<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />

# 4.7: Integration by Substitution

Difficulty Level: At Grade Created by: CK-12

## Learning Objectives

• Integrate composite functions
• Use change of variables to evaluate definite integrals
• Use substitution to compute definite integrals

## Introduction

In this lesson we will expand our methods for evaluating definite integrals. We first look at a couple of situations where finding antiderivatives requires special methods. These involve finding antiderivatives of composite functions and finding antiderivatives of products of functions.

Antiderivatives of Composites

Suppose we needed to compute the following integral:

Our rules of integration are of no help here. We note that the integrand is of the form where and

Since we are looking for an antiderivative of and we know that we can re-write our integral as

In practice, we use the following substitution scheme to verify that we can integrate in this way:

1. Let
2. Differentiate both sides so
3. Change the original integral in to an integral in :
where and
4. Integrate with respect to :

5. Change the answer back to :

While this method of substitution is a very powerful method for solving a variety of problems, we will find that we sometimes will need to modify the method slightly to address problems, as in the following example.

Example 1:

Compute the following indefinite integral:

Solution:

We note that the derivative of is ; hence, the current problem is not of the form But we notice that the derivative is off only by a constant of and we know that constants are easy to deal with when differentiating and integrating. Hence

Let

Then

Then and we are ready to change the original integral from to an integral in and integrate:

Changing back to , we have

We can also use this substitution method to evaluate definite integrals. If we attach limits of integration to our first example, we could have a problem such as

The method still works. However, we have a choice to make once we are ready to use the Fundamental Theorem to evaluate the integral.

Recall that we found that for the indefinite integral. At this point, we could evaluate the integral by changing the answer back to or we could evaluate the integral in But we need to be careful. Since the original limits of integration were in , we need to change the limits of integration for the equivalent integral in Hence,

where

Integrating Products of Functions

We are not able to state a rule for integrating products of functions, but we can get a relationship that is almost as effective. Recall how we differentiated a product of functions:

So by integrating both sides we get

or

In order to remember the formula, we usually write it as

We refer to this method as integration by parts. The following example illustrates its use.

Example 2:

Use integration by parts method to compute

Solution:

We note that our other substitution method is not applicable here. But our integration by parts method will enable us to reduce the integral down to one that we can easily evaluate.

Let and then and

By substitution, we have

We can easily evaluate the integral and have

And should we wish to evaluate definite integrals, we need only to apply the Fundamental Theorem to the antiderivative.

## Lesson Summary

1. We integrated composite functions.
2. We used change of variables to evaluate definite integrals.
3. We used substitution to compute definite integrals.

## Review Questions

Compute the integrals in problems #1–10.

### Notes/Highlights Having trouble? Report an issue.

Color Highlighted Text Notes

Show Hide Details
Description
Tags:
Subjects: