Integration by Substitution

(An archive question of the week)

Last time, we looked at a method of integration, namely partial fractions, so it seems appropriate to find something about another method of integration (this one more specifically part of calculus rather than algebra). We will look at a question about integration by substitution; as a bonus, I will include a list of places to see further examples of substitution.

I have previously written about how and why we can treat differentials (dx, dy) as entities distinct from the derivative (dy/dx), even though the latter is not really a fraction as it appears to be. One of the pages I referred to there deals more specifically with the idea of integration by substitution, and deserves special attention.

What do differentials mean here?

Reuben asked, in 2006,

Why Does Integration by Substitution Work?

I just learned the process of integration by substitution.  My textbook seems to explain it for the most part, but it says "imagine that you multiply du/dx by the dx in the derivative to get du."  This seems to work, but I've found online that technically, you can't treat differentials as numbers that can be "canceled out."

A) Is this true?  It seems to me that they're just very very very small numbers.  So why could you not cancel them out?

B) If that's the case, how do you prove that substitution works and use integration by substitution WITHOUT treating differentials as numbers?

The question about why differentials can’t, technically, be treated as numbers, was discussed in my previous post. The main question here is, why does treating them that way work here? Can we prove it, rather than just waving our hands and doing magic?

Doctor Fenton first pointed out that the differentials can be thought of just as a way to organize our work, without having to think of them as real things:

Thanks for writing to Dr. Math.  There are several ways to look at differentials, but probably the simplest is to view them as a formal bookkeeping device for keeping track of the constants when finding antiderivatives.

This bypasses the question of what they really are, which is not needed for substitution; they are mere notation.

Then he explained the basic concept of substitution as the counterpart of the chain rule in differentiation:

Many of the integration (or antidifferentiation) rules are actually counterparts of corresponding differentiation rules, and this is true of the substitution theorem, which is the integral version of the Chain Rule.

The Chain Rule says that if we have a composite function F(g(x)), and if f(x) = F'(x) is the derivative of the outer function F(x), then

   [F(g(x))]' = F'(g(x))*g'(x)
              = f(g(x))*g'(x) .

The antiderivative version of this says that if we want to find the antiderivative of f(g(x)) and we know the antiderivative F of f, then the antiderivative of f(g(x)) is just F(g(x)), and we have reduced the problem of finding the antiderivative of the complicated expression f(g(x)) to that of finding the antiderivative of f, which we usually write with a different independent variable such as u.  That is, the Chain Rule says that

   [F(g(x))]' = f(g(x))*g'(x) ,  (remember that F' = f)

so

   /
   | f(g(x))*g'(x)dx = F(g(x)) + C .
   /

On the other hand, our definition of F implies that it is an antiderivative (indefinite integral) of f, which lets us rewrite the result:

But since F' = f, 

  /
  | f(u)du = F(u)
  /

(the letter used for the variable of integration is a dummy variable, so we can use any letter we wish).  This lets us write

   /                   /
   | f(g(x))*g'(x)dx = | f(u)du
   /                   /

where we understand the right side to be F(u) with u replaced by the formula g(x).  So it appears that in the integral on the left side, we replaced g(x) by u and g'(x)dx by du (similar to the way it "appears" that the sun rises and sets).  This process doesn't have to be considered meaningful in itself, but rather just a mnemonic or aid in determining the outer function f(x) in the composition.  The differential part of this substitution can also be thought of as a mnemonic, since we can mix Leibnitz and function notation to write, if u = g(x), that

   du
   -- = g'(x)   ,
   dx

and "multiplying" by dx gives  du = g'(x)dx .  One can make this procedure logically rigorous by introducing the concept of differential forms, but that requires a lot of mathematical machinery, so that it is easier to just think of it as a formal computation.  (There are books such as Bressoud's _Second Year Calculus_, Edwards' _Advanced Calculus: A Differential Forms Approach_, and Spivak's _Calculus on Manifolds_ which go over this in detail.)

An example of the process

Doctor Fenton then gave an example, showing first how we can recognize a complicated integral as having the form \(\int f(g(x)) g'(x) dx\), and then how to do it with less insight, just making the substitution and letting the process fill in the details:

For example, to integrate

   /
   | [x^3+1]^(1/2) x^2 dx ,
   /

we can notice that the integrand has the form of the derivative of a composite function f(g(x))*g'(x), with f(x) = x^(1/2) and g(x) = x^3+1, but g'(x) = 3x^2, not just x^2, so the constant is not quite right.  However, if we multiply and divide by 3, we have

   /                             /
   | [x^3+1]^(1/2) x^2 dx = (1/3)| [x^3+1]^(1/2)*(3x^2)dx
   /                             /

and the integrand on the right is exactly a derivative now, of the composite function (2/3)[x^3+1]^(3/2) (we can see this because the outer function f(x) is x^(1/2), whose antiderivative F(x) is (2/3)x^(3/2), so the antiderivative is (1/3)F(g(x))).  This argument has computed the integral or antiderivative without substitution.

To use substitution, we let u = x^3+1, and du = 3x^2 dx, so that x^2dx = (1/3)du, and

   /                        /
   | [x^3+1]^(1/2) x^2 dx = | u^(1/2) (1/3)du
   /                        /
                            
                          = (1/3) * (2/3)u^(3/2) + C
 
                          = (2/9) [x^3+1]^(3/2) + C  .
                            
Substitution makes the process fairly mechanical so it doesn't require much thought, once you see the appropriate substitution to use, and it also automatically keeps the constants straight.

The important thing in integration is the end result:

The objective of indefinite integration is to find an antiderivative, and exactly how you do that isn't really important, at least in my opinion.  If you can just look at a formula and "see" what the antiderivative is, you have solved the problem.  Most of us can't do that, and there are a number of procedures to help, such as the substitution rule, and integration by parts (the integral version of the Product Rule of differentiation).  The real theory behind substitution is the Chain Rule, and you can look at the details of substitution as a formal process for helping you see the important parts of the composite functions involved, without worrying about their intrinsic meaning.

More examples of substitution

One example doesn’t cover all the subtleties in using the method, so let’s look at more.

The following pages, arranged from straightforward to relatively complicated, show various Math Doctors’ different styles of substitution, and contexts in which it can be applied:

Integrate Cos 2a 
[∫cos(ax) dx, Doctor Peterson]

Integral using Substitution 
[∫x/√(x2 + 1) dx, Doctor Sydney]

Integration 
[∫(2-x2)4 x dx, Doctor Jerry]

A Change of Variables 
[∫(sin(2x))5 * cos(2x) dx, Doctor Luis]

Substituting to Simplify the Integral 
[∫tan3(x) secx dx, Doctor Barrus]

Integration of Trigonometric Function by Substitution 
[∫tan(2x) dx, Doctor Sam]

A Trigonometry Integral Requiring Two Substitutions 
[∫√(1 + sin(x)) dx, Doctor Sam]

In most of these, as usual, we just took the student through the hard part and let him finish. A couple, from early in the history of Ask Dr. Math, go all the way to the answer, where that seemed appropriate. But all demonstrate how to think through the details.

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