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Mathematics LibreTexts

5.5: Substitution

  • Gilbert Strang & Edwin “Jed” Herman
  • OpenStax

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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, to help us find antiderivatives. Specifically, this method helps us 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 𝑓[𝑔(𝑥)]𝑔 (𝑥) 𝑑𝑥. For example, in the integral

(𝑥23)32𝑥𝑑𝑥.(5.5.1)

we have

𝑓(𝑥)=𝑥3

and

𝑔(𝑥)=𝑥23.

Then

𝑔(𝑥)=2𝑥.

and

𝑓[𝑔(𝑥)]𝑔(𝑥)=(𝑥23)3(2𝑥),

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 𝑢 and part of the integrand with 𝑑𝑢. 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 with Indefinite Integrals

Let 𝑢 =𝑔(𝑥), where 𝑔 (𝑥) is continuous over an interval, let 𝑓(𝑥) be continuous over the corresponding range of 𝑔, and let 𝐹(𝑥) be an antiderivative of 𝑓(𝑥). Then,

𝑓[𝑔(𝑥)]𝑔(𝑥)𝑑𝑥=𝑓(𝑢)𝑑𝑢=𝐹(𝑢)+𝐶=𝐹(𝑔(𝑥))+𝐶

Proof

Let 𝑓, 𝑔, 𝑢, and 𝐹 be as specified in the theorem. Then

𝑑𝑑𝑥[𝐹(𝑔(𝑥))]=𝐹(𝑔(𝑥))𝑔(𝑥)=𝑓[𝑔(𝑥)]𝑔(𝑥).

Integrating both sides with respect to 𝑥, we see that

𝑓[𝑔(𝑥)]𝑔(𝑥)𝑑𝑥=𝐹(𝑔(𝑥))+𝐶.

If we now substitute 𝑢 =𝑔(𝑥), and 𝑑𝑢 =𝑔(𝑥) 𝑑𝑥, we get

𝑓[𝑔(𝑥)]𝑔(𝑥)𝑑𝑥=𝑓(𝑢)𝑑𝑢=𝐹(𝑢)+𝐶=𝐹(𝑔(𝑥))+𝐶.

Returning to the problem we looked at originally, we let 𝑢 =𝑥2 3 and then 𝑑𝑢 =2𝑥 𝑑𝑥.

Rewrite the integral (Equation 5.5.1) in terms of 𝑢:

(𝑥23)3(2𝑥𝑑𝑥)=𝑢3𝑑𝑢.

Using the power rule for integrals, we have

𝑢3𝑑𝑢=𝑢44+𝐶.

Substitute the original expression for 𝑥 back into the solution:

𝑢44+𝐶=(𝑥23)44+𝐶.

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 𝑔(𝑥) within the integrand to set equal to u. Let’s select 𝑔(𝑥). such that 𝑔 (𝑥) is also part of the integrand.
  2. Substitute 𝑢 =𝑔(𝑥) and 𝑑𝑢 =𝑔 (𝑥)𝑑𝑥. into the integral.
  3. We should now be able to evaluate the integral with respect to 𝑢. If the integral can’t be evaluated we need to go back and select a different expression to use as 𝑢.
  4. Evaluate the integral in terms of 𝑢.
  5. Write the result in terms of 𝑥 and the expression 𝑔(𝑥).
Example 5.5.1: Using Substitution to Find an Antiderivative

Use substitution to find the antiderivative of 6𝑥(3𝑥2+4)4 𝑑𝑥.

Solution

The first step is to choose an expression for 𝑢. We choose 𝑢 =3𝑥2 +4 because then 𝑑𝑢 =6𝑥 𝑑𝑥 and we already have 𝑑𝑢 in the integrand. Write the integral in terms of 𝑢:

6𝑥(3𝑥2+4)4𝑑𝑥=𝑢4𝑑𝑢.

Remember that 𝑑𝑢 is the derivative of the expression chosen for 𝑢, regardless of what is inside the integrand. Now we can evaluate the integral with respect to 𝑢:

𝑢4𝑑𝑢=𝑢55+𝐶=(3𝑥2+4)55+𝐶.

Analysis

We can check our answer by taking the derivative of the result of integration. We should obtain the integrand. Picking a value for 𝐶 of 1, we let 𝑦 =15(3𝑥2+4)5 +1. We have

𝑦=15(3𝑥2+4)5+1,

so

𝑦=(15)5(3𝑥2+4)46𝑥=6𝑥(3𝑥2+4)4.

This is exactly the expression we started with inside the integrand.

Exercise 5.5.1

Use substitution to find the antiderivative of 3𝑥2(𝑥33)2 𝑑𝑥.

Hint

Let 𝑢 =𝑥3 3.

Answer

3𝑥2(𝑥33)2 𝑑𝑥 =13(𝑥33)3 +𝐶

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

Example 5.5.2: Using Substitution with Alteration

Use substitution to find the antiderivative of 𝑧𝑧25𝑑𝑧.

Solution

Rewrite the integral as 𝑧(𝑧25)1/2 𝑑𝑧. Let 𝑢 =𝑧2 5 and 𝑑𝑢 =2𝑧 𝑑𝑧. Now we have a problem because 𝑑𝑢 =2𝑧 𝑑𝑧 and the original expression has only 𝑧 𝑑𝑧. We have to alter our expression for 𝑑𝑢 or the integral in 𝑢 will be twice as large as it should be. If we multiply both sides of the 𝑑𝑢 equation by 12. we can solve this problem. Thus,

𝑢=𝑧25

𝑑𝑢=2𝑧𝑑𝑧

12𝑑𝑢=12(2𝑧)𝑑𝑧=𝑧𝑑𝑧.

Write the integral in terms of 𝑢, but pull the 12 outside the integration symbol:

𝑧(𝑧25)1/2𝑑𝑧=12𝑢1/2𝑑𝑢.

Integrate the expression in 𝑢:

12𝑢1/2𝑑𝑢=(12)𝑢3/232+𝐶=(12)(23)𝑢3/2+𝐶=13𝑢3/2+𝐶=13(𝑧25)3/2+𝐶

Exercise 5.5.2

Use substitution to find the antiderivative of 𝑥2(𝑥3+5)9 𝑑𝑥.

Hint

Multiply the du equation by 13.

Answer

𝑥2(𝑥3+5)9 𝑑𝑥 =(𝑥3+5)1030 +𝐶

Example 5.5.3: Using Substitution with Integrals of Trigonometric Functions

Use substitution to evaluate the integral sin𝑡cos3𝑡 𝑑𝑡.

Solution

We know the derivative of cos𝑡 is sin𝑡, so we set 𝑢 =cos𝑡. Then 𝑑𝑢 =sin𝑡 𝑑𝑡.

Substituting into the integral, we have

sin𝑡cos3𝑡𝑑𝑡=𝑑𝑢𝑢3.

Evaluating the integral, we get

𝑑𝑢𝑢3=𝑢3𝑑𝑢=(12)𝑢2+𝐶.

Putting the answer back in terms of t, we get

sin𝑡cos3𝑡𝑑𝑡=12𝑢2+𝐶=12cos2𝑡+𝐶.

Exercise 5.5.3

Use substitution to evaluate the integral cos𝑡sin2𝑡 𝑑𝑡.

Hint

Use the process from Example 5.5.3 to solve the problem.

Answer

cos𝑡sin2𝑡 𝑑𝑡 =1sin𝑡 +𝐶

Exercise 5.5.4

Use substitution to evaluate the indefinite integral cos3𝑡sin𝑡 𝑑𝑡.

Hint

Use the process from Example 5.5.3 to solve the problem.

Answer

cos3𝑡sin𝑡 𝑑𝑡 =cos4𝑡4 +𝐶

Sometimes we need to manipulate an integral in ways that are more complicated than just multiplying or dividing by a constant. We need to eliminate all the expressions within the integrand that are in terms of the original variable. When we are done, 𝑢 should be the only variable in the integrand. In some cases, this means solving for the original variable in terms of 𝑢. This technique should become clear in the next example.

Example 5.5.4: Finding an Antiderivative Using 𝑢-Substitution

Use substitution to find the antiderivative of 𝑥𝑥1𝑑𝑥.

Solution

If we let 𝑢 =𝑥 1, then 𝑑𝑢 =𝑑𝑥. But this does not account for the 𝑥 in the numerator of the integrand. We need to express 𝑥 in terms of 𝑢. If 𝑢 =𝑥 1, then 𝑥 =𝑢 +1. Now we can rewrite the integral in terms of 𝑢 :

𝑥𝑥1𝑑𝑥=𝑢+1𝑢𝑑𝑢=(𝑢+1𝑢)𝑑𝑢=(𝑢1/2+𝑢1/2)𝑑𝑢.

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

(𝑢1/2+𝑢1/2)𝑑𝑢=23𝑢3/2+2𝑢1/2+𝐶=23(𝑥1)3/2+2(𝑥1)1/2+𝐶=(𝑥1)1/2[23(𝑥1)+2]+𝐶=(𝑥1)1/2(23𝑥23+63)=(𝑥1)1/2(23𝑥+43)=23(𝑥1)1/2(𝑥+2)+𝐶.

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 with Definite Integrals

Let 𝑢 =𝑔(𝑥) and let 𝑔 be continuous over an interval [𝑎,𝑏], and let 𝑓 be continuous over the range of 𝑢 =𝑔(𝑥). Then,

𝑏𝑎𝑓(𝑔(𝑥))𝑔(𝑥)𝑑𝑥=𝑔(𝑏)𝑔(𝑎)𝑓(𝑢)𝑑𝑢.

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

𝑓(𝑔(𝑥))𝑔(𝑥)𝑑𝑥=𝐹(𝑔(𝑥))+𝐶.

Then

𝑏𝑎𝑓[𝑔(𝑥)]𝑔(𝑥)𝑑𝑥=𝐹(𝑔(𝑥))𝑥=𝑏𝑥=𝑎=𝐹(𝑔(𝑏))𝐹(𝑔(𝑎))=𝐹(𝑢)𝑢=𝑔(𝑏)𝑢=𝑔(𝑎)=𝑔(𝑏)𝑔(𝑎)𝑓(𝑢)𝑑𝑢

and we have the desired result.

Example 5.5.5: Using Substitution to Evaluate a Definite Integral

Use substitution to evaluate 10𝑥2(1+2𝑥3)5𝑑𝑥.

Solution

Let 𝑢 =1 +2𝑥3, so 𝑑𝑢 =6𝑥2 𝑑𝑥. Since the original function includes one factor of 𝑥2 and 𝑑𝑢 =6𝑥2 𝑑𝑥, multiply both sides of the 𝑑𝑢 equation by 1/6. Then,

𝑑𝑢=6𝑥2𝑑𝑥becomes16𝑑𝑢=𝑥2𝑑𝑥.

To adjust the limits of integration, note that when 𝑥 =0,𝑢 =1 +2(0) =1, and when 𝑥 =1, 𝑢 =1 +2(1) =3.

Then

10𝑥2(1+2𝑥3)5𝑑𝑥=1631𝑢5𝑑𝑢.

Evaluating this expression, we get

1631𝑢5𝑑𝑢=(16)(𝑢66)31=136[(3)6(1)6]=1829.

Exercise 5.5.5

Use substitution to evaluate the definite integral 01𝑦(2𝑦23)5 𝑑𝑦.

Hint

Use the steps from Example 5.5.5 to solve the problem.

Answer

01𝑦(2𝑦23)5 𝑑𝑦 =913

Exercise 5.5.6

Use substitution to evaluate 10𝑥2cos(𝜋2𝑥3) 𝑑𝑥.

Hint

Use the process from Example 5.5.5 to solve the problem.

Answer

10𝑥2cos(𝜋2𝑥3) 𝑑𝑥 =23𝜋 0.2122

Example 5.5.6: Using Substitution with an Exponential Function

Use substitution to evaluate 10𝑥𝑒4𝑥2+3𝑑𝑥.

Solution

Let 𝑢 =4𝑥3 +3. Then, 𝑑𝑢 =8𝑥 𝑑𝑥. To adjust the limits of integration, we note that when 𝑥 =0, 𝑢 =3, and when 𝑥 =1, 𝑢 =7. So our substitution gives

10𝑥𝑒4𝑥2+3𝑑𝑥=1873𝑒𝑢𝑑𝑢=18𝑒𝑢73=𝑒7𝑒38134.568

Substitution may be only one of the techniques needed to evaluate a definite integral. All of the properties and rules of integration apply independently, and trigonometric functions may need to be rewritten using a trigonometric identity before we can apply substitution. Also, we have the option of replacing the original expression for 𝑢 after we find the antiderivative, which means that we do not have to change the limits of integration. These two approaches are shown in Example 5.5.7.

Example 5.5.7: Using Substitution to Evaluate a Trigonometric Integral

Use substitution to evaluate 𝜋/20cos2𝜃𝑑𝜃.

Solution

Let us first use a trigonometric identity to rewrite the integral. The trig identity cos2𝜃 =1+cos2𝜃2 allows us to rewrite the integral as

𝜋/20cos2𝜃𝑑𝜃=𝜋/201+cos2𝜃2𝑑𝜃.

Then,

𝜋/20(1+cos2𝜃2)𝑑𝜃=𝜋/20(12+12cos2𝜃)𝑑𝜃=12𝜋/20𝑑𝜃+𝜋/20cos2𝜃𝑑𝜃.

We can evaluate the first integral as it is, but we need to make a substitution to evaluate the second integral. Let 𝑢 =2𝜃. Then, 𝑑𝑢 =2 𝑑𝜃, or 12 𝑑𝑢 =𝑑𝜃. Also, when 𝜃 =0, 𝑢 =0, and when 𝜃 =𝜋/2, 𝑢 =𝜋. Expressing the second integral in terms of 𝑢, we have

12𝜋/20𝑑𝜃+12𝜋/20cos2𝜃𝑑𝜃=12𝜋/20𝑑𝜃+12(12)𝜋0cos𝑢𝑑𝑢=𝜃2𝜃=𝜋/2𝜃=0+14sin𝑢𝑢=𝜃𝑢=0=(𝜋40)+(00)=𝜋4

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 𝑢 and 𝑑𝑢 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 𝑓[𝑔(𝑥)]𝑔(𝑥)𝑑𝑥=𝑓(𝑢)𝑑𝑢=𝐹(𝑢)+𝐶=𝐹(𝑔(𝑥))+𝐶
  • Substitution with Definite Integrals 𝑏𝑎𝑓(𝑔(𝑥))𝑔(𝑥)𝑑𝑥=𝑔(𝑏)𝑔(𝑎)𝑓(𝑢)𝑑𝑢

Glossary

change of variables
the substitution of a variable, such as 𝑢, for an expression in the integrand
integration by substitution
a technique for integration that allows integration of functions that are the result of a chain-rule derivative

This page titled 5.5: Substitution is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Gilbert Strang & Edwin “Jed” Herman (OpenStax) via source content that was edited to the style and standards of the LibreTexts platform.

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