1.4: Integration by Substitution

Learning Objectives Use substitution to evaluate indefinite integrals. Use substitution to evaluate definite integrals.

Hawk A.I. Section-Specific Tutor Please note that, to access the Hawk A.I. Tutor, you will need a (free) OpenAI account.

The Fundamental Theorem of Calculus allows us to evaluate integrals without using Riemann sums. The drawback of this method is that we must be able to find an antiderivative, which can be challenging. This section examines integration by substitution - a technique to help us find antiderivatives. Specifically, this method allows us to find antiderivatives when the integrand is the result of a Chain Rule derivative.

At first, the approach to the substitution procedure may appear obscure. 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 \(f \left( g(x) \right) g^{\prime}(x)\,dx\). For example, in the integral\[ \int (x^2−3)^3 \, 2x \, dx. \label{eq1} \]we have\[ f(x)=x^3 \nonumber \]and\[g(x)=x^2−3.\nonumber \]Then\[ g^{\prime}(x)=2x.\nonumber \]Therefore,\[ f(g(x))g^{\prime}(x)=(x^2−3)^3(2x),\nonumber \]and we see that our integrand is in the correct form. The method is called substitution, or the Substitution Method, 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 easier expression to work with for applying the integration rules.

Theorem: Substitution Method with Indefinite Integrals

Let \(u=g(x)\), where \(g^{\prime}(x)\) 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,\[ \begin{array}{rcl}
\int f\left(g(x)\right)g^{\prime}(x)\,dx & = & \int f(u)\,du \\[6pt]
& = & F(u)+C \\[6pt]
& = & F(g(x))+C \\[6pt]
\end{array} \nonumber\]

Proof

Let \(f\), \(g\), \(u\), and \(F\) be as specified in the theorem. Then\[ \dfrac{d}{dx}\big[F(g(x))\big]=F^{\prime}(g(x))g^{\prime}(x)=f[g(x)]g^{\prime}(x). \nonumber \]Integrating both sides with respect to \(x\), we see that\[ \int f\left(g(x)\right)g^{\prime}(x)\,dx=F(g(x))+C. \nonumber \]If we now substitute \(u=g(x)\), and \(du=g^{\prime}(x)\,dx\), we get\[ \int f\left(g(x)\right)g^{\prime}(x)\,dx= \int f(u)\,du=F(u)+C=F(g(x))+C. \nonumber \]

Q.E.D.

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 \(u\):\[ \int (x^2−3)^3(2x\,dx)= \int u^3\,du. \nonumber \]Using the Power Rule for integrals, we have\[ \int u^3\,du=\dfrac{u^4}{4}+C. \nonumber \]Substitute the original expression for \(x\) back into the solution:\[ \dfrac{u^4}{4}+C=\dfrac{(x^2−3)^4}{4}+C.\nonumber \]At this point, it is important to note that integration is mostly a heuristic method. That is, integration is approached more often by using a "calculated guess" derived from previous experiences. As you move through the middle part of Calculus (also known as Calculus II, or Integral Calculus), you will learn many different integration techniques. You must practice as many problems as possible to get a "natural feel" for what method to use and when to use it.

In the case of Integration by Substitution, you are often (but not always) looking for an integrand involving a function, \( g(x) \), and its approximate derivative, \( g^{\prime}(x) \). You let \( u(x) = g(x) \) so that \( du = g^{\prime}(x) dx \). At that point, you will have rewritten your original integral into a form that (hopefully) looks like\[ \int{ u \, du}. \nonumber \]

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

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

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 \(u\):\[ \int 6x(3x^2+4)^4\,dx= \int u^4\,du. \nonumber \]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 \(u\):\[ \int u^4\,du=\dfrac{u^5}{5}+C=\dfrac{(3x^2+4)^5}{5}+C.\nonumber \]

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 \(y=\frac{1}{5}(3x^2+4)^5+1\). We have\[ y=\dfrac{1}{5}(3x^2+4)^5+1,\nonumber \]so\[ \begin{array}{rcl}
y^{\prime} & = & \left(\dfrac{1}{5}\right)5(3x^2+4)^46x \\[6pt]
& = & 6x(3x^2+4)^4. \\[6pt]
\end{array} \nonumber\]This is exactly the expression we started with inside the integrand.

Checkpoint \(\PageIndex{1}\)

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

Answer

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

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 \[ \int z\sqrt{z^2−5}\,dz. \nonumber \]

Solution

Rewrite the integral as \(\displaystyle \int z(z^2−5)^{1/2}\,dz\). Let \(u=z^2−5\) and \(du=2z\,dz\). We have a problem because \(du=2z\,dz\) and the original expression has only \(z\,dz\). We must 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 \(\frac{1}{2}\). 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 \(\frac{1}{2}\) outside the integration symbol:\[ \int z(z^2−5)^{1/2}\,dz=\dfrac{1}{2} \int u^{1/2}\,du.\nonumber \]Integrate the expression in \(u\):\[ \begin{array}{rcl}
\dfrac{1}{2} \int u^{1/2}\,du & = & \left(\dfrac{1}{2}\right)\dfrac{u^{3/2}}{\dfrac{3}{2}}+C \\[6pt]
& = & \left(\dfrac{1}{2}\right)\left(\dfrac{2}{3}\right)u^{3/2}+C \\[6pt]
& = & \dfrac{1}{3}u^{3/2}+C \\[6pt]
& = & \dfrac{1}{3}(z^2−5)^{3/2}+C \\[6pt]
\end{array} \nonumber\]

Checkpoint \(\PageIndex{2}\)

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

Answer

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

Example \(\PageIndex{3}\): Using Substitution with Integrals of Trigonometric Functions

Use substitution to evaluate the integral \(\displaystyle \int \frac{\sin t}{\cos^3t}\,dt\).

Solution

We know the derivative of \(\cos t\) is \(−\sin t\), so we set \(u=\cos t\). Then \(du=−\sin t\,dt\).

Substituting into the integral, we have\[ \int \dfrac{\sin t}{\cos^3t}\,dt=− \int \dfrac{du}{u^3}.\nonumber \]Evaluating the integral, we get\[ − \int \dfrac{du}{u^3}=− \int u^{−3}\,du=−\left(−\dfrac{1}{2}\right)u^{−2}+C.\nonumber \]Putting the answer back in terms of t, we get\[ \int \dfrac{\sin t}{\cos^3t}\,dt=\dfrac{1}{2u^2}+C=\dfrac{1}{2\cos^2t}+C.\nonumber \]

Example \( \PageIndex{3} \) needs a little more explanation. How did we know to let \( u = \cos{(t)} \) rather than the sine function? To be honest, this is something that becomes more of an instinct after a while; however, there is a science to it - it's the science of considering ramifications.

If we let \( u = \sin{(t)} \), then \( du = \cos{(t)} dt \). This might not jump off the page to you and scream, "This is bad," so let's see the ramifications of this choice.

The original integrand was \( \frac{\sin{(t)}}{\cos^3{(t)}} \). Since this does not have a \( \cos{(t)} \) to "steal" in the numerator, we need to solve \( du = \cos{(t)} dt \) for \( dt \). Doing so yields \( dt = \frac{1}{\cos{(t)}} du \). If we make this substitution, the integrand would become\[ \dfrac{ \sin{(t)} }{ \cos^3{(t)} } \, dt = \dfrac{ u }{ \cos^3{(t)} } \cdot \dfrac{1}{\cos{(t)}} \, du = \dfrac{ u }{ \cos^4{(t)} } \, du. \nonumber \]There are two major issues here:

1. the integrand would have a mixture of variables, so we cannot leave it this way, and
2. the integrand would become more complex - not less.

The first issue can be overcome by using our Trigonometry knowledge.\[ \begin{array}{rclr}
\dfrac{ u }{ \cos^4{(t)} } \, du & = & \dfrac{u}{ \left( \cos^2{(t)} \right)^2 } \, du & \\[6pt]
& = & \dfrac{u}{ \left(1 - \sin^2{(t)} \right)^2 } \, du & \\[6pt]
& = & \dfrac{u}{ \left(1 - u^2 \right)^2 } \, du & \left( \text{Substitution: Recall }u = \sin{(t)} \right) \\[6pt]
\end{array} \nonumber \]However, the second issue (complexity) still exists. Instead of\[ \int{ \dfrac{ \sin{(t)} }{ \cos^3{(t)} } \, dt}, \nonumber \]we are now faced with\[ \int{ \dfrac{u}{ \left(1 - u^2 \right)^2 } \, du }. \nonumber \]In either case, we do not immediately know of an appropriate antiderivative.

Checkpoint \(\PageIndex{3}\)

Use substitution to evaluate each integral

1. \( \displaystyle \int \frac{\cos t}{\sin^2t}\,dt\)
2. \(\displaystyle \int \cos^3t\sin t\,dt \)

Answers

1. \(\displaystyle \int \frac{\cos t}{\sin^2t}\,dt = −\frac{1}{\sin t}+C\)
2. \(\displaystyle \int \cos^3t\sin t\,dt = −\frac{\cos^4t}{4}+C \)

Sometimes, we need to manipulate an integral in more complicated ways 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 done, \(u\) should be the only variable in the integrand. This sometimes means solving for the original variable in terms of \(u\). This technique should become clear in the next example.

Example \(\PageIndex{4}\): Finding an Antiderivative Using \(u\)-Substitution

Use substitution to find the antiderivative of \[ \int \dfrac{x}{\sqrt{x−1}}\,dx. \nonumber \]

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 must express \(x\) in terms of \(u\). If \(u=x−1\), then \(x=u+1\). Now we can rewrite the integral in terms of \(u\):\[ \int \dfrac{x}{\sqrt{x−1}}\,dx= \int \dfrac{u+1}{\sqrt{u}}\,du= \int \left(\sqrt{u}+\dfrac{1}{\sqrt{u}}\right)\,du= \int \left(u^{1/2}+u^{−1/2}\right)\,du.\nonumber \]Then we integrate in the usual way, replace \(u\) with the original expression, and factor and simplify the result. Thus,\[ \begin{array}{rcl}
\int (u^{1/2}+u^{−1/2})\,du & = & \dfrac{2}{3}u^{3/2}+2u^{1/2}+C \\[6pt]
& = & \dfrac{2}{3}(x−1)^{3/2}+2(x−1)^{1/2}+C \\[6pt]
& = & (x−1)^{1/2}\left[\dfrac{2}{3}(x−1)+2\right]+C \\[6pt]
& = & (x−1)^{1/2}\left(\dfrac{2}{3}x−\dfrac{2}{3}+\dfrac{6}{3}\right) \\[6pt]
& = & (x−1)^{1/2}\left(\dfrac{2}{3}x+\dfrac{4}{3}\right) \\[6pt]
& = & \dfrac{2}{3}(x−1)^{1/2}(x+2)+C. \\[6pt]
\end{array} \nonumber\]

The final example in this subsection illustrates a common tripping point for students - the dreaded substitution with resubstitution.

Example \(\PageIndex{5}\): Using Substitution with a Resubstitution

Evaluate.\[ \int 3 x^{17} \sqrt[4]{4x^9 + 8} \, dx \nonumber \]

Solution

On first inspection, there doesn't seem to be an expression whose exact derivative (or constant multiple of the derivative) is contained within the integrand. However, practicing some faith, let's try the \( u \)-substitution, \( u = 4x^9 + 8 \). This would imply that \( du = 36x^8 \, dx \). Looking at the integrand, we don't have a \( 36 x^8 \, dx \) to "steal," but we do have \( 3 x^8 \, dx \). Thus, we let \( \frac{1}{12} \, du = 3 x^8 \, dx \). Therefore,\[ \begin{array}{rcl}
\displaystyle \int 3 x^{17} \sqrt[4]{4x^9 + 8} \, dx & = & \displaystyle \int x^{9} \sqrt[4]{4x^9 + 8} \cdot 3 x^8 \, dx\\[6pt]
& = & \displaystyle \dfrac{1}{12} \int x^{9} \sqrt[4]{u} \, du \\[6pt]
\end{array} \nonumber \]At this point, I want to be clear that the notation in that final line is not traditionally acceptable. Our original integral was in terms of the variable \( x \) (and only that variable). The integral we arrived at in the last line should be in terms of only \( u \); however, we have a pesky \( x^9 \) in there. Luckily, we know that \( u = 4 x^9 + 8 \). Hence, \( \frac{u - 8}{4} = x^9 \). Substituting (also known as resubstituting) this into the last line of the computation above, we get the following.\[ \begin{array}{rcl}
\displaystyle \int 3 x^{17} \sqrt[4]{4x^9 + 8} \, dx & = & \displaystyle \dfrac{1}{12} \int x^{9} \sqrt[4]{u} \, du \\[6pt]
& = & \displaystyle \dfrac{1}{12} \int \dfrac{u - 8}{4} u^{1/4} \, du \\[6pt]
& = & \displaystyle \dfrac{1}{48} \int (u - 8) u^{1/4} \, du \\[6pt]
& = & \displaystyle \dfrac{1}{48} \int u^{5/4} - 8 u^{1/4} \, du \\[6pt]
& = & \displaystyle \dfrac{1}{48} \left( \dfrac{4}{9} u^{9/4} - \dfrac{32}{5} u^{5/4} \right) + C \\[6pt]
& = & \displaystyle \dfrac{1}{12} \left( \dfrac{1}{9} u^{9/4} - \dfrac{8}{5} u^{5/4} \right) + C \\[6pt]
& = & \displaystyle \dfrac{1}{12} \left( \dfrac{1}{9} (4x^9 + 8)^{9/4} - \dfrac{8}{5} (4x^9 + 8)^{5/4} \right) + C\\[6pt]
\end{array} \nonumber \]

Before moving away from Example \( \PageIndex{ 5 } \), it's worthwhile to mention that we should simplify expressions when possible. Unfortunately, factoring out common expressions from the answer in Example \( \PageIndex{ 5 } \) does not improve the look and feel of the solution.\[ \begin{array}{rcl}
\dfrac{1}{12} \left( \dfrac{1}{9} (4x^9 + 8)^{9/4} - \dfrac{8}{5} (4x^9 + 8)^{5/4} \right) + C & = & \dfrac{1}{12 \cdot 9 \cdot 5} (4x^9 + 8)^{5/4} \left( 5(4x^9 + 8) - 9 \cdot 8 \right) + C\\[6pt]
& = & \dfrac{1}{540} (4x^9 + 8)^{5/4} \left( 20x^9 + 40 - 72 \right) + C\\[6pt]
& = & \dfrac{1}{540} (4x^9 + 8)^{5/4} \left( 20x^9 - 32 \right) + C\\[6pt]
& = & \dfrac{4}{540} (4x^9 + 8)^{5/4} \left( 5x^9 - 8 \right) + C\\[6pt]
& = & \dfrac{1}{135} (4x^9 + 8)^{5/4} \left( 5x^9 - 8 \right) + C\\[6pt]
\end{array} \nonumber \]

Substitution for Definite Integrals

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

Theorem: Substitution with Definite Integrals

Let \(u=g(x)\) and let \(g^{\prime}\) be continuous over an interval \([a,b]\), and let \(f\) be continuous over the range of \(u=g(x)\). Then,\[ \int ^b_af(g(x))g^{\prime}(x)\,dx= \int ^{g(b)}_{g(a)}f(u)\,du. \nonumber \]

Although we will not formally prove this theorem, we justify it with some calculations. From the Substitution Method for indefinite integrals, if \(F(x)\) is an antiderivative of \(f(x)\), we have\[ \int f(g(x))g^{\prime}(x)\,dx=F(g(x))+C. \nonumber \]Then\[\begin{array}{rcl}
\int ^b_af[g(x)]g^{\prime}(x)\,dx & = & F(g(x))\bigg|^{x=b}_{x=a} \\[6pt]
& = & F(g(b))−F(g(a)) \\[6pt]
& = & F(u) \bigg|^{u=g(b)}_{u=g(a)} \\[6pt]
& = & \int ^{g(b)}_{g(a)}f(u)\,du \\[6pt]
\end{array} \nonumber\]and we have the desired result.

Example \(\PageIndex{6}\): Using Substitution to Evaluate a Definite Integral

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

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,\[ \begin{array}{crcl}
& du & = 6x^2\,dx \\[6pt]
\implies & \dfrac{1}{6}du & = & x^2\,dx. \\[6pt]
\end{array} \nonumber\]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\[ \int ^1_0x^2(1+2x^3)^5dx=\dfrac{1}{6} \int ^3_1u^5\,du. \nonumber \]Evaluating this expression, we get\[ \begin{array}{rcl}
\dfrac{1}{6} \int ^3_1u^5\,du & = & \left(\dfrac{1}{6}\right)\left(\dfrac{u^6}{6}\right)\Big|^{u =3}_{u = 1} \\[6pt]
& = & \dfrac{1}{36}\big[(3)^6−(1)^6\big] \\[6pt]
& = & \dfrac{182}{9}. \\[6pt]
\end{array} \nonumber\]

Checkpoint \(\PageIndex{6}\)

Use substitution to evaluate the definite integral.

1. \(\displaystyle \int ^0_{−1}y(2y^2−3)^5\,dy \)
2. \(\displaystyle \int ^1_0 x^2 \cos \left(\frac{ \pi }{2}x^3\right)\,dx \)

Answers

1. \(\displaystyle \int ^0_{−1}y(2y^2−3)^5\,dy = \frac{91}{3}\)
2. \(\displaystyle \int ^1_0x^2 \cos \left(\frac{ \pi }{2}x^3\right)\,dx = \frac{2}{3 \pi } \approx 0.2122\)

Example \(\PageIndex{7}\): Using Substitution with an Exponential Function

Use substitution to evaluate\[ \int ^1_0 xe^{4x^2+3}\,dx. \nonumber \]

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\[\begin{array}{rcl}
\int ^1_0xe^{4x^2+3}\,dx & = & \dfrac{1}{8} \int ^7_3e^u\,du \\[6pt]
& = & \dfrac{1}{8}e^u\Big|^{u = 7}_{u = 3} \\[6pt]
& = & \dfrac{e^7−e^3}{8} \\[6pt]
& \approx & 134.568 \\[6pt]
\end{array} \nonumber\]

Caution: Change Limits of Integration!

When performing a substitution on a definite integral, it is critical that you immediately change the limits of integration as well.

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.

Example \(\PageIndex{8}\): Using Substitution to Evaluate a Trigonometric Integral

Use substitution to evaluate \[ \int ^{ \pi /2}_0\cos^2 \theta \,d \theta. \nonumber \]

Solution

First, use a trigonometric identity to rewrite the integral. The trigonometric identity \(\cos^2 \theta =\frac{1+\cos 2 \theta }{2}\) allows us to rewrite the integral as\[ \int ^{ \pi /2}_0\cos^2 \theta \,d \theta = \int ^{ \pi /2}_0\dfrac{1+\cos2 \theta }{2}\,d \theta . \nonumber \]Then,\[ \begin{array}{rcl}
\int ^{ \pi /2}_0\left(\dfrac{1+\cos2 \theta }{2}\right)\,d \theta & = & \int ^{ \pi /2}_0\left(\dfrac{1}{2}+\dfrac{1}{2}\cos 2 \theta \right)\,d \theta \\[6pt]
& = & \dfrac{1}{2} \int ^{ \pi /2}_0\,d \theta + \int ^{ \pi /2}_0\cos2 \theta \,d \theta . \\[6pt]
\end{array} \nonumber\]We can evaluate the first integral as it is. Still, we need to make a substitution to evaluate the second integral. Let \(u=2 \theta \). Then, \(du=2\,d \theta \), or \(\frac{1}{2}\,du=d \theta \). Also, when \( \theta =0,\,u=0\), and when \( \theta = \pi /2,\,u= \pi \). Expressing the second integral in terms of \(u\), we have\[ \begin{array}{rcl}
\dfrac{1}{2} \int ^{ \pi /2}_0\,d \theta +\dfrac{1}{2} \int ^{ \pi /2}_0 \cos 2 \theta \,d \theta & = & \dfrac{1}{2} \int ^{ \pi /2}_0\,d \theta +\dfrac{1}{2}\left(\dfrac{1}{2}\right) \int ^ \pi _0 \cos u \,du \\[6pt]
& = & \dfrac{ \theta }{2}\,\bigg|^{ \theta = \pi /2}_{ \theta =0}+\dfrac{1}{4}\sin u\,\bigg|^{u= \theta }_{u=0} \\[6pt]
& = & \left(\dfrac{ \pi }{4}−0\right)+(0−0) \\[6pt]
& = & \dfrac{ \pi }{4} \\[6pt]
\end{array} \nonumber\]