# 5.5: Substitution

- Page ID
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)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

\[ ∫(x^2−3)^3 2x \, dx. \label{eq1}\]

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,

\[ \begin{align*} ∫f[g(x)]g′(x)\,dx &=∫f(u)\,du \\[5pt] &=F(u)+C \\[5pt] &= F(g(x))+C \end{align*}\]

Proof

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

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

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 \(u=x^2−3\) and then \(du=2x\,dx\).

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

\[ ∫(x^2−3)^3(2x\,dx)=∫u^3\,du.\]

Using the power rule for integrals, we have

\[ ∫u^3\,du=\dfrac{u^4}{4}+C.\]

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

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

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

Problem-Solving Strategy: Integration by Substitution

- 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.
- Substitute \(u=g(x)\) and \(du=g′(x)dx.\) into the integral.
- 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 \(u\).
- Evaluate the integral in terms of \(u\).
- 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

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

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 \(u\):

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

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

## Integration of Trigonometric Functions

The next three examples will help fill in some missing pieces of our antiderivative knowledge. We know the antiderivatives of the sine and cosine functions; what about the other standard functions tangent, cotangent, secant and cosecant? We discover these next.

Example \(\PageIndex{3}\): Integration by substitution: antiderivatives of \(\tan x\)

Evaluate \(\int \tan x\ dx.\)

**Solution**

The previous paragraph established that we did not know the antiderivatives of tangent, hence we must assume that we have learned something in this section that can help us evaluate this indefinite integral.

Rewrite \(\tan x\) as \(\sin x/\cos x\). While the presence of a composition of functions may not be immediately obvious, recognize that \(\cos x\) is "inside" the \(1/x\) function. Therefore, we see if setting \(u = \cos x\) returns usable results. We have that \(du = -\sin x\ dx\), hence \(-du = \sin x\ dx\). We can integrate:

\[\begin{align}\int \tan x \ dx &= \int \frac{\sin x}{\cos x}\ dx \\ &= \int \frac1{\underbrace{\cos x}_u}\underbrace{\sin x\ dx}_{-du} \\ &= \int \frac {-1}u \ du\\ &= -\ln |u| + C \\ &= -\ln |\cos x| + C.\end{align}\]

Some texts prefer to bring the \(-1\) inside the logarithm as a power of \(\cos x\), as in:

\[\begin{align} -\ln |\cos x| + C &= \ln |(\cos x)^{-1}| + C\\ &= \ln \left| \frac{1}{\cos x}\right| + C\\&= \ln |\sec x| + C.\end{align}\]

Thus the result they give is \(\int \tan x \ dx = \ln|\sec x| + C\). These two answers are equivalent.

Example \(\PageIndex{4}\): Integrating by substitution: antiderivatives of \(\sec x\)

Evaluate \(\int \sec x\ dx\).

**Solution**

This example employs a wonderful trick: multiply the integrand by "1" so that we see how to integrate more clearly. In this case, we write "1" as

$$1 = \frac{\sec x + \tan x}{\sec x + \tan x}.$$

This may seem like it came out of left field, but it works beautifully. Consider:

\[\begin{align} \int \sec x\ dx &= \int \sec x\cdot \frac{\sec x + \tan x}{\sec x + \tan x}\ dx \\ &= \int \frac{\sec^2 x + \sec x\tan x}{\sec x + \tan x}\ dx.\end{align}\]

Now let \(u = \sec x+\tan x\); this means \(du = (\sec x\tan x+ \sec^2 x)\ dx\), which is our numerator. Thus:

\[\begin{align} &= \int \frac{du}{u} \\ &= \ln |u| + C \\ &= \ln |\sec x+\tan x| + C.\end{align}\]

We can use similar techniques to those used in Examples \(\PageIndex{3}\) and \(\PageIndex{4}\) to find antiderivatives of \(\cot x\) and \(\csc x\) (which the reader can explore in the exercises.) We summarize our results here.

Theorem \(\PageIndex{1}\): Antiderivatives of Trigonometric Functions

- \( \int \sin x \ dx = -\cos x +C\)
- \(\int \cos x\ dx = \sin x + C\)
- \(\int \tan x\ dx = -\ln|\cos x|+C\)
- \( \int \csc x \ dx = -\ln|\csc x+\cot x| +C\)
- \(\int \sec x\ dx = \ln|\sec x+\tan x| + C\)
- \( \int \cot x\ dx = \ln|\sin x|+C\)

Note

Although we have these formulas, at this point it will be best to show supporting work for all but the first two (the antiderivatives of sine and cosine).

Example \(\PageIndex{5}\): Using Substitution with Integrals of Trigonometric Functions

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

**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

\[ ∫\dfrac{\sin t}{\cos^3t}\,dt=−∫\dfrac{du}{u^3}.\nonumber\]

Evaluating the integral, we get

\[ −∫\dfrac{du}{u^3}=−∫u^{−3}\,du=−\left(−\dfrac{1}{2}\right)u^{−2}+C.\nonumber\]

Putting the answer back in terms of t, we get

\[ ∫\dfrac{\sin t}{\cos^3t}\,dt=\dfrac{1}{2u^2}+C=\dfrac{1}{2\cos^2t}+C.\nonumber\]

Exercise \(\PageIndex{3}\)

Use substitution to evaluate the integral \( \displaystyle ∫\dfrac{\cos t}{\sin^2t}\,dt.\)

**Hint**-
Use the process from Example \(\PageIndex{5}\) to solve the problem.

**Answer**-
\(\displaystyle ∫\dfrac{\cos t}{\sin^2t}\,dt = −\dfrac{1}{\sin t}+C\)

Exercise \(\PageIndex{4}\)

Use substitution to evaluate the indefinite integral \(\displaystyle ∫\cos^3t\sin t\,dt. \)

**Hint**-
Use the process from Example \(\PageIndex{5}\) to solve the problem.

**Answer**-
\(\displaystyle ∫\cos^3t\sin t\,dt = −\dfrac{\cos^4t}{4}+C \)

We explore one more common trigonometric integral.

Example \(\PageIndex{6}\): Integration by substitution: powers of \(\cos x\) and \(\sin x\)

Evaluate \(\int \cos^2x\ dx\).

**Solution**

We have a composition of functions as \(\cos^2x = \big(\cos x\big)^2\).

However, setting \(u = \cos x\) means \(du = -\sin x\ dx\), which we do not have in the integral. Another technique is needed.

The process we'll employ is to use a Power Reducing formula for \(\cos^2x\) (perhaps consult the back of this text for this formula), which states

$$\cos ^2x = \frac{1+\cos(2x)}{2}.$$

The right hand side of this equation is not difficult to integrate. We have:

\[\begin{align} \int \cos^2x\ dx &= \int \frac{1+\cos(2x)}2\ dx \\ &= \int \left( \frac12 + \frac12\cos(2x)\right)\ dx. \end{align} \]

Integrating, we obtain:

\[\begin{align} &= \frac12x + \frac12\frac{\sin(2x)}{2} + C\\&= \frac12x + \frac{\sin(2x)}4 + C.\end{align}\]

We'll make significant use of this power--reducing technique in future sections.

## A u-Substitution with a Twist

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 \(\PageIndex{7}\): Finding an Antiderivative Using u-Substitution

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

**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 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*:

\[ ∫\dfrac{x}{\sqrt{x−1}}\,dx=∫\dfrac{u+1}{\sqrt{u}}\,du=∫\sqrt{u}+\dfrac{1}{\sqrt{u}}\,du=∫(u^{1/2}+u^{−1/2})\,du.\nonumber\]

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

\[ \begin{align*} ∫(u^{1/2}+u^{−1/2})\,du &=\dfrac{2}{3}u^{3/2}+2u^{1/2}+C \\[5pt] &= \dfrac{2}{3}(x−1)^{3/2}+2(x−1)^{1/2}+C \\[5pt] &= (x−1)^{1/2}\left[\dfrac{2}{3}(x−1)+2\right]+C \\[5pt] &= (x−1)^{1/2}\left(\dfrac{2}{3}x−\dfrac{2}{3}+\dfrac{6}{3}\right) \\[5pt] &= (x−1)^{1/2}\left(\dfrac{2}{3}x+\dfrac{4}{3}\right) \\[5pt] &= \dfrac{2}{3}(x−1)^{1/2}(x+2)+C. \end{align*}\]

## 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,

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

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

\[\begin{align*} ∫^b_af[g(x)]g′(x)\,dx &= F(g(x))\bigg|^{x=b}_{x=a} \\[5pt] &=F(g(b))−F(g(a)) \\[5pt] &= F(u) \bigg|^{u=g(b)}_{u=g(a)} \\[5pt] &=∫^{g(b)}_{g(a)}f(u)\,du \end{align*}\]

and we have the desired result.

Example \(\PageIndex{8}\): 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 \(du=6x^2dx\). Since the original function includes one factor of \(x^2\) and \(du=6x^2dx\), multiply both sides of the \(du\) equation by \(1/6.\) Then,

\[ \begin{align*} du &=6x^2\,dx \\[5pt] \text{becomes}\quad \dfrac{1}{6}du &=x^2\,dx. \end{align*}\]

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

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

Exercise \(\PageIndex{5}\)

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

**Hint**-
Use the steps from Example \(\PageIndex{8}\) to solve the problem.

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

Exercise \(\PageIndex{6}\)

Use substitution to evaluate \(\displaystyle ∫^1_0x^2 \cos \left(\dfrac{π}{2}x^3\right)\,dx. \)

**Hint**-
Use the process from Example \(\PageIndex{8}\) to solve the problem.

**Answer**-
\(\displaystyle ∫^1_0x^2 \cos \left(\dfrac{π}{2}x^3\right)\,dx = \dfrac{2}{3π}≈0.2122\)

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 \(\PageIndex{9}\).

Example \(\PageIndex{9}\): Using Substitution to Evaluate a Trigonometric Integral

Use substitution to evaluate \[∫^{π/2}_0\cos^2θ\,dθ. \nonumber \]

**Solution**

Let us first use a trigonometric identity to rewrite the integral. The trig identity \(\cos^2θ=\dfrac{1+\cos 2θ}{2}\) allows us to rewrite the integral as

\[∫^{π/2}_0\cos^2θ\,dθ=∫^{π/2}_0\dfrac{1+\cos2θ}{2}\,dθ. \nonumber\]

Then,

\[ \begin{align*} ∫^{π/2}_0\left(\dfrac{1+\cos2θ}{2}\right)\,dθ &=∫^{π/2}_0\left(\dfrac{1}{2}+\dfrac{1}{2}\cos 2θ\right)\,dθ \\[5pt] &=\dfrac{1}{2}∫^{π/2}_0\,dθ+∫^{π/2}_0\cos2θ\,dθ. \end{align*}\]

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 \(\dfrac{1}{2}\,du=dθ\). Also, when \(θ=0,\,u=0,\) and when \(θ=π/2,\,u=π.\) Expressing the second integral in terms of \(u\), we have

\[ \begin{align*}\dfrac{1}{2}∫^{π/2}_0\,dθ+\dfrac{1}{2}∫^{π/2}_0 \cos 2θ\,dθ &=\dfrac{1}{2}∫^{π/2}_0\,dθ+\dfrac{1}{2}\left(\dfrac{1}{2}\right)∫^π_0 \cos u \,du \\[5pt] &=\dfrac{θ}{2}\,\bigg|^{θ=π/2}_{θ=0}+\dfrac{1}{4}\sin u\,\bigg|^{u=\pi}_{u=0} \\[5pt] &=\left(\dfrac{π}{4}−0\right)+(0−0)=\dfrac{π}{4} \end{align*}\]

## 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

## Contributors

Gilbert Strang (MIT) and Edwin “Jed” Herman (Harvey Mudd) with many contributing authors. This content by OpenStax is licensed with a CC-BY-SA-NC 4.0 license. Download for free at http://cnx.org.

- Apex Calculus: the subsection on integrating trigonometric functions is mostly from Apex Calculus, Section 6.1.
- Edited by Paul Seeburger (Monroe Community College)