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# 17.4 Integration by Substitution

We will now use the chain rule to find some antiderivatives. Let be a real valued function that is continuous on some interval and differentiable on the interior of , and let be a function such that has an antiderivative on some interval . We will suppose that and . It then follows that is continuous on and differentiable on , and

 (17.32)

for all in the interior of ; i.e., is an antiderivative for on . Thus
 (17.33)

There is a standard ritual for using (17.33) to find when an antiderivative can be found for . We write:

Let . Then (or ), so

 (17.34)

In the first equality of (17.34) we replace by and by , and in the last step we replace by . Since we have never assigned any meaning to  " or  ", we should think of (17.34) just as a mnemonic device for remembering (17.33).

17.35   Example. Find .

Let . Then , so

Suppose we want to find . If we had a in the denominator, this would be a simple problem. (In fact we just considered this problem in the previous example.) We will now discuss a method of introducing the missing .

Suppose is a function on an interval such that is never zero on the interior of , and suppose that is an inverse function for . Then

for all in the interior of , so

We now apply the ritual (17.34): Let . Then , so

If we can find an antiderivative for , then

We have shown that if is an inverse function for , then
 (17.36)

There is a ritual associated with this result also. To find :

Let . Then so .

Hence

 (17.37)

17.38   Example. To find .

Let . Then so .

Thus

We can now use integration by parts to find . Let

Then

Hence

17.39   Example. To find .

Let . Then so .

17.40   Example. To find .

Let . Then so .

Hence

17.41   Example. To find .

Let . Then , and

Thus

17.42   Exercise. A Find the following antiderivatives:

a)
.

b)
.

c)
.

d)
.

e)
.

f)
.

g)
.

Next: 17.5 Trigonometric Substitution Up: 17. Antidifferentiation Techniques Previous: 17.3 Integration by Parts   Index
Ray Mayer 2007-09-07