# 1.5 Substitution

### Learning Objectives

• Use substitution to evaluate indefinite integrals.
• Use substitution to evaluate definite integrals.

The Fundamental Theorem of Calculus gave us a method to evaluate integrals without using Riemann sums. The drawback of this method, though, is that we must be able to find an antiderivative, and this is not always easy. In this section, we examine a technique, called integration by substitution, that helps finding antiderivatives. Specifically, this method allows to find antiderivatives when the integrand is the result of a chain-rule derivative.

At first, the approach to the substitution procedure may not appear very obvious. However, it is primarily a visual task—that is, the integrand shows you what to do; it is a matter of recognizing the form of the function. So, what are we supposed to see? We are looking for an integrand of the form

For example, in the integral we have and Then, and we see that our integrand is in the correct form.

The method is called substitution because we substitute part of the integrand with the variable and part of the integrand with . It is also referred to as change of variables because we are changing variables to obtain an expression that is easier to work with for applying the integration rules.

### Substitution for Indefinite Integrals

Let where is continuous, let be continuous over the range of , and let be an antiderivative of Then,

### Proof

Let , , , and be as specified in the theorem. Then

This means that is an antiderivative of and hence

Since and is an antiderivative of , we have that , which completes the proof. □

In practice, when we perform a substitution, we might not know what is right away, and we use a change of variable to replace with a simpler integral that we then evaluate. This step can be viewed as substituting and . For us, and are just parts of integral notation. However, in a more rigorous mathematical analysis, they are called differentials that have a proper definition. Without going into details, we notice that if , then the formula agrees with the Leibniz’s notation for the derivative of with respect to . Indeed, , and the above formula can be obtained by “multiplying” both sides by . This, one more time, confirms a deep connection between the differential calculus and the integral calculus.

Returning to the problem we looked at originally, we let and then Rewriting the integral in terms of , we obtain:

Using the power rule for integrals, we have

Substituting the original expression for back into the solution, we get

We can generalize the procedure in the following Problem-Solving Strategy.

### Problem-Solving Strategy: Integration by Substitution

1. Look carefully at the integrand and select an expression within the integrand to set equal to . Quite often, we select so that is also part of the integrand.
2. Substitute and into the integral.
3. We should now be able to evaluate the integral with respect to . If the integral can’t be evaluated we need to go back and select a different expression to use as .
4. Evaluate the integral in terms of .
5. Replace with to write the result in terms of .

### Using Substitution to Evaluate an Indefinite Integral

Use substitution to evaluate

#### Solution

The first step is to choose an expression for . We choose because then and we already have in the integrand. Write the integral in terms of :

Now we can evaluate the integral with respect to and then return to the variable :

## Analysis

As usual, we can check our answer by taking the derivative of the result of integration to see if we really obtain the integrand. We have

This is exactly the expression in the integrand we started with, which means that our answer is correct.

Use substitution to evaluate

Sometimes we need to adjust the constants in our integral if they don’t match up exactly with the expressions we are substituting.

### Using Substitution with Alteration

Use substitution to evaluate the indefinite integral

#### Solution

Let and Now we have a problem because and the original expression has only We have to alter our expression for or the integral in will be twice as large as it should be. Multiplying both sides of the equation by solves this problem:

We can then write the integral in terms of as follows:

Integrating the expression in by rewriting as and using the power rule for integrals, we obtain:

Use substitution to find the antiderivative of

#### Hint

Multiply the equation by

### Using Substitution with Integrals of Trigonometric Functions

Use substitution to evaluate the integral

#### Solution

First, we note that we can rewrite the integral as , that is, we can combine the numerator with . Since the denominator is a power of and the derivative of is it is natural to take Then and hence . Substituting into the integral, we have

Evaluating the integral, we get

Putting the answer back in terms of , we get

Use substitution to evaluate the integral

In a similar way, one can use a sibstitution to derive a formula for . Indeed, rewriting as and taking , we get that and hence

Going back to the variable , we get

The answer is also commonly given in the form , which is, in fact, the same since

and hence by the properties of logarithms (),

.

Likewise, a substituion of , allows to establish the formula

Other trigonometric integrals that can be evaluated using a substitution are and . However, their evaluation is not straightforward and requires using a trick. To find , we multiply and divide the integrand by :

As crazy as it might seem, it helps because the derivative of the denominator is precisely equal to the numerator. That is, if we set , then , and hence

To evaluate one can either multiply the integrand by or by , and then make the corresponding change of variable. This results in seemingly different answers of and . To see that they are, in fact, the same, one needs to use the trigonometric formula together with the rules of logarithms ( and ):

and hence .

Because of the significance of the above formulas, especially in Section 3.2 Trigonometric Integrals, we gather all of them in a single statement below.

### Basic Trigonometric Integrals That Use a Substitution

Sometimes we need to manipulate an integral in ways that are more complicated than just multiplying or dividing by a constant. After the substitution is completed, should be the only variable in the integrand. In some cases, to achieve this, we need to solve for the original variable in terms of . We illustrate how it works in the next example.

### Evaluating an Indefinite Integral Using a Substitution

Use substitution to find the antiderivative of

#### Solution

If we let then But this does not account for the in the numerator of the integrand. We need to express in terms of to complete the substitution. If then Now we can rewrite the integral in terms of :

Then we integrate in the usual way, replace with the original expression, and factor and simplify the result. Thus,

Use substitution to evaluate the indefinite integral

### Substitution for Definite Integrals

Substitution can be used with definite integrals, too. However, using substitution to evaluate a definite integral requires a change to the limits of integration. If we change variables in the integrand, the limits of integration change as well.

### Substitution for Definite Integrals

Let , be continuous over an interval , and let be continuous over the range of Then,

Although we will not formally prove this theorem, we justify it with some calculations here. From the substitution rule for indefinite integrals, if is an antiderivative of we have

Then

and we have the desired result.

### Using Substitution to Evaluate a Definite Integral

Use substitution to evaluate

#### Solution

Take . Then and hence . To adjust the bounds of integration, note that corresponds to and corresponds to . We then obtain

Use substitution to evaluate

Take .

### Using Substitution with a Trigonometric Function

Use substitution to evaluate

#### Solution

Let Then, and we have that . To adjust the limits of integration, we note that when and when So our substitution gives

## Analysis

Note that in the integral in terms of that we obtained, the lower limit of integration was bigger than the upper limit. This happens a lot when substitution is used and we do not need to worry about that or fix anything – just proceed with the regular integration process.

Use substitution to evaluate

Take .

### Evaluating a Definite Integral using Substitution

Use substitution to evaluate

#### Solution

Let and then Since the original function has , we need to “split” from it to combine with , that is, we write , and then . Multiplying both sides of  by , we obtain that , and so to proceed with the chosen substitution, we need to express in terms of . Since we defined as , we have that , and hence . Finally, to adjust the limits of integration, note that when and when Then

Evaluating the integral, we obtain

Use substitution to evaluate the definite integral

Take .

### Key Concepts

• Substitution is a technique that simplifies the integration of functions that are the result of a chain-rule derivative. The term ‘substitution’ refers to changing variables or substituting the variable and for appropriate expressions in the integrand.
• When using substitution for a definite integral, we also have to change the limits of integration.

### Key Equations

• Substitution with Indefinite Integrals
• Substitution with Definite Integrals

### Exercises

In the following exercises, evaluate the indefinite integral using the indicated substitution.

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In the following exercises, use a suitable change of variables to evaluate the indefinite integral.

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In the following exercises, use a change of variables to evaluate the definite integral.

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(Hint: Write in terms of and and make a substitution )

In the following exercises, evaluate the indefinite integral with constant using a substitution. Then, graph the function and the antiderivative over the indicated interval. If possible, estimate a value of that would need to be added to the antiderivative to make it equal to the definite integral with being the left endpoint of the given interval.

38. [T] over

39. [T] over

The antiderivative is Since the antiderivative is not continuous at one cannot find a value of that would make work as a definite integral.

40. [T] over

41. [T] over

The antiderivative is You should take so that

42. [T] over

43. [T] over

The antiderivative is One should take

44. If in what can you say about the value of the integral?

45. Is the substitution in the definite integral okay? If not, why not?

No, because the integrand is discontinuous at

In the following exercises, use a change of variable to show that each definite integral is equal to zero.

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the integral becomes

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the integral becomes

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the integral becomes

since the integrand is odd.

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53. Show that the average value of over an interval is the same as the average value of over the interval for

Setting and we obtain

54. Find the area under the graph of between and where , is fixed, and evaluate the limit as

55. Find the area under the graph of between and where and is fixed. Evaluate the limit as

As the limit is if and the limit does not exist and has a trend of if

56. The area of a semicircle of radius 1 can be expressed as Use the substitution to express the area of a semicircle as the integral of a trigonometric function. You do not need to compute the integral.

57. The area of the top half of an ellipse with a major axis that is the -axis from to and with a minor axis that is the -axis from to can be written as Use the substitution to express this area in terms of an integral of a trigonometric function. You do not need to compute the integral.

58. [T] The following graph is of a function of the form Estimate the coefficients and , and the frequency parameters and . Use these estimates to approximate

59. [T] The following graph is of a function of the form Estimate the coefficients and and the frequency parameters and . Use these estimates to approximate