Section 4.6: Making Substitutions in Algebraic Expressions

This content comes directly from Larry Musolino’s open textbook Techniques of Calculus 1 Section 6.2 Integration by Substitution .

Access this resource for free at https://psu.pb.unizin.org/math110/

Learning Objectives

In this section, you will:

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

Let’s Get Started…

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 [latex]f\left[g(x)\right]{g}^{\prime }(x)dx.[/latex] For example, in the integral [latex]\int {({x}^{2}-3)}^{3}2xdx,[/latex] we have [latex]f(x)={x}^{3},g(x)={x}^{2}-3,[/latex] and [latex]g\text{'}(x)=2x.[/latex] Then,

[latex]f\left[g(x)\right]{g}^{\prime }(x)={({x}^{2}-3)}^{3}(2x),[/latex]

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 [latex]u[/latex] 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 [latex]u=g(x),[/latex] where [latex]{g}^{\prime }(x)[/latex] is continuous over an interval, let [latex]f(x)[/latex] be continuous over the corresponding range of [latex]g[/latex], and let [latex]F(x)[/latex] be an antiderivative of [latex]f(x).[/latex] Then,

[latex]\begin{array}{cc}\int f\left[g(x)\right]{g}^{\prime }(x)dx\hfill & =\int f(u)du\hfill \\ & =F(u)+C\hfill \\ & =F(g(x))+C.\hfill \end{array}[/latex]

Proof

Let [latex]f[/latex], [latex]g[/latex], [latex]u[/latex], and [latex]F[/latex] be as specified in the theorem. Then

[latex]\begin{array}{cc}\frac{d}{dx}F(g(x))\hfill & ={F}^{\prime }(g(x)){g}^{\prime }(x)\hfill \\ & =f\left[g(x)\right]{g}^{\prime }(x).\hfill \end{array}[/latex]

Integrating both sides with respect to [latex]x[/latex], we see that

[latex]\int f\left[g(x)\right]{g}^{\prime }(x)dx=F(g(x))+C.[/latex]

If we now substitute [latex]u=g(x),[/latex] and [latex]du=g\text{'}(x)dx,[/latex] we get

[latex]\begin{array}{cc}\int f\left[g(x)\right]{g}^{\prime }(x)dx\hfill & =\int f(u)du\hfill \\ & =F(u)+C\hfill \\ & =F(g(x))+C.\hfill \end{array}[/latex]

Returning to the problem we looked at originally, we let [latex]u={x}^{2}-3[/latex] and then [latex]du=2xdx.[/latex] Rewrite the integral in terms of [latex]u[/latex]:

[latex]{\int \underset{u}{\underbrace{({x}^{2}-3)}}}^{3}\underset{du}{\underbrace{(2xdx)}}=\int {u}^{3}du.[/latex]

Using the power rule for integrals, we have

[latex]\int {u}^{3}du=\frac{{u}^{4}}{4}+C.[/latex]

Substitute the original expression for [latex]x[/latex] back into the solution:

[latex]\frac{{u}^{4}}{4}+C=\frac{{({x}^{2}-3)}^{4}}{4}+C.[/latex]

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 [latex]g(x)[/latex] within the integrand to set equal to [latex]u[/latex]. Let’s select [latex]g(x).[/latex] such that [latex]{g}^{\prime }(x)[/latex] is also part of the integrand.
  2. Substitute [latex]u=g(x)[/latex] and [latex]du={g}^{\prime }(x)dx.[/latex] into the integral.
  3. We should now be able to evaluate the integral with respect to [latex]u.[/latex] If the integral can’t be evaluated we need to go back and select a different expression to use as [latex]u[/latex].
  4. Evaluate the integral in terms of [latex]u.[/latex]
  5. Write the result in terms of [latex]x[/latex] and the expression [latex]g(x).[/latex]

 

EXAMPLE 1

Using Substitution to Find an Antiderivative

Use substitution to find the antiderivative of [latex]\int 6x{\left(3{x}^{2}+4\right)}^{4} \,dx[/latex].

 

Show/Hide Solution

[latex]\int 6x{\left(3{x}^{2}+4\right)}^{4} \,dx=\int {u}^{4} \, du[/latex]

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

[latex]\begin{array}{lll} \int{u}^{4} \, du & = & \frac{{u}^{5}}{5} + C \hfill \\ \phantom{\int{u}^{4} \, du} & = & \frac{{\left(3{x}^{2}+4\right)}^{5}}{5}+C. \hfill \end{array}[/latex]

 

image Analysis

We can check our answer by taking the derivative of the result of integration. We should obtain the integrand. Picking a value for [latex]C[/latex] of 1, we let [latex]y=\frac{1}{5}{\left(3{x}^{2}+4\right)}^{5}+1.[/latex] We have

[latex]\begin{array}{lll} {y}^{\prime} & = & \frac{1}{5} \cdot 5 \cdot {\left(3{x}^{2}+4\right)}^{4} \cdot 6x\hfill \\ \phantom{{y}^{\prime}} & = & 6x{\left(3{x}^{2}+4\right)}^{4}. \hfill \end{array}[/latex]

This is exactly the expression we started with inside the integrand [latex]\checkmark[/latex].

 

image  Try It #1

Use substitution to find the antiderivative of [latex]\int 3{x}^{2}{\left({x}^{3}-3\right)}^{2} \,dx[/latex].

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

 

EXAMPLE 2

Using Substitution with Alteration

Use substitution to find [latex]\int z\sqrt{{z}^{2}-5} \,dz[/latex].

 

Show/Hide Solution

[latex]\begin{array}{ccc} \hfill u & = & {z}^{2}-5 \hfill & \\ \hfill du & = & 2z \, dz \hfill \\ \hfill & = & z \,dz \hfill \end{array}[/latex]

Write the integral in terms of [latex]u[/latex], but pull the [latex]\frac{1}{2}[/latex] outside the integration symbol:

[latex]\int z{\left({z}^{2}-5\right)}{\frac{1}{2}} \,dz=\frac{1}{2}\int {u}^{\frac{1}{2}} \, du.[/latex]

Integrate the expression in [latex]u[/latex] :

[latex]\begin{array}{lll} \frac{1}{2}\int {u}^{\frac{1}{2}} \, du & = & \frac{1}{2} \cdot \frac{{u}^{\frac{3}{2}}}{\frac{3}{2}} + C\hfill \\ \phantom{\frac{1}{2}\int {u}^{\frac{1}{2}}} & = & \frac{1}{2} \cdot \frac{2}{3} \cdot {u}^{\frac{3}{2}} + C \hfill \\ \phantom{\frac{1}{2}\int {u}^{\frac{1}{2}}} & = & \frac{1}{3} {u}^{\frac{3}{2}} + C \hfill  \\ \phantom{\frac{1}{2}\int {u}^{\frac{1}{2}}} & = & \frac{1}{3}{\left({z}^{2}-5\right)}^{\frac{3}{2}} + C. \hfill\end{array}[/latex]

 

image  Try It #2

Use substitution to find [latex]\int {x}^{2}{\left({x}^{3}+5\right)}^{9} \,dx[/latex].

EXAMPLE 3

Using Substitution with Integrals of Trigonometric Functions

Use substitution to find [latex]\int \frac{{sin}\left(t\right)}{{cos}^{3}\left(t\right)} \,dt[/latex].

 

Show/Hide Solution

 

image  Try It #3

Use substitution to find [latex]\int \frac{{cos}\left(t\right)}{{sin}^{2}\left(t\right)} \,dt[/latex].

Section 4.6 Exercises

This content comes directly from the OpenStax textbook Calculus Volume 1 Section 5.5 Substitution

Access this resource for free at https://openstax.org/books/calculus-volume-1/pages/5-5-substitution

 

[Answers to odd problem numbers are provided at the end of the problem set.  Just scroll down!]

 

Use substitution to find each integral below.

1.  [latex]\int x{\left(1-x\right)}^{99} \, dx[/latex]

 

2.  [latex]\int t{\left(1-{t}^{2}\right)}^{10} \, dt[/latex]

 

3.  [latex]\int {\left(11x-7\right)}^{-3} \, dx[/latex]

 

4.  [latex]\int {\left(7x-11\right)}^{4} \, dx[/latex]

 

5.  [latex]\int {cos}^{3}\left(\theta\right){sin}\left(\theta\right) \, d\theta[/latex]

 

6.  [latex]\int {sin}^{7}\left(\theta\right){cos}\left(\theta\right) \, d\theta[/latex]

 

7.  [latex]\int {cos}^{2}\left(\pi t\right){sin}\left(\pi t\right) \, dt[/latex]

 

8.  [latex]\int {sin}^{2}\left(x\right){cos}^{3}\left(x\right) \, dx[/latex] [Hint:  [latex]{sin}^{2}\left(x\right)+{cos}^{2}\left(x\right)=1[/latex]]

 

9.  [latex]\int t\,\,{sin}\left({t}^{2}\right){cos}\left({t}^{2}\right) \, dt[/latex]

 

10. [latex]\int {t}^{2}\,\,{cos}^{2}\left({t}^{3}\right){sin}\left({t}^{3}\right) \, dt[/latex]

 

11.  [latex]\int \frac{{x}^{2}}{{\left(x-3\right)}^{2}} \, dx[/latex]

 

12.  [latex]\int \frac{{x}^{3}}{\sqrt{1-{x}^{2}}} \, dx[/latex]

 

13.  [latex]\int \frac{{y}^{5}}{{\left(1-{y}^{3}\right)}^{\frac{3}{2}}} \, dx[/latex]

 

Answers to Section 4.6 Odd Problems

1.  [latex]-\frac{{1-x}^{100}}{100}+C[/latex]

 

3.  [latex]-\frac{1}{22{\left(11x-7\right)}^{2}}+C[/latex]

 

5.  [latex]-\frac{{cos}^{4}\left(\theta\right)}{4}+C[/latex]

 

7.  [latex]-\frac{{cos}^{3}\left(\pi t\right)}{3\pi}+C[/latex]

 

9.  [latex]-\frac{1}{4}{cos}^{2}\left({t}^{2}\right)+C[/latex]

 

11.  [latex]-\frac{1}{3\left({x}^{3}-3\right)}+C[/latex]

 

13.  [latex]\frac{2\left({y}^{3}-2\right)}{3\sqrt{1-{y}^{3}}}+C[/latex]

 

 

License

Icon for the Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License

Baylor University's Co-requisite Supplement for Calculus I Copyright © 2023 by Amy Graham Goodman is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License, except where otherwise noted.

Share This Book