$

Altogether, this demonstrates what you can accomplish by “playing with possibilities” rather than following standard methods.

I answered with approval and curiosity:

Hi, Amia.

Your work looks

good, and creative.I did it more routinely, by partial fractions, and got the same result.

May I ask,

what led you to think of doing that?I generally tend to change negative exponents to positive, rather than vice versa, because it’s easy to make mistakes with negatives; but I wonder if this approach could be generalized.

Here is the routine method using partial fractions:

$$\frac{x^2-x+1}{x^4+x^2}=\frac{x^2-x+1}{x^2(x^2+1)}=\frac{A}{x}+\frac{B}{x^2}+\frac{Cx+D}{x^2+1}\\x^2-x+1=Ax(x^2+1)+B(x^2+1)+x^2(Cx+D)\\x^2-x+1=Ax^3+Ax+Bx^2+B+Cx^3+Dx^2\\x^2-x+1=(A+C)x^3+(B+D)x^2+Ax+B$$

Equating coefficients, we get the system of equations

$$\left\{\begin{matrix}\begin{aligned}A+C&=0\\B+D&=1\\A&=-1\\B&=1\end{aligned}\end{matrix}\right.$$

From which we find that $$\left\{\begin{matrix}\begin{aligned}A&=-1\\B&=1\\C&=1\\D&=0\end{aligned}\end{matrix}\right.$$

(Below, we’ll see an example using a different approach, plugging in four values for \(x\) to obtain four equations to solve.)

SoĀ our integral becomes $$\int\left(\frac{-1}{x}+\frac{1}{x^2}+\frac{x}{x^2+1}\right)dx=-\ln|x|-\frac{1}{x}+\frac{1}{2}\ln|x^2+1|\\=\frac{1}{2}\ln\left|\frac{x^2+1}{x^2}\right|-\frac{1}{x}+C$$ just as we got above.

Amia replied:

The shape of this question is similar to questions I solved before

Like

1/(x^3+x)

x/(x^2+x^4)

So I tried here the same procedure, and it works.

Can we generalize this approach?

This wasn’t elaborated, but my understanding is that he is rewriting \(\frac{1}{x^3+x}\) by multiplying the numerator and denominator by *x*, perhaps observing that this will make a substitution (\(u=x^2\)) possible: $$\int\frac{1}{x^3+x}dx=\int\frac{x}{x^4+x^2}dx=\int\frac{\frac{1}{2}du}{u^2+u}\\=\frac{1}{2}\int\frac{du}{u(u+1)}=\frac{1}{2}\int\left(\frac{1}{u}-\frac{1}{u+1}\right)du\\=\frac{1}{2}\left(\ln|u|-\ln|u+1|\right)=\frac{1}{2}\ln\left|\frac{u}{u+1}\right|=\frac{1}{2}\ln\left|\frac{x^2}{x^2+1}\right|+C$$

We still had to use substitution and partial fractions, but the degree was reduced. Using partial fractions directly on the original integral, we would have to do this:

$$\int\frac{1}{x^3+x}dx=\int\frac{1}{x(x^2+1)}dx=\int\left(\frac{A}{x}+\frac{Bx+C}{x^2+1}\right)dx\\=\int\left(\frac{1}{x}-\frac{x}{x^2+1}\right)dx=\ln|x|-\frac{1}{2}\ln\left|x^2+1\right|=\frac{1}{2}\ln\left|\frac{x^2}{x^2+1}\right|+C$$

This doesn’t seem like a great simplification, so there may be another trick I’m missing; but the fact that a multiplication initially appears to complicate the integrand, but then makes a different approach possible, is the link between the two problems.

As for generalizing, this trick works only when it happens to work (if I’m understanding it correctly); so the generalization is just “look for possible shortcuts, rather than being locked into a routine”. That’s good advice once you’ve mastered the routines.

Amia had another problem with another trick:

I have this solution for another question.

This integrand is another proper rational function. I’ll discuss below what he is doing, adding and subtracting \(x^2\) in order to be able to do some impressive cancelling – in both terms, not just in one as above.

For comparison, here is the work for this problem using partial fractions:

$$\frac{7x^2-16x-2}{(x^2+2)(x-2)}=\frac{Ax+B}{x^2+2}+\frac{C}{x-2}\\7x^2-16x-2=(Ax+B)(x-2)+C(x^2+2)\\7x^2-16x-2=(A+C)x^2+(-2A+B)x+(-2B+2C)$$

Equating coefficients, $$\left\{\begin{matrix}\begin{aligned}A+C&=7\\-2A+B&=-16\\-2B+2C&=-2\end{aligned}\end{matrix}\right.$$

This yields $$\left\{\begin{matrix}\begin{aligned}A&=8\\B&=0\\C&=-1\end{aligned}\end{matrix}\right.$$

So the integral becomes $$\int\left(\frac{8x}{x^2+2}-\frac{1}{x-2}\right)dx=4\ln\left|x^2+2\right|-\ln\left|x-2\right|\\=\ln\left|\frac{(x^2+1)^4}{x-2}\right|+C$$

The shortcut obtained exactly the results of the partial fractions.

I responded, applying this idea to the first problem:

I had considered mentioning a third method I had used for your problem that is essentially the idea you use in this new one, which might be called “partial cancellation”:

$$\int\frac{x^2-x+1}{x^4+x^2}dx=\int\frac{x^2-x+1}{x^2(x^2+1)}dx\\=\int\frac{(x^2+1)-x}{x^2(x^2+1)}dx=\int\left(\frac{x^2+1}{x^2(x^2+1)}-\frac{x}{x^2(x^2+1)}\right)dx\\=\int\left(\frac{1}{x^2}-\frac{x}{x^2(x^2+1)}\right)dx=\int\frac{1}{x^2}dx-\int\frac{x}{x^2(x^2+1)}dx$$

and so on.

This saves some of the work of partial fractions.

Here I saw that one of the factors of the denominator is found in the numerator, so that splitting the fraction in that way permitted cancellation in each of the resulting fractions.

Let’s finish the work, avoiding the temptation to cancel in the second fraction, and instead taking advantage of an opportunity for substitution (\(u=x^2\)), just as I did in what I called “the inspiration” above:

$$\int\frac{1}{x^2}dx-\int\frac{x}{x^2(x^2+1)}dx=\int x^{-2}dx-\frac{1}{2}\int\frac{du}{u(u+1)}\\=-x^{-1}-\frac{1}{2}\int\left (\frac{1}{u}-\frac{1}{u+1}\right)du=-\frac{1}{x}-\frac{1}{2}\left(\ln\left|u\right|-\ln\left|u+1\right|\right)\\=-\frac{1}{x}-\frac{1}{2}\ln\left|\frac{u}{u+1}\right|=-\frac{1}{x}-\frac{1}{2}\ln\left|\frac{x^2}{x^2+1}\right|=\frac{1}{2}\ln\left|\frac{x^2+1}{x^2}\right|-\frac{1}{x}$$

The aspect of your original solution that caught my attention was the use of

negative exponents, which seems highly unusual; but this aspect, breaking the numerator into two parts that can becanceled separately, is also present, and can definitely be used for many problems.In your work this time, you’ve taken that idea one step further, using a technique I’ve seen used in factoring (e.g.Ā Factoring Tricks from an Old Textbook).

In general, integration is an art, and setting yourself free to consider non-standard options when you see they might be useful is a valuable skill. Both aspects of your solution are potentially useful.

In the page referred to, I called this trick “wishful thinking” (in problems A and D), adding and subtracting a term (\(x^2\) for Amia) that put one part into a desirable form:

Back in March of last year, Amia had a similar question, this time not evading partial fractions, but trying an alternative detail:

Hi Dr math,

I have a question about integration by partial fractions.

The question is attached:

Here he is using partial fractions, but chooses to take a linear numerator for the term with denominator \((x-1)^2\), rather than split it into two terms, one with denominator \((x-1)\), and the other \((x-1)^2\), as we traditionally do. We mentioned this possibility (and showed why it is not preferred) in Partial Fractions: How and Why; the example there has denominator \((x-2)^3\) and we consider using \(Bx^2+Cx+D\) as the numerator, but point out that that is not useful for integration. Here Amia is demonstrating that it can be done, and what happens when you do it, which will add to what we said there.

He uses the alternative method for solving for *A*, *B*, and *C* that I mentioned above. When it comes to integrating this term, he uses the “partial cancelling” method we saw above, splitting it again into two fractions (which, as we’ll see, are the same as we’d get the other way).

Unfortunately, he makes a small error or two, so I’ll redo his work in order to have a correct answer to compare:

$$\frac{3x^2+2}{x(x-1)^2}=\frac{A}{x}+\frac{Bx+C}{(x-1)^2}\\3x^2+2=A(x-1)^2+(Bx+C)x$$

Setting *x* to 1, 0, and 2 as he did, we get $$\left\{\begin{matrix}\begin{aligned}&x=1:5=B+C\\&x=0:2=A\\&x=2:14=A+4B+2C\end{aligned}\end{matrix}\right.$$

Solving for the constants, $$\left\{\begin{matrix}\begin{aligned}&A=2\\&B=1\\&C=4\end{aligned}\end{matrix}\right.\\$$

so that the new form is $$\frac{3x^2+2}{x(x-1)^2}=\frac{2}{x}+\frac{x+4}{(x-1)^2}$$

(You can check that this is true.)

Now we integrate, using Amia’s trick:

$$\int\frac{3x^2+2}{x(x-1)^2}dx=\int\frac{2}{x}dx+\int\frac{x+4}{(x-1)^2}dx\\=\int\frac{2}{x}dx+\int\frac{(x-1)+5}{(x-1)^2}dx\\=\int\frac{2}{x}dx+\int\frac{1}{x-1}dx+\int\frac{5}{(x-1)^2}dx\\=2\ln|x|+\ln|x-1|-\frac{5}{x-1}\\=2\ln\left|\frac{x}{x-1}\right|-\frac{5}{x-1}+C$$

We’ll compare this to the standard method below.

Doctor Fenton answered, showing that the process is equivalent:

Hi Amia,

Yes, that is valid, but you are not really doing anything differently.Ā Note that

$$\frac{Bx+C}{(x-1)^2}=\frac{B(x-1+1)+C}{(x-1)^2}=\frac{B(x-1)+B+C}{(x-1)^2}\\=\frac{B}{x-1}+\frac{B+C}{(x-1)^2}=\frac{B}{x-1}+\frac{D}{(x-1)^2}$$

where D = B + C.

Instead of solving for B and D, you solve for B and the sum B + C, from which you can determine D.

This shows that his work produces the same form as the usual partial fractions method. Let’s actually do the latter and see that we get the same fractions:

$$\frac{3x^2+2}{x(x-1)^2}=\frac{A}{x}+\frac{B}{(x-1)}+\frac{C}{(x-1)^2}\\

3x^2+2=A(x-1)^2+Bx(x-1)+Cx\\

3x^2+2=Ax^2-2Ax+A+Bx^2-Bx+Cx\\

3x^2+2=(A+B)x^2+(-2A-B+C)x^2+A$$

Equating coefficients, we get the system of equations

$$\left\{\begin{matrix}A+B=3\\-2A-B+C=0\\A=2\end{matrix}\right.$$

From which we find that $$\left\{\begin{matrix}A=2\\B=1\\C=5\end{matrix}\right.$$

This gives the fractions $$\frac{2}{x}+\frac{1}{(x-1)}+\frac{5}{(x-1)^2}$$

Integrating, we get

$$\int\frac{2}{x}dx+\int\frac{1}{x-1}dx+\int\frac{5}{(x-1)^2}dx\\=2\ln|x|+\ln|x-1|-\frac{5}{x-1}\\=2\ln\left|\frac{x}{x-1}\right|-\frac{5}{x-1}+C$$

just as we did above.

Amia replied:

Thank you Dr,

But I see it is easier when we do the integrations:

A/(x-1), B/(x-1)Ā².Ā ??

I’m right?

Doctor Fenton responded:

Yes.Ā When you integrate

(Bx + C)/(x – 1)

^{2}Ā ,unless C = -B, in which case it becomes B(x – 1)/(x – 1)

^{2}= B/(x – 1), then you will have to break up the integrand in the same way.

As we said in the older page, “Our motivation for partial fractions is their use in calculus, and that drives the required form of the fractions. What I suggested above would otherwise be perfectly valid, and we would not need all the specific rules that are taught. But then the calculus would be harder. (I will not go into the details, but you might like to try integrating an expression of the form above, to see why.)”

So despite the fact that we can sometimes use special tricks to simplify part of the work, on the whole, the method of partial fractions has been refined over the years to a nearly-optimal method.

]]>

We’ll start with this 2001 question about two definitions:

Rate vs. Ratio In a class with other math teachers, we were discussing the relation of rate versus ratio.Are all rates ratiosandare all ratios rates?What is your take on this and how would you explain this?

That is, would a Venn diagram look like this (all rates are ratios):

Or like this (all ratios are rates):

or something else?

I answered with some research (some links have been redirected to Archive.org):

Hi, Stacy. I've seen different ways to distinguish these concepts, so I looked around to find some samples. This page Math League, Ratio and Proportion http://www.mathleague.com/help/ratio/ratio.htm says "Aratiois a comparison of two numbers. "Arateis a ratio that expresses how long it takes to do something, such as traveling a certain distance. To walk 3 kilometers in one hour is to walk at the rate of 3 km/h. The fraction expressing a rate has units of distance in the numerator and units of time in the denominator." This approach is more or less what I use in real life;ratio is the general concept, and rate refers _usually_ to time (though not always, as I'll mention below). Soall rates are ratios.

That is, “rate” is a **specific kind of ratio**; but this makes it a little *too* specific.

But it's common to make a different distinction: Jim Reed, Rate and Ratio http://dev1.epsb.edmonton.ab.ca/math14_Jim/math7/strand1/1208.htm (which, ironically, has a link to the page above) says "Rate: A comparison of 2 measurementswith different units." "Ratio: A comparison of numberswith the same unitsso units are not required." This makes ratio and rate both specific terms, so thatneither is a special case of the other. That seems very orderly, but I don't like it, because (a)generic terms are very useful, in order to be able to talk about how we work with ANY entity of this sort, and (b) thisdoesn't fit common usage, as in "tax rate" or "interest rate," where both numbers have the same unit. Of course, math doesn't generally worry too much about common usage, but since I see no real mathematical advantage in this distinction, we might as well be realistic.

This would make them distinct concepts:

But inclusive definitions are generally preferred in math. We need a word that covers both!

The next, MathSteps, Rates http://www.eduplace.com/math/mathsteps/6/e/ says "Aratiois a comparison of two numbers or measurements. The numbers or measurements being compared are called the terms of the ratio. Arateis a special ratio in which the two terms are indifferent units. For example, if a 12-ounce can of corn costs 69 cents, the rate is 69 cents for 12 ounces. The first term of the ratio is measured in cents; the second term in ounces." This makes ratio generic again, satisfying one of my objections, but requires rates to have different units. We could allow that to be the mathematical definition, and not worry about common usage; but it doesn't quite stand up to the sorts of things a mathematician thinks about. If you have a "rate" of one ounce per pound, does that become a "ratio" of 1:16 when you change units to ounces per ounce? In my mind, a change of units should not change what the ratio is called; it's aratio between two quantities, not between the numbers currently being used to represent them. If we use one ounce of pigment per pound of paint, is that a rate or a ratio? I might call it aratewhen I'm mixing it, because I think of it as "per" or "for each," but aratiowhen I'm using it, because it just describes the mixture!

The distinction can become very cloudy!

Finally, let's check a dictionary. According to Merriam-Webster, http://m-w.com/ we have "Ratio1 a : theindicated quotientof two mathematical expressions b : the relationship in quantity, amount, or size between two or more things : PROPORTION" "Rate3 a : afixed ratio between two thingsb : a charge, payment, or price fixed according to a ratio, scale, or standard: as (1) : a charge per unit of a public-service commodity (2) : a charge per unit of freight or passenger service (3) : a unit charge or ratio used by a government for assessing property taxes (4) British : a local tax 4 a : a quantity, amount, or degree of something measured per unit of something else b : an amount of payment or charge based on another amount; specifically : the amount of premium per unit of insurance" Though this isn't necessarily a mathematician's or teacher's definition, it supports my sense that "ratio" is generic, while"rate" applies to several slightly different kinds of ratios. A rate generally involves a "something else," eithertwo different kinds of units(such as distance per time), or justtwo distinct thingsmeasured with the same unit (such as interest money per loaned money). I don't know whether that helps, but at least it may let you move on to more significant issues.

My final statement here seems most reasonable. The answer is

In the questions below, we will be implicitly restricted to ratios of the non-rate sort, ratios between either two quantities of the same type (e.g. pounds of butter and pounds of sugar) that could be combined into a single whole, or, even more so, that whole and a part of it. Rates will not fit any of this discussion.

Next, from 2003:

Definition of Ratio I am a copy editor for a public-relations firm that works with clients in high technology. Recently, I lost a battle with other editors on the staff who insisted that a construction such as"eight out of ten" is a ratio. I said I didn't think it met the mathematical criteria for ratio, but since I don't know precisely how "ratio" is defined by mathematicians, I was unable to argue persuasively - and don't know for a fact that my co-editors aren't right, that "eight out of ten" (or 8 out of 10) IS a ratio. Can you help?

Suzanne presumably sees ratios as necessarily “part to part” rather than “part to whole” (like a fraction). Can a fraction be called a ratio?

I answered:

Hi, Suzanne. This is as much an English language question as a math question, and that makes it very confusing.Words like this are not used as consistently as you might expect, even among math teachers or mathematicians.(That's partly because mathematicians today don't tend to pay much attention to ratios, and therefore don't have to define them carefully.) My first impression is that we TEND to think of ratios as comparisons of, say, the number of boys to the number of girls, rather than ofa part to a whole, but that the term "ratio" does not necessarily exclude the latter. So it might not be technically wrong to use it that way, but depending on the context there might be clearer ways to phrase what you want to say, so as to avoid suggesting that the comparison ispart to part.

So, if you just mentioned a ratio, I would probably expect the ratio of two parts of a whole; but you might not mean that.

As in the question above, though, I like to confirm my opinion on questions like this:

Then I did a little searching. For word questions, I like to see what a dictionary says, since lexicographers know how to make distinctions between words (though they often don't understand the mathematical distinctions). Merriam-Webster (m-w.com) says "ratio1 a : the indicated quotient of two mathematical expressions b : the relationship in quantity, amount, or size betweentwo or more things: PROPORTION" This agrees with my general sense of the word:anyquotient can be called a ratio, but in particular ittendsto compare two distinct things. But, of course, we make a distinction between ratio andproportion, and they call them synonymous. How do they define the latter?

Here is another ambiguity: “Proportion” can be used either as a synonym for “ratio”, or as an equation of ratios:

"proportion3 : the relation ofone part to anotheror to the wholewith respect to magnitude, quantity, or degree : RATIO 4 : SIZE, DIMENSION 5 : a statement ofequality between tworatiosin which the first of the four terms divided by the second equals the third divided by the fourth (as in 4/2=10/5)" Their definition 5 is the one we use technically, in distinction to ratio. But look at their definition 3: When they consider this synonymous with ratio, they specifically include relations ofpart to wholeas well aspart to part. That would say that your example IS a ratio.

So much for the dictionary …

How about math sites? Mathematicians, as I said, don't deal with this much, but math teachers do. Here is one reference I found: Ratio. Fraction. What's the Difference? http://www.sci.tamucc.edu/txcetp/cr/math/rf/RatioFraction.pdf "Due to common notation, students often improperly interchange the ideas ofratioandfraction. Through this lesson students will learn why the two aredifferentideas and when they actually canoverlap. ... "Discuss the fact thatfractionsalways illustrate a "part to whole" relationship whileratioscan be used to illustrate a much larger set of relationships; such aspart to partandwhole to part." This seems to say that a fraction always refers to part of a whole, but a ratio can indicate a variety of relationships. There are some examples in a worksheet, but no answers.

This agrees with my impression.

The next page I found happened to be the answer sheet, and illustrates what they mean: "2.Two out of every fivestudents in this class plan to be middle school teachers. Circle One: Ratio Fraction[Both]"Reasoning: "This is definitely aratioof future middle school teachers to total students in the class of 2:5. In addition, since the ratio ispart to whole, it canalsobe thought of as afractionalrelationship. 2/5 of the students in the class plan to be middle school teachers." So "two of every five" is considered a ratio.

I added a little more, and closed:

I found lots of pages that talk aboutratio of part to whole, as well as part to part. So it looks like you've lost - but I'd still like to see the context in which you don't think the word "ratio" fits, because even if something CAN be called a ratio, there may be reasons not to use the word "ratio" in a particular sentence.

Suzanne responded:

Wow, thank you for the incredibly thorough explanation you so quickly provided. I admit to be being math-phobic, but I found your examples illuminating. Thanks again, and I'll quit arguing with my co-workers. Or find something else to argue about, more likely.

A question from 2013 questioned that:

Ratios and Fractions: A Reconciliation I am emailing you for support. Three math teachers were having a discussion today about howfractionsshow a part of a whole, butdo not express part to part. Here is an example. In a class of 25 students, there are 15 boys and 10 girls. This represents the ratio of boys to girls: 15:10 But this cannot translate to 15/10. Saying 15 boys out of 10 girls would be like comparing apples to oranges! You can say 15/25 (15 boys out of 25 students) 10/25 (10 girls out of 25 students) One teacher offers this website below as her support that fractions can be "part to part," and thatyou can express a "part to part" as a fraction. https://www.khanacademy.org/math/pre-algebra/rates-and-ratios/ Ā Ā Ā ratios_and_proportions/v/introduction-to-ratios-new-hd-version I disagree. But when I consult Dr. Math -- my "go to" site for math questions -- I find this explanation: http://mathforum.org/library/drmath/view/63884.htmlCan a fraction be part to part?

The Khan Academy link is dead, so we can’t be sure exactly what it said; I can’t find a current video there that shows such ratios as fractions.

That last reference is our previous question, where I, too, supported “part-to-whole” ratios. But is this the same as a **fraction**?

I answered:

Hi, Anne. The last page you referred to (by me) says essentially thata fraction is a ratio-- that is, a ratio can compare part to whole. You are looking at the opposite question:Can a fraction compare part to part?One thing that gets in the way of this sort of discussion is that in practice we find thatwe can WRITE most ratios as fractions, unifying the two concepts so that we don't need to learn two entirely separate concepts (as they did in the old days). So even though something isfundamentally a part-to-part ratio of 3:2, and not a fraction, we canWRITE it as 3/2and do all the same simplifying we'd do with any fraction.

That is, it is common to **write** ratios using the slash (“/”); but does that mean such a ratio **is** a fraction?

There are major differences, of course --a ratio can be 4:0whilea fraction can't be 4/0; you can havecompound ratios like 2:3:4, which don't correspond to any sort of fraction; and you don't have"improper" ratiosor "mixed" ratios. But we do often blur the distinction in cases where it doesn't affect how we work with a ratio. This, I think, is what Khan is doing: not saying that the ratios he's working with ARE fractions, but thatwe can use fraction NOTATION as a way to write and work with ratios. Anyone with experience in math will probably do just what he does -- develop a habit of writing ratios as fractions when it makes sense to do so -- because that simplifies our thinking.

Most of the work we do with ratios is identical to what we do with fractions, so we tend to borrow the notation. But not all ratios make sense that way.

Where the fraction notation becomes especially helpful is with algebra. In the old days,proportionshad to be written using ratio notation (and couldn't even use "="): 3:2 :: x:10 To solve this, they would say "the product of themeans[the middle two numbers] equals the product of theextremes[the outside numbers]," and write 2*x = 3*10 Today, we tend to write the ratios in a proportion as fractions, so that we can use algebra (and/or "cross multiplication") to solve: 3 x --- = --- ---> 2x = 3*10 2 10 So, while there is value in teaching adistinctionbetween ratios and fractions in terms of their applications and meanings, there is also value in learning totransformone into the other in order to work with them conveniently. An overly rigid distinction in the elementary grades might even get in the way of learning later on, and can at least lead to deprecating good techniques used by adults.

The old notation and terminology is discussed in Many Ways to Solve a Proportion.

Notice that in my example, the “fractions” are improper, but that doesn’t matter because we are just working with *numbers*, not with *parts and wholes*.

In fact, a ratio is closely related to a fraction even conceptually. If the ratio of boys to girls is 3:2, then the number of boys is 3/2 of the number of girls; there are 1 1/2 boys for every girl. So the boundary between ratios and fractions is not impermeable. What is my answer to your question, you ask? In my opinion, both sides in your disagreement are right: You're right thatpart-to-part ratios are not really fractions, but they're right thatwriting them as fractions can be appropriate.

We’ll close with a 2014 question:

Fractions as Ratios: Wholly or Partially? How wouldratio and fractionbe placed in aVenn diagram? Would fraction fall totally within the ratio circle, or only partially within it? I guess another way to ask this question is, "Are ALL fractions ratios?" I have asked this question at the university level and there is disagreement. What is your opinion?

Which is right?

Or maybe this?

I answered:

Hi, Donna. I would expect some differences of opinion, particularly because each word is used in variouscontexts, and one might give a different answer depending on what one is thinking about when the question is asked. It would be illuminating to hear what was said by the people you asked, not just as a true/false question, but in terms of thereasonsthey give. I myself gave a generally affirmative answer to this question here: Definition of Ratio http://mathforum.org/library/drmath/view/63884.html Here we're thinking of both ratios and fractions as representations of a relationship between numbers, where ratios can be "anything to anything," whilefractions are always "part to whole."

This again is our second question above, where I seem to make fractions a subset of ratios.

But from a different perspective, I would consider a fraction to be anentirely different kind of thingfrom a ratio: a fraction is a way of writing a NUMBER -- namely as an "indicated quotient" -- whereasa ratio is not a number at all, but a comparison of two numbers. In this sense, we write fractions all the time without thinking at all of "part-to-whole" relationships or any such application of the concept. For example, when I write the equation 2/3 x - 1/2 = 5/6, I would not normally think of those fractions as ratios. On the other hand, you could choose to represent them as ratios if you had reason to.

You may notice that we quoted above a definition of a *ratio* as an “indicated quotient”! I didn’t fully approve of that; “quotient” means division, and a ratio is not actually a division, but a relationship. (But note that the colon we use in ratios is used in many countries as the division sign, so the distinction is even more confused.)

I wrote from this perspective here, emphasizing different senses in which the word "fraction" is used: What Exactly is a Fraction? http://mathforum.org/library/drmath/view/56034.html So, as arelationshipbetween numbers,fractions can be considered to be a subset of ratios, only written differently. But as a way ofwritinga number,a fraction is not even in the same universe as a ratio, so you wouldn't put them in the same Venn diagram.

That question is the basis of What is a Fraction, Really?

In summary, when asked specifically about the relationship between ratios and fractions, I would say the latter are asubset, and would make the Venn diagram; but when asked what a fraction IS, without reference to a Venn diagram,I might not mention ratios at all. And if you then asked me whether all fractions are ratios, I'd say, "Of course not." That may be what your respondents are doing -- but I'd still like to hear some of their explanations for their opinions, which might reveal even more perspectives I haven't thought of.

Here is an attempt at a Venn diagram including all that we’ve said here:

A **part-to-part ratio** can be *represented as* a fraction, but does not *represent* a fraction. When seen as a ratio at all, a **fraction** might represent a **part-to-whole ratio**, or a **rate** (which, again, involves different sorts of quantities that are not part of one whole).

If we think of fractions as numbers and not ratios, we get this alternative:

Now a fraction is just a way to write a number. We could say that a rate is actually a number, and so belongs on the right side; but I won’t try get everything exactly right!

]]>

First, a question about carpet from 2008:

Finding the Length of Carpet Left on a Roll I can see the end of a roll of carpet and figure out the approximate square yardage by taking theinside circumference,outside circumferenceandnumber of total layers left. I need to see if there is a formula. EX: 12"=inside roll circumference, 24"=outside roll circumference, 12 layers on end of roll. The difference between 24" and 12" is 12", so each layer must be about 1" more than the previous layer (i.e. first layer length = 12", 2nd = 13", 3rd = 14" . . . 24th = 24"). Now, when I add all layers together I get a certain total length. What is the equation for such a problem?

Last time we got a formula using the number of turns and diameter; here we have circumferences rather than diameters. Lee suggests using series, which we also tried last time.

I answered Lee:

We have many answers to this kind of question; you might try searching our archives for the phrase "roll of carpet" to find those that happen to be about carpets. Here is one, with references to others: Determining Length of Material Remaining on a Roll http://www.mathforum.org/library/drmath/view/67031.html My formula there isL = pi/4 (Do^2 - Di^2)/twhere Do = outer diameter Di = inner diameter t = thickness of one layer

Now we have to adapt it:

You give thecircumferences, which we'll have to divide by pi, and since you know the number of layers, we can make a different formula based on a number of layers N. Let's say we have Co = outer circumference [so Do = Co/pi] Ci = core circumference [so Di = Ci/pi] N = number of layers [so t = (Do-Di)/(2N) = (Co-Ci)/(2 pi N)] Then my formula becomes L = pi/4 (Co^2 - Ci^2)/(pi^2) * (2 pi N)/(Co - Ci) = (Co + Ci)*N/2 For your numbers this gives L = (24 + 12)*12/2 = 216 inches

This is essentially the same as the last answer given last time, but using circumference rather than diameter: $$L=\frac{1}{2}N(C_o+C_i)$$

What about his series approach? There’s a little problem there, that will lead to a new idea:

This is exactly what I would get as an approximation using your idea of thelinear increase in circumferencewith each layer. The sum of an arithmetic series like your 12 + 13 + 14 + ... + 23 (note thatyou counted 13 layersby including both 12 and 24) is the average of the first and last, times the number of terms. In your example, this is 12*(12+23)/2 =210inches. But what about the last layer? We only got to a length of 23. If wesplit the differenceand treat the series as 12.5 + 13.5 + ... + 23.5, our length turns out to be 12*(12.5+23.5)/2 =216. And in fact, we could derive the formula above starting with this idea of the arithmetic series.

He had too many layers, by counting both the inside and outside circumferences. But counting only the 12 layers this way leads to a too-small answer – or does it?

Here I made the adjustment we made in our series derivation last time, by treating the length of a layer as the circumference in its middle. This approach amounts to assuming the carpet bends in such a way that the inside compresses and the outside expands (as if there were an incompressible layer in the middle):

This might apply to, say, a foam mat being rolled up. But this led me to a new speculation:

Actually, I suspect that in the case of carpet it is possible thatthe backing would be more rigid than the top, so that the actual length would be better estimated by the 210 inch figure; my 216 assumes that the outside of each layerexpandsand the insidecontractsequally when it is rolled up. The formula using your idea turns out to be L = (Co + Ci - (Co - Ci)/N)*N/2 =((N-1)Co + (N+1)Ci)/2I've never looked at it from this perspective before and obtained this particular formula! It's not quite as simple, and considering that the carpet really forms a spiral rather than neat layers, it's still just an approximation; but it has an interesting "elegance"!

Picture the carpet this time with an incompressible backing on the inside of the roll, the rest being expandable fibers:

To get this formula, we canĀ take the total length as the sum of the inner circumference of each layer. The first layer has length \(C_i\), and the last isĀ \(C_i+(N-1)\frac{(C_o-C_i)}{N}=\frac{NC_i+(N-1)(C_o-C_i)}{N}=\frac{C_i+(N-1)C_o}{N}\). The sum of these *N* terms is N times the average of the first and last terms: $$L=\frac{N}{2}\left(C_i+\frac{C_i+(N-1)C_o}{N}\right)\\=\frac{N}{2}\left(\frac{(N+1)C_i+(N-1)C_o}{N}\right)\\=\frac{(N+1)C_i+(N-1)C_o}{2}$$

This formula amounts to averaging the inner and outer circumferences (as in our final formula last time), but weighting the former more heavily than the latter. Using Lee’s numbers, it givesĀ $$\frac{(11)24+(13)12}{2}=210$$ as we found before.

On the other hand, in editing this, I realized that carpet is commonly rolled with the backing on the *outside*! That makes sense for several reasons. In this case, the formula becomes $$L=\frac{(N-1)C_i+(N+1)C_o}{2},$$ with the outside given greater weight.

Now consider this question from 2001:

Calculating the Length of String on a Reel What is the formula for calculating thelength of wirethat can go on a reel (spindle) where the reels may have differentdiameters,hub diameters, andwidths, and thewires may have different diameters?

This is like the carpet problem, except that each layer is made up of many wires side by side.

I answered:

Hi, Larry. The answer will be similar to that for a similar question we have answered many times, about the amount of material (such as carpet) on a roll: How Much Carpet is Left? http://mathforum.org/dr.math/problems/gillette.03.27.99.html We find the length by equating the volume of material on the roll to its volume laid out flat.

There we talked about the cross-section area, but we could just as well multiply everything by the width to get volumes. With wire, we need to.

There is always a caveat in this calculation, however: we have toassume that the material retains the same thickness as it bends around the roll, and is not either compressed or wound loosely. The calculation is verysensitive to the actual thicknessof the material on the roll. The same is even more true in your case. The wire may be wound in different ways that will incorporate different amounts of air space between turns, so it is hard to be sure how much of the volume on the reel consists of wire. The best thing to do would be tomeasure an actual reel of known lengthin order to determine this ratio and adjust the formula, rather than blindly trust theory.

This is something we did in the last couple explanations last time. It’s easy to picture winding up the wire very carefully and tightly, or very haphazardly and loosely, giving different amounts in the same diameter. Here, we’ll assume the tightest possible case, and then multiply by a factor to account for reality.

Having said that, let's look at the problem. For simplicity, I'm going topretend that the wire has a square cross section, since then there would be no need for space between turns. Your measurements will determine how closely this corresponds to reality. Suppose the reel hashub diameter Di,outside diameter Do, andwidth W. Then the volume of wire on it is V = pi(Do^2 - Di^2)W/4 Suppose you have a wire withlength Land square cross-section with sideDw (the diameter of the wire). Its volume is V = LDw^2 Setting these equal, we get LDw^2 = pi(Do^2 - Di^2)W/4 and we can solve to find L = pi(Do^2 - Di^2)W/(4Dw^2)

This is the ideal case: $$L=\frac{\pi(D_o^2-D_i^2)W}{4D_w^2}$$

Now suppose youmeasure an actual reel, and the actual length is X times this calculated length (let's hope not too far from 1). Then you can take X as the typicalcorrection factor, and multiply any calculated length by X to estimate the actual length when you don't know it ahead of time. This correction factor may well be more or less constant for all reels wound by the same method.

If you have questions, they may be answered below.

A 2003 question asks about a different formula:

Length of Cable on a Reel What is the formula used to determine thelength of cable on a reel? Specifically the way to determine the "k" factor if I use the formulaL = A*(A+B)*C*kwhere L = length of wire (in feet) A = diameter of wire (in inches) B = diameter of drum (in inches) C = width of drum (in inches) k = someempirical factor. If I were going to just run a bunch of iterations (can't find k value), and if I were using a 30-inch drum 30 inches wide and winding 2.5 inch cable, would I determine thelength of the first layerusing a diameter of 30 inches or 31.25 inches (half of the wire diameter)? What do I use for a k value? (My reference book only goes to 1.5" diameter.) What do I use for the initial diameter? If I use a 30"x30" drum and 2.5" cable and run three iterations starting with the first "layer" diameter of 30", I can get 329.87' of cable in three layers. But if I start with a diameter of 31.25" I can get 353.43' of cable in three layers. Which method is correct?

It will turn out that the definition of *A* is misleading, so we’ll have to clear that up. Skip down to the bottom to see what he really means!

But note that he is doing some good thinking, akin to what we did above, in working out a series derivation of the carpet formula, wisely pondering whether to use the inner radius or the middle radius for each layer.

I answered again:

Hi, Joe. Your formula seems very odd; for example, if you increase the diameter of the cable, that should decrease the amount that will fit on the reel, yet your formula shows an increase. Did you copy the formula and definitions right? Ah! Looking closer at your discussion, I realize that your "diameter of wire" is apparently not the diameter of the cable itself, but theoutside diameter of the full reel, while "diameter of reel" is the size of thecorearound which it is wound.Your "k" apparently incorporates the diameter of the cable, so you have to look it up in a table rather than include the cable diameter in the formula.

This suggests that Joe’s *A*, *B*, and *C* are our \(D_o\), \(D_i\), and \(W\), respectively, and *k* is a function of \(D_w\). But it will turn out to be a little different.

I referred to the answer above:

The formula I came up with there is L = pi(Do^2 - Di^2)W/(4Dw^2) where L = length of wire (all measurements in the same unit) W = width of reel Di = diameter of core Do = diameter of full reel Dw = diameter of wire As I suggest there, you will probably have to multiply by an arbitrary factor that depends on the nature of the wire and the way it is wound, but will not vary (much) for different sizes of reel and perhaps of wire.

This repeats what we said above. Can this be put into Joe’s form?

Using your variables, plus "d" for the diameter of the wire itself, this would be L = length of wire (in feet, requiring division by 12 below) A = diameter of full reel (in inches) B = diameter of drum [core] (in inches) C = width of drum (in inches) L = pi/12 (A^2 - B^2)C/(4d^2) = pi/(48d^2) (A - B)(A + B)C Comparing this to your L = A*(A + B)*C*k it seems that you are not subtracting B from A (or elseI am still misunderstanding what A is), and that k = pi/(48d^2) times whatever adjustment factor is necessary to account for space between turns.

Comparing the forms, you might be able to guess that \(A-B\) is somehow incorporated into the actual meaning of *A*. We’ll find that to be true.

Using your example, my formula would give, for a 30-inch by 30-inch reel holding three layers of 2.5-inch cable (so that the outer diameter A is 30+6*2.5 = 45), L = pi/(48*2.5^2) (45-30)(45+30)30 = 0.01047*15*75*30 = 353.4 ft If that is not reasonably close to what you actually measure, I would recommend winding several layers, and then use the ratio of the actual length to the calculated length as an additional factor in the equation, in order to make it more realistic.

Not quite.

Joe replied:

Thanks for your help. To clear up what A is.A = 1/2 thickness of all of the wire from the outside of the hub, or the radius of thickness of the wire wound on the drum, or theoutside radius minus the drum radius. I got this formula from a pocket reference book; however, the k values they use don't go high enough for my application.

I’ll incorporate this into the formula soon, but Joe had a further question that I answered first:

I understand your formula and agree with your conclusions. However, I have one more question. When you set the volume of the wire equal to the square of the wire thickness dw^2, (assuming a square cross section) the formula works. But if you use the actual circular cross sectional area (pi*r^2)*length of wire and set that volume equation equal to the first equation the answer is wrong.Why do you assume a square cross-sectionwhen in actuality it isn't square but rather a circle?

I responded:

Hi, Joe. I assumed a square cross-section in orderto include both the wire itself and the air around it, supposing that the wires lie in this pattern: ***** ***** ***** ** ** ** ** ** ** * * * * * * * * * * * * * * * * ** ** ** ** ** ** ***** ***** ***** ** ** ** ** ** ** * * * * * * * * * * * * * * * * ** ** ** ** ** ** ***** ***** ***** ** ** ** ** ** ** * * * * * * * * * * * * * * * * ** ** ** ** ** ** ***** ***** ***** and not, say, this: ***** ***** ***** ** ** ** ** ** ** * * * * * * * * * * * * * * * * ** ******* ******* ** ****** ******* ****** * * * * * * * * * * * ****** ******* ****** ** ******* ******* ** * * * * * * * * * * * * * * * * ** ** ** ** ** ** ***** ***** *****

That is, the assumption we made is that the wires are arranged like this:

while they could be arranged more tightly as

In the former case, each circle can be thought of asincluded in a squarethat fits with those around it, filling the whole space. In the second picture, we would have to usehexagons, and then we would still bemissing some airat the top, bottom, and sides.

There’s also another effect here.Ā In the second form, the distance between centers of wires in successive layers, rather than \(D_w\), is \(\frac{\sqrt{3}}{2}D_w\approx0.866\):

This makes the layers closer together.

So, ignoring the effects at the ends of layers, our answer would be multiplied by about \(\frac{4}{3}\):

$$L=\frac{\pi(D_o^2-D_i^2)W}{4\left(\frac{\sqrt{3}}{2}D_w\right)^2}=\frac{\pi(D_o^2-D_i^2)W}{3D_w^2}$$

But in reality, wires probably lie *less* neatly than the square arrangement, rather than *more*, so the actual length would be smaller.

In any case, apart from edge effects that would make our formula inaccurate for small amounts of wire, we are really justmultiplying the circular cross-section by some factor related to how the wires lie, which we really have to adjust at the end anyway, since nothing works as well in reality as in the mathematical approximations we're using. The square assumption just keeps things simple.

So the factor *k* will adjust for whatever the reality is, and, as Joe stated, has to be determined empirically (that is, by measuring actual reels of wire).

Now, let’s get the formula right:

I was wondering if your A might be my (Do - Di); but your calling it a "diameter" kept me from considering that it might be, as you now say it is, (Do - Di)/2. Thenmy Do - Di is your 2Aandmy Do + Di is your 2(A+B)[that is, Do + Di = (Do-Di)+2Di = 2A + 2B], so my formula using your variables is L = length of wire (in feet, requiring division by 12 below) A =thickness of wire on reel(in inches) B = diameter of drum [core] (in inches) C = width of drum (in inches) L = pi/12 (Do - Di)(Do + Di)W/(4Dw^2) = pi/12 2A 2(A+B) C/(4d^2) = pi/(12d^2) A(A+B)C and your k is pi/(12d^2), times any adjustment factor you need. I'm curious to see how their k's agree with this, and whether they adjust it using more than just a constant factor. In any case, it sounds like we've got it figured out.

So the formula is now $$L=\frac{\pi}{12}\cdot(D_o-D_i)\cdot(D_o+D_i)\cdot\frac{W}{4D_w^2}\\=\frac{\pi}{12}\cdot2A\cdot2(A+B)\cdot\frac{C}{4d^2}\\=\frac{\pi}{12d^2}A(A+B)C$$ So *k* is some multiple of \(\frac{\pi}{12d^2}\).

Joe added:

Thanks for all your help, I agree with your version of using a square cross section to include the air spaces. If you're still interested, the k values from the table are .25" = 3.29 .5" = 0.925 .75" = 0.428 1" = 0.239 1.25" = 0.152 1.5" = 0.107 1.75" = 0.0770 Thanks.

Putting this into a spreadsheet to compare it to my “*k*“, we see this:

I closed:

Thanks. It looks as iftheir k is about 0.9 times mine, slightly more for the smaller wire sizes. It makes sense that this is less than 1, since you would want to allow for non-optimal layering, which would fit less cable in the same space. You may want to multiply my formula by about 0.9 to be consistent, and then check the results against reality.

So our final formula, at least for larger wire, is $$L\approx\frac{0.075\pi}{d^2}A(A+B)C$$ where, again, *A* is the thickness of the wire around the reel, *B* is the diameter of the reel, and *C* is the width of the reel, while *d* is the diameter of the wire. Actual results depend on how the wire is wrapped, and would probably vary according to the width and number of layers, not just the diameter of the wire.

And, to repeat our original version of the formula with this 0.9 multiplier, $$L\approx0.9\frac{\pi(D_o^2-D_i^2)W}{4D_w^2}\approx\frac{0.7(D_o^2-D_i^2)W}{D_w^2}$$

In editing this, it occurred to me to search for a source of this formula and the table of *k*; surprisingly, after 21 years, I found one. Here is the equivalent formula and table from this site:

The numbers agree with Joe’s table, where the diameters match. Here is the table comparing these with my *k*:

But why would the thinnest wire require a smaller multiplier, so that less wire fits in a given space than we expect in theory? I would expect that there is less of an edge effect when more strands fit in one layer, but that would imply it can be wrapped *better*, and *more* would fit. Math can’t always explain the details of reality!

We’ll start with this reverse question from 1995, to establish the concept:

Calculating the Diameter of a Carpet Roll How do youcalculate the diameterof a carpet roll when you have thelengthand thethickness?

Doctor Andrew answered with some useful ideas:

Hi! Great question. For a darn good approximation you could find thecross-sectional area(the area of the circular end of the roll) needed for the roll, assume that the roll is aperfect cylinder, and then calculate thediameterrequired for that cross-sectional area. If a carpet haslength Landthickness t, the cross-sectional area of the roll is the same as the cross-sectional area of the edge of the carpet when it is lying on the floor, which is the very long rectangle with length L and thickness t. See if you can take it from here to find the diameter.

Here are the dimensions he has in mind for such a perfect cylinder:

It has two diameters (the core, *C*, and the outer diameter, *D*), thickness *t*, and a length when rolled out, *L*:

(Not to scale!)

Neither of us has ever seen a carpet roll shaped perfectly cylindrically since it is aspiralwith an edge that usually sticks out off the rest of the roll. I imagine that trying to describe the shape of the spiral could be pretty tough. You could probably come up with a number of approximations, the most accurate of which would take into account how the spiral begins and what kind of space would be left in the center of the roll due to physical limitations on how the carpet can bend. You might notice that the approximation in the above paragraph becomes less and less accurate as the ratio of L to t (the fraction L/t) becomes smaller and smaller. If you think more about it, you'll realize thatwhen a carpet bends it is already changing its shape(stretching on the outside, compressing on the inside) in a way that makes describing it exactly with mathematics very difficult.

Here is a slightly better image of what the roll might look like, though the bend would not really be this smooth:

The more layers (and the thinner the carper), the closer we can approximate it with a cylinder.

In what follows, we’ll find the length given the dimensions of the roll, reversing what was said here. Most of the answers will be based on the area idea. Next time we’ll consider the compression issue.

Next, from 1996, we get an actual formula:

Length of Material on a Cylindrical Roll Is there aformulafor calculating the length of material on a roll knowing theroll diameter, thecore diameter, and the materialthickness? I recently tried using a formula I saw in a newsletter, but it was not even close when I tried a real life example. The formula was something like(64.54*(D*D)-(C*C))/T*1000where D is the roll diameter, C is the core diameter, and T is the thickness, all in inches. The answer was supposedly in feet.

His formula, which is not quite written right, can be written correctly as $$\frac{64.54(D^2-C^2)}{1000T}$$

Doctor Anthony derived that formula, using the area approach and then converting to inches by dividing by 12:

The area of cross-section of the roll can be expressed in two ways. When stretched out in a straight line it is arectangleof area L*T where L is the length in inches, and T the thickness viewed edge on. Whenrolledon the drum, this same cross-section will be pi*R^2 - pi*r^2 where R = outer radius (=D/2), and r = inner radius (=C/2) So we have L*T = pi{D^2/4 - C^2/4} L = pi/(4T){D^2 - C^2} inches Length = pi/(48T){D^2 - C^2} feet =(0.06545/T){D^2 - C^2)feet (T,D and C in inches) Try this in real life.It will be sensitive to the value of T you use, so measure that carefully.

That’s exactly the formula the anonymous questioner had seen. Without the unit conversion (so that the length is in whatever units you use for the other measurements), the formula is $$L=\frac{\pi(D^2-C^2)}{4T}$$

The last few answers here will look at substitutes for the thickness, making it easier to be accurate.

For an alternative derivation of the same formula, consider this 2001 question:

Fabric Left on a Roll I am going to write a program for my TI-83+ (I can also use a TI89 if necessary) to calculatehow many feet of material are left on a roll. I know the outsidediameter of the core(this is the tube around which the film is wrapped, I know the outsidediameter of the roll, and I know thethickness(mil thickness) of the material. What I don't know is the formula to use to calculate the feet of material left on the roll. What I thought I could do is start out with thedistance around the core. In this particular case, that would be pi*3.75 (the outside diameter). Theneach additional layerof material would be the previous outside diameter plus twice the thickness (1 mil = 0.001"). I'd have a field in which I'd keep a running total of these inches. I'd set up another field to keep track of the total number of "wraps" of material, which would be a calculated field (outside diameter of the roll/2 - 3.75/2 [this removes the core's dimension] divided by the mil thickness of the film). I'd put a loop in the program to keep adding to the running total of inches until the number of wraps = 0. This would seem to be the hard way, but as I said, I don't know what formula to use. I know there has to be a formula I can use for this calculation to make things a lot easier. Does anybody out there know the formula I'm looking for?

Doctor Robert answered, using the suggested method and finding the same formula we’ve seen:

I'll take a whack at this one. No guarantee of success. Let me define thediameter of the tube as Doand thediameter of the roll as D. Lett be the thicknessof the material. Then I figure that the number of wraps, n, on the tube, is:n = (D-Do)/(2*t)and we need to find the total length of these n wraps. I'll figure the length of a wrap to be the circumference of a circle whose radius is from the center of the tube to the center of the wrap I'm dealing with. In other words, the radius of the first wrap is Do/2 + (1/2)t; the radius of the second wrap is Do/2 + (3/2)t; the radius of the third wrap is Do/2 + (5/2)t.

Each layer add *t* to the radius, and thus \(2t\) to the diameter, so we divide the difference in diameters by \(2t\) to count the layers.

He is using the radius to the middle of each layer (figuring that each layer is compressed on the inside and stretched on the outside), so he takes the first layer as having radius \(\frac{t}{2}\) more than the core radius, and adds *t* to the radius for each layer. This makes an arithmetic series:

Now the circumference is 2pi times the radius and we need tosum a bunch of circumferences. Generalizing on the formula above, the radius of the kth wrap is: radius of kth wrap =Do/2 + t(2k-1)/2where k= 1,2,3,... Length of material = sum from k=1 to n of 2*pi*[Do/2 + t*(2k-1)/2] = n*2*pi*Do/2 + sum from k=1 to n of 2*pi*t(2k-1)/2) = n*pi*Do + pi*t times sum from k=1 to n of (2k-1) Now the sum of 1+3+5 + ...+2n-1 = n^2 So, length = n*pi*Do + pi*t*n^2 But length = n*pi(Do + n*t) But n = (D-Do)/(2t) so the formula is Length = (D-Do)*pi/(2t)*[Do + (D-Do)/2] = (D-Do)*pi/(4t)[Do+D] =(D^2-Do^2)*pi/(4t)I hope that this formula is correct and serves you well. There is no guarantee.

The summation is $$\sum_{k=1}^n 2\pi\left[\frac{D_o}{2}+\frac{t(2k-1)}{2}\right]=n\pi D_o+\pi t\sum_{k=1}^n (2k-1)\\=n\pi D_o+\pi tn^2=\pi n(D_o+nt)$$ This uses the fact that the sum of *n* consecutive odd numbers is \(n^2\), which we proved in Inductive Proofs: Four Examples.

So the formula becomes $$L=\pi\cdot\frac{(D-D_o)}{2t}\cdot\left[D_o+\frac{D-D_o}{2}\right]\\=\frac{\pi(D-D_o)}{2t}\cdot\frac{D+D_o}{2}=\frac{\pi(D^2-D_o^2)}{4t}$$

Note that this gives the same formula as before, by adding all the layers in our idealized cylinder.

This question from 2003 provides a slightly different approach equivalent to the same idea:

Determining Length of Material Remaining on a Roll Many apologies if this is not what you do, but I'm at a loss for where to find this answer. I need to know the mathematical formula to determine the length of material as it is on a roll. Specifically; my company supplies rubber and tape, both in rolls on a core. Without unrolling the roll, we'd like to determine the length of the material with the following information:Outside diameter of the core,outside diameter of the whole rollandthicknessof the material (determined by a micrometer). Do you have any such formula?

We do, indeed!

I answered, starting with links to the answers above and below:

Hi, Laura. We get this question a lot; here are some of the explanations in our archive: Length of Material on a Cylindrical Roll http://mathforum.org/library/drmath/view/51723.html How Much Carpet is Left? http://mathforum.org/library/drmath/view/56481.html Fabric Left on a Roll http://mathforum.org/library/drmath/view/54357.html Briefly, the length of the material will be the product of thenumber of layersand theaverage length of one layerin the roll. Both are easy to find: Number of layers = (Do-Di)/(2t) [total thickness/one layer] Average layer = pi(Do+Di)/2 [circumference at average diameter] whereDo and Di are the outer and inner diameters, andt is the thicknessof the material. So the length is L = pi(Do+Di)(Do-Di)/(4t) = pi/4 (Do^2 - Di^2)/t

We get the same formula; the idea is that the sum of an arithmetic series is the number of terms times the average of all the terms, which is the average of the first and last.

Asmall errorin the thickness measurementcan make a big difference; I would recommend measuring it not with a micrometer, but by measuring theoutside diameter of a roll of known lengthand calculating t from this formula in reverse. That will ensure that the number you usereflects the way the material lies on the roll. But try both measurements to see how they compare.

This last comment has great practical importance. Not only is it hard to accurately measure a small thickness, but that may be changed by compression or by loose wrapping. Measuring the roll itself makes sure we use the real thickness. So let’s do that!

We’ll turn that last idea into a formula with this 1999 question:

How Much Carpet is Left? We use rolled carpet in our business and would like to know how much is left on a rollgiven the remaining diameter. Given:Beginning OD,core OD, number oflayers,length of roll,New ODFind: Length of carpet used or remaining.

There’s more information here than we need. We won’t be using the number of layers, but we’ll do that in the next answer.

Doctor Rick answered:

Hi, Zene, welcome to Ask Dr. Math. I enjoy solving problems from the working world - students often ask us, "Will I ever have a use for this stuff?" The answer is, you'd be surprised how often math comes in handy at work. Here is a method that will give an approximate length of the carpet roll; the thinner the carpet relative to the diameter of the core, the better the approximation. Consider across-sectionof the roll. It is anannulus(like a washer - a disk with a hole in the middle). Its area is the area of the outer circle minus the area of the inner circle (the core).

Here is an annulus (yellow), whose area is \(\pi R^2-\pi r^2\):

LetC be the core diameter,F the diameter of a full roll, andL the lengthof carpet on afull roll. Let's also defineT, the thicknessof the carpet. The cross-section area of the full carpet is Area = (pi/4)(F^2 - C^2) (Read F^2 as F squared.) When you unroll the carpet, its cross-sectional area is LT (L times T). This area should be (approximately) the same as the area of the roll, so we can find the thickness T: T = (pi/4)(F^2 - C^2)/L

You can check the accuracy of the calculation so far by counting the layers of carpet on the full roll; this should be n = (F - C)/(2T) = (F - C)L/((pi/2)(F^2 - C^2)) = 2L/(pi(F + C))

You can also think of this calculation as dividing the length by the average circumference of a turn.

Now, if thediameter of a partial roll is D, we can use the same area calculations as above, but use the known T to find thelength Pof the carpet on thepartial roll: P = Area/T = (pi/4)(D^2 - C^2)/((pi/4)(F^2 - C^2)/L) =L(D^2 - C^2)/(F^2 - C^2)So there you have it. The length of carpet on a partial roll is the length on a full roll times the difference between the diameter squared and the core diameter squared, divided by the same calculation for the full roll.

This is $$P=\frac{L(D^2-C^2)}{F^2-C^2}$$ That’s nice: Divide the current difference of squared diameters, by the original, and you get the fraction of the roll that is left.

But there’s another way to avoid the sensitivity to the measured thickness, seen in this 2003 question:

Length of Coiled Belt Working with conveyor belts we use a formula that gives a very close approximation of the length of a coil of belt, but I would like to know the reasoning of it. Theoutside and inside diametersof the coil (in inches) are added and thenmultiplied by the number of wraps, this is thenmultiplied by a constant(.1309), and the result is to be read in feet. It works, but I would like to knowwhy. I tried to reason it out but I am stumped.

This turns out to be a valid formula: $$0.1309N(D+d)$$

I answered, starting with the formula we’ve seen:

Hi, Manny. We have a number of answers to questions about the length on a roll of some material; one of them may give a formula like this based on the number of turns. The cross sectional area of the coil is the difference between the circle formed by the outer diameter, D, and that formed by the inner diameter, d: A = pi (D/2)^2 - pi (d/2)^2 = pi/4 (D^2 - d^2) The cross-sectional area of the belt straightened out is A = TL where T is the thickness and L is the length. Setting these equal, TL = pi/4 (D^2 - d^2)

This is our basic formula.

ButT can be found by dividing (D-d)/2 by the number of turns N: (D-d)/(2N) L = pi/4 (D^2 - d^2) Solving for L, L = pi/4 2N (D^2-d^2)/(D-d) Since D^2 - d^2 = (D-d)(D+d), this is L = pi/2 N (D+d) That is your formula; to convert inches to feet we have to divide by 12, and (pi/2)/12 = 0.1309.

The calculation of *T* here is equivalent to the calculation of *N* in the previous answer. The new formula is $$L=\frac{\pi}{2}N(D+d)$$ using consistent units, and $$L=\frac{\pi}{2}N(D+d)\cdot\frac{1\text{ foot}}{12\text{ inches}}\approx0.1309N(D+d)$$ to get feet of length given diameters in inches.

Next time we’ll look at further variations in the question, focusing on how reality can change the formula.

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Here are some examples, from 1999:

Locus and Equations of Lines I'm currently trying to help my niece who is in 10th grade with locus and line equations. The concept isn't coming back to me. She has the following homework:Describethe locus of points that are3 units from the linex = -1 and give theequationof the locus. Describe the locus of points that are6 units from the point(3,-1) and give the equation of the locus. Describe the locus of pointsequidistant from the x and y axisand give the equation. Describe the locus of pointsequidistant from the pointsF (-2,1) and G (6,3) and give the equation. Describe the locus of pointsequidistant from linesy = 4 and y = -2 and give the equation of the line. We can graph the points on the x and y axis but we don't understand how to find the locus of points. We know the formulas for slope, the line equation, and the distance formula, but when do you use each? What is the process for these problems so I can explain them to her? Thank you for your help. Marguerite

Since the goal here is to write both descriptions (in words) and equations, we are going a little beyond the questions last time, though it is really the same concept.

I answered:

Hi, Marguerite. The basic concept of alocus in geometryis very closely related to that of anequation in algebra. We want to describeall points that satisfy some condition, by choosing an unknown point (x, y) and seeing what it means, algebraically, to satisfy the condition.

Recall that a locus is **a set of points, thought of as a single entity**. The graph of an equation is the **set of all points** represented by ordered pairs that satisfy the equation; so the *equation* is just a way to describe symbolically the condition for the locus, and its *graph* is the locus. We are to describe each locus both in words and as an equation.

In my reply, I led Marguerite close to the answers, but left her some thinking to do. Here, we’ll finish them up.

Let's look at your problems one at a time. 1. Describe the locus of points that are3 units from the line x = -1and give the equation of the locus. Pick any point (x, y). What does it mean to be 3 units from the line x = -1? Since this is avertical line, the distance is just thehorizontal distancefrom x to -1: ^ ^ | | | y+ *(x,y) | | | | | | ----------+---+-------+-- -1| 0| x | | | | |<--------->| d=|x-(-1)| (We use the absolute value because x could be on either side of -1.) So the equation of the locus is: |x - (-1)| = 3

We’ve translated the condition given in words into an equation.

Here is a typical point shown above, 3 units to the right of the blue line:

But the point could also be to the left:

Recall the comment last time that a locus must consist of *all* such points, and *only* such points; if we’d omitted the absolute value, we wouldn’t have the entire locus:

Every pointthat is 3 units from the line satisfies this equation, andevery pointthat satisfies this equation is 3 units from the line. So this is the locus we are looking for. (Incidentally, don't worry that y isn't in this equation - that's how vertical lines work.) Can you see what this locus is geometrically? Think in terms ofparallel lines, and remember what I said about beingon either side! The equation can be rewritten as two equations, in the form "x = this OR x = that."

The equation can be written as \(|x+1|=3\); this means that \(x+1\) must be either \(3\) or \(-3\). And the equations \(x+1=3\) and \(x+1=-3\) are equivalent to $$x=2\text{ or }x=-4$$ So here is our locus, \(|x+1|=3\), consisting of two parallel lines:

Last time we saw a circle as the locus of a dog tethered to a stake; this is the locus of a dog tethered to a “zip line”! It’s a **pair of lines parallel to the given line**, one on each side of it.

2. Describe the locus of points that are6 units from the point (3,-1)and give the equation of the locus. This is similar, except here the distance formula you want will be thePythagorean theorem. Write an expression for the distance of point (x, y) from (3, -1), and set it equal to 6 to make the equation of the locus. To describe the locus, think about how you would draw it, by making a string of length 6 and tacking one end of it at (3, -1). Where can the other end be? Could you also draw this with a compass?

Here is a point 6 units from \(3, -1)\):

We want an equation that will be true whenever a point \((x,y)\) is a distance of 6 units from \((3,-1)\). The distance *d* between two points \((x_1,y_1)\) and \((x_2,y_2)\) satisfies \(d^2=(x_2-x_1)^2+(y_2-y_1)^2\), so our locus is defined by $$(x-3)^2+(y-(-1))^2=6^2$$ which we can simplify to $$(x-3)^2+(y+1)^2=36$$ What shape is it? A **circle** with center \((3,-1)\) and radius 6:

We saw this idea, last time – but without the equation.

3. Describe the locus of pointsequidistant from the x and y axisand give the equation. This requires two expressions: one for the distance of point (x, y) from the x axis, and one for the distance from the y axis. Those are pretty simple, but remember those absolute values! Now set them equal, and you have an equation describing all points for which the two distances are equal, which is your locus. To describe this, either look at the equation in terms of quadrants, or think geometrically about angle bisectors.

The distance of a point \(P(x,y)\) from the *x* axis is \(|y|\); the distance from the *y* axis is \(|x|\):

But either *x* or *y* can be negative, so we have to be careful not to ignore points in other quadrants:

So the equation of our locus is \(|y|=|x|\). But what does that look like? Algebraically, we can rewrite it as \(y=\pm|x|\), and since this means to remove the sign from *x* and then put either sign back on, it is merely \(y=\pm x\). This is a pair of lines, \(y=x\) and \(y=-x\):

What we’ve done here is to **bisect** all four angles at the origin (because the congruent right triangles shown imply congruent angles).

4. Describe the locus of pointsequidistant from the points F (-2,1) and G (6,3)and give the equation. Again, write an expression for the distance of (x, y) from (-2, 1), and one for the distance of (x, y) from (6, 3), and set them equal. You can simplify this equation quite a lot; squaring both sides will be the first step.

The two distances are \(\sqrt{x+2)^2+(y-1)^2}\) and \(\sqrt{(x-6)^2+(y-3)^2}\). Setting them equal, and then squaring, we get $$\sqrt{(x+2)^2+(y-1)^2}=\sqrt{x-6)^2+(y-3)^2}\\(x+2)^2+(y-1)^2=(x-6)^2+(y-3)^2\\x^2+4x+4+y^2-2y+4=x^2-12x+36+y^2-6y+9\\16x+4y=40\\4x+y=10$$

That looked bad at first, but worked out nicely. It is clearly a line, and we’ll see in a moment that it is just the line it needs to be.

You may prefer to do the hard work in geometry rather than algebra. There is a theorem that tells you that the locus of points equidistant from two points is theperpendicular bisectorof the segment they define. You will have to find the coordinates of themidpointof this segment (think of averaging their coordinates), and write the equation of the line through this pointperpendicularto the segment (the slope of this line will be the negative reciprocal of the slope between the two points).

This theorem was mentioned last time.

The slope of line FG is \(\frac{1}{4}\); the midpoint M is \((\frac{-2+6}{2},\frac{1+3}{2})=(2,2)\). The slope of the perpendicular line is \(-4\), so the equation of the line is \(y-2=-4(x-2)\), which simplifies to \(4x+y=10\), the same answer as before.

You can prove that the line is perpendicular by considering congruent triangles PFM and PGM.

It's fun to do this both ways and see that geometry and algebra agree! But it does take some work either way.

5. Describe the locus of pointsequidistant from lines y = 4 and y = -2and give the equation of the line. Now we're back to something like the first problem. Write expressions for the distance of (x, y) from each line, and set them equal. The absolute values get in the way, but if you think carefully you can simplify this into the equation for a single line. You may be able to avoid the absolute values entirely if you think about what is happening geometrically first. Where can a point be if it is the same distance from two parallel lines?

This is like the first example we saw last time, except that we need to give an equation.

Here are the two lines:

Here is a point equidistant from them:

The distance from \(P(x,y)\) to \(y=-2\) is \(|y-(-2)|\); the distance from \(P(x,y)\) to \(y=4\) is \(|y-4|\). So our equation is $$|y+2|=|y-4|$$

One way to solve this is to observe that the absolute values change behavior when \(y=-2\) and when \(y=4\), so we can consider three cases: $$y\ge4: y+2=y-4\Rightarrow 2=-4\text{ (impossible)}\\

-2<y<4: y+2=-(y-4)\Rightarrow 2y=2\Rightarrow y=1\\

y\le-2: -(y+2)=-(y-4)\Rightarrow -2=4\text{ (impossible)}$$

So the equation of the locus is simply \(y=1\). This is just the average of the two *y*-coordinates, which makes sense: The locus is the parallel line halfway between the given lines:

I hope these examples demonstrate the meaning of locus and the methods that can be used to find the equation. If you need more, try searching for the word "locus" in our archives. If some of my specific steps didn't make sense, or I went too fast, write back and I'll try again.

A 2008 question takes us deeper into the process, with harder examples:

Drawing a Locus of Points I was doing my geometry homework, and some of the questions are rather difficult and some are easy. For example: All points in the plane that areequidistant from the rays of an angleandequidistant from two points on one side of the angle. The directions say to drew a diagram to find the locus of points that satisfy the conditions. I got the first part before the word "and." My drawing: / / / /_______ Okay, so there's the angle, and then I drew in anangle bisectorwhich I was unable to do on here.

(I’ll hold onto the rest of the question, which includes another example and a general question, until later.)

This is, indeed, a trickier question, because it combines two separate conditions. Erin knows the first: The locus of points equidistant from two rays is the angle bisector. (We touched on this in example 3 above.)

I answered:

Hi, Erin. You've asked a big question; but let's see if I can help one part at a time, then come to some broader conclusions. You are given two conditions: A: equidistant from the rays (sides) of the angle B: equidistant from two (given) points on one side (ray) To find all points that satisfy BOTH A and B, you can draw each locus (A and B) separately, and theintersectionof those will be your combined locus--the only point(s) for which both A and B are true.

A point that satisfies the conditions for two loci is in their intersection.

You've drawn locus A, a ray from the vertex. Now let's add in two given points on the horizontal side: / / / / / / <--- locus A / / / / / / O-----P-------Q--- What is the locus B, the points that are the same distance from P as from Q? You're probably as familiar with this one as with the angle bisector. Draw it, and where that line crosses locus A is the single point in the locus "A and B".

This is what we saw in example 4 above: the perpendicular bisector of segment PQ. So the solution will the intersection of the two lines, which is a single point. Here is an example:

This would be harder to assign as an algebraic problem, because finding the equation of the angle bisector requires either trigonometry or harder algebra. Give it a try if you like.

Here is part 2 of the question, which is our first (and only) problem in space rather than on a plane:

Another problem I had trouble with was: All points in spaceequidistant from two intersecting planes. I drew two planes intersecting perpendicularly just to make it easier to visualize. ______ | | ___o | |o____ | | | | |____| |____| o |____|o The dots are the four points equidistant from each of the planes, but the book shows that where the dots are, are two other planes that intersect each other. And the four planes intersect at one point.

The idea of four points suggests that Erin is visualizing this at least a little correctly.

I answered:

It can be quite difficult todrawsomething like this in the first place; I wouldn't say that your planes look like they intersect, but perhaps you can visualize what you have in mind anyway. Here's how I might draw it (the best I can in plain text, and I'm pretty much an expert at it!): + /| / | + | | | +------| +--------+ / | /: / / |/ : / +---------+--------+ | | | + | / |/ + I drew the planes sothey look perpendicular, which they will not generally be; but it's hard enough to draw them this way!

Some geometry classes spend a little time learning to draw 3D figures like this; probably most don’t! I gave a little lesson here. It might be easier to make a physical model using sheets of cardboard – or just pages in a book!

Now, where can you be and be thesame distance from both planes? The situation is much like the angle bisector of an angle in a plane; in fact, this figure represents a "dihedral angle" (between two planes). Aside view(still making them look perpendicular) would be: ^ . | . . | . . | . . | . <---------+--------> . | . . | . . | . . | . v The two dotted lines represent all points in the plane that are equidistant from the two solid lines.

This looks like our example 3 above.

Now imagine this as just aside viewof the planes; the dotted lines, too, are side views of planes, which pass through the same line of intersection. I'll just barely try to draw this: + . /| . . . / | .. . .+ | . . . | | . . +.-----| +--.-----+ / . | /:. / / . |/.: / +---------+--------+ . . | .| . . . | +. . .. | / . . . |/ . + Note that, since it's hard both to draw and to visualize three dimensions,I used two dimensionsin two ways: first, using an ANALOGUE (intersecting lines in a plane areLIKE intersecting planes in space, so they give a sense of what the result might be like); and second, using aSIDE VIEW (or cross section)in order to draw part of what I'm imagining, still keeping things simpler than trying to picture the whole thing. We do the same thing to try to visualize four-dimensional objects, which is quite a trick!

Here is a version of the graphs above, from GeoGebra:

As in other figures, the blue planes are the given, and the red planes are the locus. But with software, we don’t need to start with perpendicular planes. Here is an example where they are not:

She closed with some general questions:

I think my major problem with these locus problems, is reading the problem.I just don't think I understand what I'm reading, or I may understand part of it and when the second part comes, I don't know what to do. I think I'm having problems channeling all the information it gives, so I try to break it up. When I break it up, I sometimes have problems putting it together. My teacher told me tosplit the problem into two, like do the first part before "and" then do the second part, which would be after "and." He also told us to draw the given in one color, then the points or whatever it is that the problem says in another color, so where the non-given things that meet is what we describe.

I answered the first paragraph:

That's not a bad analysis. Reading mathematicians' twisted grammar just takes time and experience! A big part of math is being able tobreak a large problem into pieces, and then be able toput them back together. Often that skill is developed by being even more orderly than you need to be once you understand the topic more fully; you may need to write down more of what you are doing (sort of like making notes to yourself to remind yourself where you left your glasses when you took them off, so you can find them again later). For example, I gave names to the two parts, A and B, so I could clearly communicate with myself (and you) about those parts. I first summarized the overall strategy (draw A, draw B, intersect), and then filled in the outline. Suchself-communication(naming things and writing down what they mean) andorganization(outlining a plan before doing it) is very helpful. You won't always see ahead of time what all the steps will be, but you can at least keep good records of what you have done while you explore. Think of yourself as Lewis and Clark,not knowing just how they'd get to the Pacific, but making their own maps so they'd know where they'd beenonce they found where they were! Problem-solving in general requires that sort of skill. But I'm digressing...

We’ve talked about similar problem-solving ideas in How Can I Stop Making Careless Mistakes?

As to the teacher’s ideas in the second paragraph:

All of that is good advice. (I used dotted lines for a similar purpose.) It just takes time and practice to make it all work!

Erin wrote back:

Thank you very much for answering my question. I understand it better, and I think I'll do more problems over the weekend and try to get better at it. Thanks for the Lewis and Clark advice and organization advice. I'll start being more organized again. I think organization helps a lot in math! Thanks again!]]>

We’ll start with this question from 2003:

The Meaning of Locus What is thelocus of points equidistant from two parallel lines8 meters apart? I have trouble understanding themeaning of locus.

Since it was assumed in the answer, I’ll point out that “equidistant” means “the same distance from both lines”.

I answered:

Hi, Marina. The idea of locus is very simple, but a little subtle.You can almost leave the word out of many problems; this one asks you simply to describe all points that are equidistant from the two lines. The point of the word "locus" is merely that you are tothink of all the points that fit that description as a single entity, which might be a curve or line or a set of discrete points.

We’ll look further into the origin of the concept at the end. You could consider it an old word for “set”; in Latin the word *locus* means “place”. Our problem, then, is simply “Describe the *set consisting of all* points equidistant from two parallel lines 8 meters apart?” or “*Where* are all the points equidistant from two parallel lines 8 meters apart?”

Here are our two parallel lines:

Now we can answer the question:

So picture it one point at a time. What sort of point is the same distance from both lines? Imagine a point somewhere;how do you measure the distancefrom each line? You draw a perpendicular from the point to each line, and their lengths are the two distances. Now,what does it mean if they are the same?You will find that the points you are looking for are exactly between the two lines.

Here is one such point:

Once you get that idea, think about how you can describeall such points. Imagine that you were forced somehow to walk only in places that are the same distance from the two lines, perhaps because there is a rope from you to each of the lines and some mechanism keeps them the same length.Where will the grass be trampled down?That is the locus.

We’ll see this idea of trampled grass again later! Here is where all those points are (that is, the locus):

Notice that we can’t just think about *some* points that fit; we have to find them *all*, and to be sure *no other* points do. Once we see that, we need to describe what we see: the **parallel line halfway between** the two given lines.

Although the question doesn’t require it, we could *prove* that all points on this new line are equidistant from the given lines, and that every point equidistant from the two lines is on the new line. The first step, though, is to *see* it as we have done, so we can make a *conjecture*, and then prove that.

For another example, consider this question from 1999:

Locus What do you have to do to get theperpendicular bisector of all the loci of a triangle? What does it mean when it asks you about equidistance from certain points?

The question this time is about the meaning of “equidistant”; but it’s not stated quite right, so we’ll have to rephrase the question as it was presumably given, explaining the terms that made it easy to misstate.

But one thing Jo got right: The plural of “**locus**” (which, as I said, is a Latin word), is “**loci**” (pronounced “low-sigh”, or, if you learned Latin as I did, “low-key”).

I answered:

Hi, Jo. Thanks for writing. Your first question doesn't make a lot of sense, because there is no "locus of a triangle." But I think I can see what you're asking about. A "locus" is the set of all points that satisfy some rule or description, and a perpendicular bisector is one of the simplest examples of a locus. If I asked you "what is the locus of all points equidistant from two given points?", the answer would be "the perpendicular bisector of the segment determined by the two points." Here's what the question means: What is the locus of all points ... (What geometrical object consists of all points X ...) that are equidistant from A and B? (...for which the distances AX and BX are equal?) That is, "equidistant from two points" means "the same distance from both points."

Here is a point that is equidistant from points A and B:

Here's what the answer means: The perpendicular bisector ... (the linethrough the midpointof the segment,perpendicularto the segment) of the segment determined by A and B. (you make the segment by connecting A and B with a straight line)

That is, this line **bisects** segment AB (cuts it exactly in half) and is **perpendicular** to it (at right angles):

You can prove (and probably your text did this) thatany point on the perpendicular bisectorof a segment AB is the same distance from both endpoints; you can see this by drawing in the segments and seeing that you have an isosceles triangle, so that AX = BX: | + X /|\ / | \ / | \ / | \ / | \ / | \ / | \ / | \ / | \ / | \ / | \ / | \ / |_ \ / | | \ +--------------+--------------+ A | B | | This tells us that any point on the perpendicular bisector is equidistant from A and B, so it is part of the locus we are looking for.

This is only half of the requirement: Every point on the locus must meet the requirements, and also every point that meets the requirements must be part of the locus:

Conversely, you can show thatif any point is equidistant from A and B, it must be on the perpendicular bisector, so there are no other points in the locus: the perpendicular bisector IS the entire locus we're looking for. That's what a locus is:the set of ALL points that fit some description(in this case, being equidistant from A and B). In our case, the perpendicular bisector is "the locus, the whole locus, and nothing but the locus" of equidistant points.

The word “all” is essential!

There are many other examples of a locus. For example, the locus of points 1 inch from point A is thecircleof radius 1 centered at A. Every point of the circle is 1 inch from A, and every point one inch from A is part of the circle. I hope that helps a little to clarify the idea of alocus, and the meaning ofequidistant. If you have problems you still have trouble with, maybe you can write back with a specific problem so I can see the wording of it.

Here is a point 1 unit away from point A:

Here is that circle:

We’ll see more examples next time.

I want to close for now with this question, from Navneet in 2016:

Locus Locution Which of these is the correct definition of a circle? 1. A circle is thelocus of *a point*which moves in a plane in such a way that its distance from a given fixed point is always constant. OR 2. A circle is thelocus of *points*which moves in a plane in such a way that its distance from a given fixed point is always constant.

Is one of these incorrect? Or are they both acceptable?

I answered, yet again (yes, this is a favorite topic of mine):

Hi, Navneet. Theproper (older) terminologyis "locus ofa point." We define a constraint that applies to a point, and imagine following ONE POINT as itmovesaround subject to that constraint. We can also describe this as "the locus ofall pointsthat ...," thinking not of a single point moving, but just of theset of pointssatisfying the condition. This is the more modern usage that emerged with the development of the set concept. Your second version uses the idea of moving, which is not appropriate when talking about a set of points, so it is not good wording.

The **first** definition, “A circle is the **locus of a point** which moves in a plane in such a way that its distance from a given fixed point is always constant,” envisions the **process** of drawing a circle, in which a point, namely the point of a compass, is **moving around**, keeping a constant distance from the center. The locus is the circle it draws, which is the path it followed. We can call this the **dynamic** definition.

The **second** definition, which would have been better written as, “A circle is the **locus of points** in a plane in whose distance from a given fixed point is a given constant,” looks at the **result** of drawing the circle and sees that it consists of **all points** a given distance from the center. It doesn’t consider the motion, just the resulting figure. We can call this the **static** definition.

See this Wikipedia article, which explains the two usages (key phrases capitalized by me for emphasis): https://en.wikipedia.org/wiki/Locus_(mathematics) "Until the beginning of the 20th century, a geometrical shape (for example, a curve) was not considered as an infinite set of points; rather, it was consideredas an entity on which a point may be located or on which it moves. Thus a circle in the Euclidean plane was defined asTHE LOCUS OF A POINTthat is at a given distance of a fixed point, the center of the circle. In modern mathematics, similar concepts are more frequently reformulated by describing shapes assets; for instance, one says that the circle isTHE SET OF POINTSthat are at a given distance of the center."

The first perspective sees a locus as “where a point lives”, as a river is the “locus” of fish; the second, as the collection of such points themselves.

Thus, the common wording today appears in MathWorld: http://mathworld.wolfram.com/Locus.html "For example, THELOCUS OF POINTSin the plane equidistant from a given point is a circle, andTHE SET OF POINTSin three-space equidistant from a given point is a sphere."

You can see that the word “locus” can be replaced by the word “set”, as we see also here:

See also the Math Open Reference: http://www.mathopenref.com/locus.html "We can say "THE LOCUS OF ALL POINTS on a plane at distance R from a center point is a circle of radius R." In other words, we tend to use the word locus to meanthe shape formed by a set of points. An odd thing is that you can often just drop the word locus, and it still makes sense: "The set of all coplanar points distance R from a central point forms a circle." "... Sometimes the idea of locus has a slightly different explanation. If you think ofa point moving along some path, we sometimes say that the path is the locus of the point. So for example a point that moves a fixed distance from another point draws out a circle. So we could say "THE LOCUS OF A POINT moving at a fixed distance from a center point is a circle.

One problem is that a locus is not necessarily a **path**; it can be a plane in space, or a region such as the interior of a circle; these probably didn’t occur in the original uses of the concept.

Your reference books may well use theold form("locus of a point") with or without explicit mention of motion. Other sources you find, especially online, are likely to use thenew form("locus of points") with no reference to motion at all. Does that help?

Navneet wasn’t quite sure about the “old form”:

What should bethe appropriate and precise definitionof the word "locus" in mathematics? We know that the definition of locus is "the *set of all points* (usually forming a curve or surface) satisfying some condition." Buthow can we use it for one point, as in "A circle is the locus of *a point* which...."?

I answered:

You aren't really asking about the DEFINITION, but about how the older WORDING fits that definition. I'll try to explain that in more detail. In the old wording, "the locus of a point MOVING under some condition" refers to the set of all places that that point will reach as it moves. Think of the point as a pen point moving around, and the locus as the ink mark it leaves behind. The pen is a MOVING point, not a single location; the locus (drawn curve) consists of all the points (fixed locations) where it ever traveled. In particular,a circle is the locus of the pencil point of a compass as it turns. I imagine that may be exactly what was in the mind of whoever first used the word "locus" in this way.

Again, the locus can be thought of as the path traced by the pen, or as the set of ink molecules on the paper. They both represent the same static entity.

Incidentally, we can describe a curve as a function of time (the path of a point), or we can eliminate time from the equation(s) and be left with just an equation in *x* and *y* (the equation of a curve). These are two views of the same locus. But the time factor is ignored in thinking about the locus.

Another example I've given is this: suppose youtether a dog to a ten-foot ropethat you stake in the middle of your yard.What part of your yard will it trample?The dog can move around, but is constrained by the rope to stay within ten feet of the stake; so all the places that the dog could trample will lie in aten-foot diskwith that center. This is the "locus of the dog": the set of all points where the dog will ever travel. (A disk is the locus of points whose distance from the center is less than or equal to a radius.) Or, if the dog is like some I have seen, the actual part of the lawn it tramples may be not the interior of the circle, butthe circle itself, since it will spend all its time at the end of its rope trying to get away. And in fact this is how we define the circle as a locus: the set of all points a fixed distance from the center. But we are describing it asthe locus of "a point" (the dog) as it moves.

Long ago I had a neighbor whose dog behaved this way. Today, another neighbor sometimes attaches his dog to a “zip line”, producing a locus consisting of a pair of parallel lines if the dog is on a similar mood, or a rectangular region otherwise.

In the more modern wording, we don't think of a single point moving, but go directly to the idea of a set of points. Perhaps this change in common usage was made in part to prevent the confusion you are expressing due to the use of the word "point" in two different ways (moving and fixed). The meaning is the same whether we speak of many points or a single one.

Next time we’ll have more examples to illustrate these concepts, and will also move on to *algebraic* descriptions of loci.