4.6: 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, 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 \(f\big[g(x)\big]g′(x)\,dx\). For example, in the integral

\[ ∫(x^2−3)^3 \, 2x \, dx. \label{eq1} \]

we have

\[ f(x)=x^3 \nonumber \]

and

\[g(x)=x^2−3.\nonumber \]

Then

\[ g'(x)=2x.\nonumber \]

and

\[ f[g(x)]g′(x)=(x^2−3)^3(2x),\nonumber \]

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,

\[ \begin{align*} ∫f[g(x)]g′(x)\,dx &=∫f(u)\,du \\[4pt] &=F(u)+C \\[4pt] &= F(g(x))+C \end{align*}\]

Proof

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

\[ \dfrac{d}{dx}\big[F(g(x))\big]=F′(g(x))g′(x)=f[g(x)]g′(x). \nonumber \]

Integrating both sides with respect to \(x\), we see that

\[ ∫f[g(x)]g′(x)\,dx=F(g(x))+C. \nonumber \]

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

\[ ∫f[g(x)]g′(x)\,dx=∫f(u)\,du=F(u)+C=F(g(x))+C. \nonumber \]

□

Returning to the problem we looked at originally, we let \(u=x^2−3\) and then \(du=2x\,dx\).

Rewrite the integral (Equation \ref{eq1}) in terms of \(u\):

\[ ∫(x^2−3)^3(2x\,dx)=∫u^3\,du. \nonumber \]

Using the power rule for integrals, we have

\[ ∫u^3\,du=\dfrac{u^4}{4}+C. \nonumber \]

Substitute the original expression for \(x\) back into the solution:

\[ \dfrac{u^4}{4}+C=\dfrac{(x^2−3)^4}{4}+C.\nonumber \]

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 \(\PageIndex{1}\): Using Substitution to Find an Antiderivative

Use substitution to find the antiderivative of \(\displaystyle ∫6x(3x^2+4)^4\,dx.\)

Solution

The first step is to choose an expression for \(u\). We choose \(u=3x^2+4\) because then \(du=6x\,dx\) and we already have \(du\) in the integrand. Write the integral in terms of \(u\):

\[ ∫6x(3x^2+4)^4\,dx=∫u^4\,du. \nonumber \]

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 \(u\):

\[ ∫u^4\,du=\dfrac{u^5}{5}+C=\dfrac{(3x^2+4)^5}{5}+C.\nonumber \]

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 \(y=\dfrac{1}{5}(3x^2+4)^5+1.\) We have

\[ y=\dfrac{1}{5}(3x^2+4)^5+1,\nonumber \]

so

\[ \begin{align*} y′ &=\left(\dfrac{1}{5}\right)5(3x^2+4)^46x \\[4pt] &=6x(3x^2+4)^4.\end{align*}\]

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

Exercise \(\PageIndex{1}\)

Use substitution to find the antiderivative of \(\displaystyle ∫3x^2(x^3−3)^2\,dx.\)

Hint

Let \(u=x^3−3.\)

Answer

\(\displaystyle ∫3x^2(x^3−3)^2\,dx=\dfrac{1}{3}(x^3−3)^3+C \)

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

Example \(\PageIndex{2}\): Using Substitution with Alteration

Use substitution to find the antiderivative of \[ ∫z\sqrt{z^2−5}\,dz. \nonumber \]

Solution

Rewrite the integral as \(\displaystyle ∫z(z^2−5)^{1/2}\,dz.\) Let \(u=z^2−5\) 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 \(\dfrac{1}{2}\). we can solve this problem. Thus,

\[ u=z^2−5\nonumber \]

\[ du=2z\,dz \nonumber \]

\[ \dfrac{1}{2}du=\dfrac{1}{2}(2z)\,dz=z\,dz. \nonumber \]

Write the integral in terms of \(u\), but pull the \(\dfrac{1}{2}\) outside the integration symbol:

\[ ∫z(z^2−5)^{1/2}\,dz=\dfrac{1}{2}∫u^{1/2}\,du.\nonumber \]

Integrate the expression in \(u\):

\[ \begin{align*} \dfrac{1}{2}∫u^{1/2}\,du &= \left(\dfrac{1}{2}\right)\dfrac{u^{3/2}}{\dfrac{3}{2}}+C \\[4pt] &= \left(\dfrac{1}{2}\right)\left(\dfrac{2}{3}\right)u^{3/2}+C \\[4pt] &=\dfrac{1}{3}u^{3/2}+C \\[4pt] &=\dfrac{1}{3}(z^2−5)^{3/2}+C \end{align*}\]

Exercise \(\PageIndex{2}\)

Use substitution to find the antiderivative of \(\displaystyle ∫x^2(x^3+5)^9\,dx.\)

Hint

Multiply the du equation by \(\dfrac{1}{3}\).

Answer

\(\displaystyle ∫x^2(x^3+5)^9\,dx = \dfrac{(x^3+5)^{10}}{30}+C \)

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,

\[∫^b_af(g(x))g′(x)\,dx=∫^{g(b)}_{g(a)}f(u)\,du. \nonumber \]

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

\[ ∫f(g(x))g′(x)\,dx=F(g(x))+C. \nonumber \]

Then

\[\begin{align*} ∫^b_af[g(x)]g′(x)\,dx &= F(g(x))\bigg|^{x=b}_{x=a} \\[4pt] &=F(g(b))−F(g(a)) \\[4pt] &= F(u) \bigg|^{u=g(b)}_{u=g(a)} \\[4pt] &=∫^{g(b)}_{g(a)}f(u)\,du \end{align*}\]

and we have the desired result.

Example \(\PageIndex{3}\): Using Substitution to Evaluate a Definite Integral

Use substitution to evaluate \[ ∫^1_0x^2(1+2x^3)^5\,dx. \nonumber \]

Solution

Let \(u=1+2x^3\), so \(du=6x^2\,dx\). Since the original function includes one factor of \(x^2\) and \(du=6x^2\,dx\), multiply both sides of the \(du\) equation by \(1/6.\) Then,

\[ \begin{align*} du &=6x^2\,dx \\[4pt] \text{becomes}\quad \dfrac{1}{6}du &=x^2\,dx. \end{align*}\]

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

\[ ∫^1_0x^2(1+2x^3)^5dx=\dfrac{1}{6}∫^3_1u^5\,du. \nonumber \]

Evaluating this expression, we get

\[ \begin{align*} \dfrac{1}{6}∫^3_1u^5\,du &=\left(\dfrac{1}{6}\right)\left(\dfrac{u^6}{6}\right)\Big|^3_1 \\[4pt] &=\dfrac{1}{36}\big[(3)^6−(1)^6\big] \\[4pt] &=\dfrac{182}{9}. \end{align*}\]

Exercise \(\PageIndex{3}\)

Use substitution to evaluate the definite integral \(\displaystyle ∫^0_{−1}y(2y^2−3)^5\,dy. \)

Hint

Use the steps from Example \(\PageIndex{3}\) to solve the problem.

Answer

\(\displaystyle ∫^0_{−1}y(2y^2−3)^5\,dy = \dfrac{91}{3}\)

Exercise \(\PageIndex{4}\)

Use substitution to evaluate \(\displaystyle ∫^4_0\sqrt{2x+1}\,dx. \)

Hint

Use the process from Example \(\PageIndex{3}\) to solve the problem.

Answer

\(\displaystyle ∫^4_0\sqrt{2x+1}\,dx = \dfrac{26}{3}\)

Example \(\PageIndex{5}\): Using Substitution with an Exponential Function

Use substitution to evaluate \[ ∫^1_0xe^{4x^2+3}\,dx. \nonumber \]

Solution

Let \(u=4x^3+3.\) 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

\[\begin{align*} ∫^1_0xe^{4x^2+3}\,dx &= \dfrac{1}{8}∫^7_3e^u\,du \\[4pt] &=\dfrac{1}{8}e^u\Big|^7_3 \\[4pt] &=\dfrac{e^7−e^3}{8} \\[4pt] &≈134.568 \end{align*}\]

Just a quick note: It is possible to evaluate a definite integral using substitution without actually changing the bounds to values for "u." To do this, begin by evaluating the indefinite integral. After you have evaluated the indefinite integral using u-substitution, and expressed the result in terms of the original variable, evaluate that final result with the original bounds. This process will yield the same result. In the video below, a definite integral is evaluated using both methods, so that you can compare.

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 \[∫f[g(x)]g′(x)\,dx=∫f(u)\,du=F(u)+C=F(g(x))+C \nonumber \]
• Substitution with Definite Integrals \[∫^b_af(g(x))g'(x)\,dx=∫^{g(b)}_{g(a)}f(u)\,du \nonumber \]

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