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, that helps finding antiderivatives. Specifically, this method allows to 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 \ds f\left(g\left(x\right)\right){g}^{\prime }(x)\,dx . For example, in the integral \ds \int {\left({x}^{2}-3\right)}^{3}(2x)\,dx , we have \ds f\left(u\right)={u}^{3},g\left(x\right)={x}^{2}-3, and \ds g ' \left(x\right)=2x. Then,

\ds f\left(g\left(x\right)\right){g}^{\prime }(x)={\left({x}^{2}-3\right)}^{3}\left(2x\right),

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 for Indefinite Integrals

Let \ds u=g\left(x\right), where \ds {g}^{\prime }(x) is continuous, let \ds f\left(x\right) be continuous over the range of g, and let \ds F\left(x\right) be an antiderivative of \ds f\left(x\right). Then,

\ds \begin{array}{cc}\ds \int f\left(g\left(x\right)\right){g}^{\prime }(x)\,dx \hfill &\ds =\int f\left(u\right)du\hfill \\[5mm] &\ds =F\left(u\right)+C\hfill \\[5mm] &\ds =F\left(g\left(x\right)\right)+C.\hfill \end{array}

Proof

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

\ds \begin{array}{cc}\ds \frac{d}{\,dx }\Big(F\left(g\left(x\right)\right)\Big)\hfill &\ds ={F}^{\prime }(g\left(x\right)){g}^{\prime }(x)\hfill \\[5mm] &\ds =f\left(g\left(x\right)\right){g}^{\prime }(x).\hfill \end{array}

This means that F(g(x)) is an antiderivative of f\left(g\left(x\right)\right){g}^{\prime }(x) and hence

\ds \int f\left(g\left(x\right)\right){g}^{\prime }(x)\,dx =F\left(g\left(x\right)\right)+C.

Since u=g(x) and F is an antiderivative of f, we have that F(g(x))+C=F(u)+C=\ds\int f(u)\,du, which completes the proof.

In practice, when we perform a substitution, we might not know what F is right away, and we use a change of variable u=g(x) to replace \ds \int f\left(g\left(x\right)\right){g}^{\prime }(x)\,dx with a simpler integral \ds\int f(u)\, du that we then evaluate. This step can be viewed as substituting \ds u=g\left(x\right) and \ds du=g ' \left(x\right)\,dx. For us, dx and du are just parts of integral notation. However, in a more rigorous mathematical analysis, they are called differentials that have a proper definition. Without going into details, we notice that if u=g(x), then the formula \ds du=g ' \left(x\right)\,dx agrees with the Leibniz’s notation \dfrac{du}{dx} for the derivative u' of u with respect to x. Indeed, \dfrac{du}{dx}=u'=g'(x), and the above formula can be obtained by “multiplying” both sides by dx. This, one more time, confirms a deep connection between the differential calculus and the integral calculus.

Returning to the problem we looked at originally, we let \ds u={x}^{2}-3 and then \ds du=2x\,dx . Rewriting the integral in terms of u, we obtain:

\ds {\int \underset{u}{\underbrace{\left({x}^{2}-3\right)}}^3}\underset{du}{\underbrace{\left(2x\,dx \right)}}=\int {u}^{3}du.

Using the power rule for integrals, we have

\ds \int {u}^{3}du=\frac{{u}^{4}}{4}+C.

Substituting the original expression for x back into the solution, we get

\ds \frac{{u}^{4}}{4}+C=\frac{{\left({x}^{2}-3\right)}^{4}}{4}+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 \ds g\left(x\right) within the integrand to set equal to u. Quite often, we select \ds g\left(x\right) so that \ds {g}^{\prime }(x) is also part of the integrand.
  2. Substitute \ds u=g\left(x\right) and \ds du={g}^{\prime }(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. Replace u with g(x) to write the result in terms of x.

Using Substitution to Evaluate an Indefinite Integral

Use substitution to evaluate \ds \int 6x{\left(3{x}^{2}+4\right)}^{4}\,dx .

Solution

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

\ds \int 6x{\left(3{x}^{2}+4\right)}^{4}\,dx =\int {u}^{4}du.

Now we can evaluate the integral with respect to u and then return to the variable x:

\ds \begin{array}{ll}\ds \int {u}^{4}du\hfill &\ds =\frac{{u}^{5}}{5}+C\hfill \\[5mm] &\ds =\frac{{\left(3{x}^{2}+4\right)}^{5}}{5}+C.\hfill \end{array}

Analysis

As usual, we can check our answer by taking the derivative of the result of integration to see if we really obtain the integrand. We have
\ds \begin{array}{ll}\ds \\[5mm] \hfill \left(\frac{{\left(3{x}^{2}+4\right)}^{5}}{5}+C\right)'&\ds =\left(\frac{1}{5}\right)5{\left(3{x}^{2}+4\right)}^{4}(3x^2+4)'\hfill \\[5mm] &\ds ={\left(3{x}^{2}+4\right)}^{4}(6x).\hfill \end{array}

This is exactly the expression in the integrand we started with, which means that our answer is correct.

Use substitution to evaluate \ds \int 3{x}^{2}{\left({x}^{3}-3\right)}^{2}\,dx .

Answer

\ds \int 3{x}^{2}{\left({x}^{3}-3\right)}^{2}\,dx =\frac{1}{3}{\left({x}^{3}-3\right)}^{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.

Using Substitution with Alteration

Use substitution to evaluate the indefinite integral \ds \int z\sqrt{{z}^{2}-5}\,dz.

Solution

Let \ds u={z}^{2}-5 and \ds du=2z\phantom{\rule{0.2em}{0ex}}dz. Now we have a problem because \ds du=2z\phantom{\rule{0.2em}{0ex}}dz and the original expression has only \ds z\phantom{\rule{0.2em}{0ex}}dz. We have to alter our expression for du or the integral in u will be twice as large as it should be. Multiplying both sides of the du equation by \ds \frac{1}{2} solves this problem:

\ds \begin{array}{ll}\ds \hfill \ds\frac{1}{2}du&\ds =\frac{1}{2}\left(2z\right)dz=z\phantom{\rule{0.2em}{0ex}}dz.\hfill \end{array}

We can then write the integral in terms of u as follows:

\ds\int z\sqrt{z^2-5}\,dz=\int \sqrt{u}\cdot\frac12\, du=\frac12\int\sqrt u\, du

Integrating the expression in u by rewriting \sqrt u as u^{1\text{/}2} and using the power rule for integrals, we obtain:

\ds \begin{array}{ll}\ds \frac12\int\sqrt u\, du\hfill &\ds =\frac{1}{2}\int {u}^{1\text{/}2}\,du\hfill \\[5mm]&\ds =\left(\frac{1}{2}\right)\frac{{u}^{3\text{/}2}}{\frac{3}{2}}+C\hfill  \\[5mm] &\ds =\left(\frac{1}{2}\right)\left(\frac{2}{3}\right){u}^{3\text{/}2}+C\hfill \\[5mm] &\ds =\frac{1}{3}{u}^{3\text{/}2}+C\hfill \\[5mm] &\ds =\frac{1}{3}{\left({z}^{2}-5\right)}^{3\text{/}2}+C.\hfill \end{array}

Use substitution to find the antiderivative of \ds \int {x}^{2}{\left({x}^{3}+5\right)}^{9}\,dx .

Answer

\ds \frac{{\left({x}^{3}+5\right)}^{10}}{30}+C

Hint

Multiply the du equation by \ds \frac{1}{3}.

Using Substitution with Integrals of Trigonometric Functions

Use substitution to evaluate the integral \ds \int \frac{\text{sin}\phantom{\rule{0.1em}{0ex}}(t)}{{\text{cos}}^{3}(t)}\,dt.

Solution

First, we note that we can rewrite the integral as \ds \int \frac1{\cos^3(t)}\cdot\sin(t)\,dt, that is, we can combine the numerator with dt. Since the denominator is a power of \cos(t) and the derivative of \ds \text{cos}\phantom{\rule{0.1em}{0ex}}(t) is \ds - \text{sin}\phantom{\rule{0.1em}{0ex}}(t), it is natural to take \ds u=\text{cos}\phantom{\rule{0.1em}{0ex}}(t). Then \ds du= - \text{sin}\phantom{\rule{0.1em}{0ex}}(t)dt, and hence \sin(t)\,dt=-du. Substituting into the integral, we have

\ds \int \frac{\text{sin}\phantom{\rule{0.1em}{0ex}}(t)}{{\text{cos}}^{3}(t)}\,dt=\int \frac1{\cos^3(t)}\cdot\sin(t)\,dt=\int \frac{1}{{u}^{3}}(-du)= - \int \frac{1}{{u}^{3}}\,du.

Evaluating the integral, we get

\ds \begin{array}{ll}\ds - \int \frac{1}{{u}^{3}}\,du\hfill &\ds = - \int {u}^{-3}\,du\hfill \\[5mm] &\ds = - \left(-\frac{1}{2}\right){u}^{-2}+C=\frac12u^{-2}+C.\hfill \end{array}

Putting the answer back in terms of t, we get

\ds \begin{array}{cc}\ds \int \frac{\text{sin}\phantom{\rule{0.1em}{0ex}}(t)}{{\text{cos}}^{3}(t)}\,dt\hfill &\ds =\frac{1}{2} u^{-2}+C\hfill \\[5mm] &\ds =\frac{1}{2}\big(\cos(t)\big)^{-2}+C.\hfill \end{array}

Use substitution to evaluate the integral \ds \int {\text{cos}\phantom{\rule{0.1em}{0ex}}(t)}{2^{\text{sin}(t)}}dt.

Answer

\ds \frac{2^{\sin(t)}}{\ln(2)}+C

In a similar way, one can use a sibstitution to derive a formula for \ds\int \tan(x)\,dx. Indeed, rewriting \tan(x) as \dfrac{\sin(x)}{\cos(x)} and taking u=\cos(x), we get that du=-\sin(x)\,dx and hence

\ds \begin{array}{cc}\ds \int \tan(x)\,dx &\ds=\int\dfrac{\sin(x)}{\cos(x)}\,dx=\int\dfrac{1}{\cos(x)}\sin(x)\,dx\\[5mm] &\ds=\int \frac1u(-du)=-\int \frac1u\,du=-\ln|u|+C.\hfill\end{array}

Going back to the variable x, we get

\ds \begin{array}{cc}\ds \int \tan(x)\,dx &\ds=-\ln|u|+C=-\ln|\cos(x)|+C.\hfill\end{array}

The answer is also commonly given in the form \ln|\sec(x)|+C, which is, in fact, the same since

\ds\sec(x)=\frac1{\cos(x)}=\big(\cos(x)\big)^{-1},

and hence by the properties of logarithms (\ln(a^p)=p\ln(a)),

\ln|\sec(x)|=\ln\left|\big(\cos(x)\big)^{-1}\right|=-\ln|\cos(x)|.

Likewise, a substituion of u=\sin(x), allows to establish the formula

\ds\int \cot(x)\,dx=\int \frac{\cos(x)}{\sin(x)}\,dx=\ln|\sin(x)|+C.

Other trigonometric integrals that can be evaluated using a substitution are \ds\int \sec(x)\, dx and \ds\int \csc (x)\, dx. However, their evaluation is not straightforward and requires using a trick. To find \ds\int \sec(x)\, dx, we multiply and divide the integrand by \big(\sec(x)+\tan(x)\big):

\ds\int \sec(x)\, dx=\int\frac{\sec(x)\big(\sec(x)+\tan(x)\big)}{\sec(x)+\tan(x)}\,dx=\int\frac{\sec^2(x)+\sec(x)\tan(x)}{\sec(x)+\tan(x)}.

As crazy as it might seem, it helps because the derivative of the denominator is precisely equal to the numerator. That is, if we set u=\sec(x)+\tan(x), then du=\big(\sec(x)\tan(x)+\sec^2(x)\big)\,dx, and hence

\begin{array}{ll}\ds\int \sec(x)\, dx&\ds=\int\frac{\sec^2(x)+\sec(x)\tan(x)}{\sec(x)+\tan(x)}=\int \frac{du}{u}=\ln|u|+C\\[5mm]&\ds=\ln|\sec(x)+\tan(x)|+C.\end{array}

To evaluate \ds\int \csc(x)\, dx, one can either multiply the integrand by \big(\csc(x)+\cot(x)\big) or by \big(\csc(x)-\cot(x)\big), and then make the corresponding change of variable. This results in seemingly different answers of -\ln|\csc(x)+\cot(x)|+C and \ln|\csc(x)-\cot(x)|+C. To see that they are, in fact, the same, one needs to use the trigonometric formula \csc^2(x)-\cot^2(x)=1 together with the rules of logarithms (\ln(a)+\ln(b)=\ln(ab) and \ln(1)=0):

\begin{array}{ll}\ds\ln|\csc(x)+\cot(x)|+\ln|\csc(x)-\cot(x)|&\ds=\ln\left|\big(\csc(x)+\cot(x)\big)\big(\csc(x)-\cot(x)\big)\right|\\[5mm]&\ds=\ln\left|\csc^2(x)-\cot^2(x)\right|=\ln(1)=0, \end{array}

and hence \ln|\csc(x)-\cot(x)|=-\ln|\csc(x)+\cot(x)|.

Because of the significance of the above formulas, especially in Section 3.2 Trigonometric Integrals, we gather all of them in a single statement below.

Basic Trigonometric Integrals That Use a Substitution

\begin{array}{ll} \ds\int \tan(x)\,dx &\ds=\ds-\ln|\cos(x)|+C=\ln|\sec(x)|+C\\[5mm] \ds\int \cot(x)\,dx &\ds=\ds\ln|\sin(x)|+C\\[5mm] \ds \int \sec(x)\,dx &\ds=\ds\ln|\sec(x)+\tan(x)|+C\\[5mm] \ds \int \csc(x)\,dx &\ds=\ds-\ln|\csc(x)+\cot(x)|+C=\ln|\csc(x)-\cot(x)|+C\end{array}

Sometimes we need to manipulate an integral in ways that are more complicated than just multiplying or dividing by a constant. After the substitution is completed, u should be the only variable in the integrand. In some cases, to achieve this, we need to solve for the original variable in terms of u. We illustrate how it works in the next example.

Evaluating an Indefinite Integral Using a Substitution

Use substitution to find the antiderivative of \ds \int \frac{x}{\sqrt{x-1}}\,dx .

Solution

If we let \ds u=x-1, then \ds 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 to complete the substitution. If \ds u=x-1, then \ds x=u+1. Now we can rewrite the integral in terms of u:

\ds \begin{array}{ll}\ds \int \frac{x}{\sqrt{x-1}}\,dx \hfill &\ds =\int \frac{u+1}{\sqrt{u}}du\hfill \\[5mm] &\ds =\int \left(\sqrt{u}+\frac{1}{\sqrt{u}}\right)du\hfill \\[5mm] &\ds =\int \left({u}^{1\text{/}2}+{u}^{-1\text{/}2}\right)du.\hfill \end{array}

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

\ds \begin{array}{cc}\ds \int \left({u}^{1\text{/}2}+{u}^{-1\text{/}2}\right)du\hfill &\ds =\frac{2}{3}{u}^{3\text{/}2}+2{u}^{1\text{/}2}+C\hfill \\[5mm]  &\ds =\frac{2}{3}{\left(x-1\right)}^{3\text{/}2}+2{\left(x-1\right)}^{1\text{/}2}+C\hfill \end{array}

Use substitution to evaluate the indefinite integral \ds \int t(1-2t)^7dt.

Answer

\ds \frac{(1-2t)^9}{36}-\frac{(1-2t)^8}{32}+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 for Definite Integrals

Let \ds u=g\left(x\right), \ds {g}'(x) be continuous over an interval \ds \left[a,b\right], and let f be continuous over the range of \ds u=g\left(x\right). Then,

\ds \int\limits_{a}^{b}f\left(g\left(x\right)\right){g}^{\prime }(x)\,dx =\int\limits_{g\left(a\right)}^{g\left(b\right)}f\left(u\right)du.

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

\ds \int f\left(g\left(x\right)\right){g}^{\prime }(x)\,dx =F\left(g\left(x\right)\right)+C.

Then

\ds \begin{array}{cc}\ds \int\limits_{a}^{b}f\left(g\left(x\right)\right){g}^{\prime }(x)\,dx \hfill &\ds ={F\left(g\left(x\right)\right)}\Big|_{x=a}^{x=b}\hfill \\[5mm] &\ds =F\left(g\left(b\right)\right)-F\left(g\left(a\right)\right)\hfill \\[5mm] &\ds ={F\left(u\right)}\Big|_{u=g\left(a\right)}^{u=g\left(b\right)}\hfill \\[5mm] &\ds =\int\limits_{g\left(a\right)}^{g\left(b\right)}f\left(u\right)du,\hfill \end{array}

and we have the desired result.

Using Substitution to Evaluate a Definite Integral

Use substitution to evaluate \ds \int\limits_{0}^{1} (x^3+1)e^{x^4+4x} \,dx .

Solution

Take u=x^4+4x. Then du=(4x^3+4)\,dx =4(x^3+1)\,dx and hence \ds(x^3+1)\,dx =\frac14du. To adjust the bounds of integration, note that x=0 corresponds to u=0^4+4\cdot0=0 and x=1 corresponds to u=1^4+4\cdot1=5. We then obtain

\begin{array}{ll}\ds \int\limits_0^1 (x^3+1) e^{x^4+4x} \,dx \hfill&\ds =\int\limits_1^5 \frac14 e^u du\hfill\\[5mm]&\ds =\frac14 e^u\Big|_1^5\hfill\\[5mm]&\ds =\frac{e^5-e}{4}.\end{array}

Use substitution to evaluate \ds \int\limits_{1}^{e}\frac{\ln(x)}{x}\,dx .

Answer

\ds \frac12

Hint

Take u=\ln(x).

Using Substitution with a Trigonometric Function

Use substitution to evaluate \ds \int\limits_{1\text{/}2}^{1}\frac{\sin\left({\frac1x}\right)}{x^2}\,dx .

Solution

Let \ds u=\frac1x=x^{-1}. Then, \ds \ds du=(-1)x^{-2}\,dx =-\frac1{x^2}\,dx , and we have that \ds\frac1{x^2}\,dx =-du. To adjust the limits of integration, we note that when \ds \ds x=\frac12,u=2, and when \ds x=1,u=1. So our substitution gives

\ds \begin{array}{cc}\ds \int\limits_{1\text{/}2}^{1}\frac{\sin\left({\frac1x}\right)}{x^2}\,dx \hfill &\ds =\int\limits_{2}^{1}\sin(u)\cdot (-1)du\hfill \\[5mm] &\ds =(\cos(u)){}\Big|_{2}^{1}\hfill \\[5mm] &\ds =\cos(1)-\cos(2)\hfill \end{array}

Analysis

Note that in the integral in terms of u that we obtained, the lower limit of integration was bigger than the upper limit. This happens a lot when substitution is used and we do not need to worry about that or fix anything – just proceed with the regular integration process.

Use substitution to evaluate \ds \int\limits_{\pi^2\text{/}16}^{\pi^2\text{/}9}\frac{\sec^2({\sqrt x})}{\sqrt x}\,dx .

Answer

\ds 2\sqrt3-2

Hint

Take u=\sqrt{x}.

Evaluating a Definite Integral using Substitution

Use substitution to evaluate \ds \int\limits_{0}^{1}{x}^{5}{\left(1-{x}^{3}\right)}^{4}\,dx .

Solution

Let \ds u=1-{x}^{3}, and then \ds du=-3{x}^{2}\,dx . Since the original function has x^5, we need to “split” x^2 from it to combine with \,dx, that is, we write x^5=x^2\cdot x^3, and then x^5 \,dx =x^3\cdot x^2\,dx. Multiplying both sides of  \ds du=-3{x}^{2}\,dx by \ds-\frac13, we obtain that \ds x^2\,dx =-\frac13 du, and so to proceed with the chosen substitution, we need to express x^3 in terms of u. Since we defined u as \ds u=1-{x}^{3}, we have that x^3=1-u, and hence \ds x^5\,dx =x^3\cdot x^2\,dx =(1-u)\left(-\frac13\right)du. Finally, to adjust the limits of integration, note that when \ds x=0,u=1-\left(0\right)=1, and when \ds x=1,u=1-1=0. Then

\ds \int\limits_{0}^{1}{x}^{5}{\left(1-{x}^{3}\right)}^{2}\,dx =\int\limits_{1}^{0}{u}^{4}(1-u)\left(-\frac13\right)du=-\frac13\int\limits_1^0 u^4(1-u)du.

Evaluating the integral, we obtain

\ds \begin{array}{ll}\ds -\frac13\int\limits_{1}^{0}u^4(1-u)du\hfill &\ds =-\frac13\int\limits_1^0(u^4-u^5)du\hfill\\[5mm]&\ds =\left(-\frac{1}{3}\right)\left(\frac{{u}^{5}}{5}-\frac{u^6}6\right){}\Big|_{1}^{0}\hfill \\[5mm] &\ds =-\frac{1}{3}\left[(0-0)-\left(\frac15-\frac16\right)\right]\hfill \\[5mm] &\ds =\frac{1}{90}.\hfill \end{array}

Use substitution to evaluate the definite integral \ds \int\limits_{-1}^{0}\frac{y^3}{y^2+1}dy.

Answer

\ds \frac{\ln(2)-1}{2}

Hint

Take u=y^2+1.

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
    \ds \int f\left(g\left(x\right)\right){g}^{\prime }(x)\,dx =\int f\left(u\right)du=F\left(u\right)+C=F\left(g\left(x\right)\right)+C
  • Substitution with Definite Integrals
    \ds \int\limits_{a}^{b}f\left(g\left(x\right)\right)g ' \left(x\right)\,dx =\int\limits_{g\left(a\right)}^{g\left(b\right)}f\left(u\right)du

Exercises

In the following exercises, evaluate the indefinite integral using the indicated substitution.

1. \ds \int {\left(x+1\right)}^{4}\,dx ;\ u=x+1

Answer

\ds \frac{1}{5}{\left(x+1\right)}^{5}+C

2. \ds \int \frac2{2x-1}\,dx ;\ u=2x-1

3. \ds \int \frac{x}{(x^2+3)^7}\,dx ;\ u=x^2+3

Answer

\ds -\frac{1}{12{\left(x^2+3\right)}^{6}}+C

4. \ds \int \sqrt{5-3x}\,dx ;\ u=5-3x

5. \ds \int (x^2+1) \sin(x^3+3x)\,dx ;\ u=x^3+3x

Answer

\ds -\frac13\cos(x^3+3x)+C

6. \ds \int \frac{2-(\ln(x))^3}{x}\,dx ;\ u=\ln(x)

7. \ds \int \frac{2^{\sqrt x}}{\sqrt x}\,dx ;\ u=\sqrt{x}

Answer

\ds \frac{2}{\ln(2)}2^{\sqrt{x}}+C

8. \ds \int \frac{x+\cos(x)}{2\sin(x)+x^2}\,dx ;\ u=2\sin(x)+x^2

9. \ds \int x\sqrt{x+1}\,dx ;\ u=x+1

Answer

\ds \frac25(x+1)^{5\text{/}2}-\frac23(x+1)^{3\text{/}2}+C

10. \ds \int \frac{x}{\sqrt{2x-3}}\,dx ;\ u=2x-3

In the following exercises, use a suitable change of variables to evaluate the indefinite integral.

11. \ds \int e^{-2x}\,dx

Answer

\ds -\frac12 e^{-2x}+C

12. \ds \int \frac{1}{3t+1}dt

13. \ds \int {\left(1-5x\right)}^{99}\,dx

Answer

\ds -\,\frac{{\left(1-5x\right)}^{100}}{500}+C

14. \ds \int \frac{t}{{\left(1-{t}^{2}\right)}^{10}}dt

15. \ds \int \frac{x^5+x^2}{x^6+2x^3+3}\,dx

Answer

\ds \frac{1}{6}\ln|x^6+2x^3+3|+C

16. \ds \int \frac{1}{\sqrt{4x+7}}\,dx

17. \ds \int {\text{cos}}^{3}(\theta) \phantom{\rule{0.1em}{0ex}}\text{sin}\phantom{\rule{0.1em}{0ex}}(\theta) d\theta

Answer

\ds -\frac{{\text{cos}}^{4}\theta }{4}+C

18. \ds \int {\text{sin}}^{7}(\theta) \phantom{\rule{0.1em}{0ex}}\text{cos}\phantom{\rule{0.1em}{0ex}}(\theta) d\theta

19. \ds \int \frac{\ln(x)+\sec^2(\ln(x))}{x}dt

Answer

\ds \frac12(\ln(x))^2+\tan(\ln(x))+C

20. \ds \int \frac{e^{1\text{/}x}}{x^2}\,dx

21. \ds \int \frac{t^2}{\sqrt{1-t}}dt

Answer

\ds -2\sqrt{1-t}+\frac43(1-t)^{3\text{/}2}-\frac25(1-t)^{5\text{/}2}+C

22. \ds \int (t^3+2t)(t^2-1)^5dt

23. \ds \int \frac{{x}^{5}}{{\left({x}^{3}-3\right)}^{2}}\,dx

Answer

\ds \frac{1}{3}\ln|x^3-3|-\frac1{x^3-3}+C

24. \ds \int \sec(x)\tan(x) 3^{\sec(x)}\,dx

25. \ds \int \frac{e^{2x}+2e^x}{e^x+1}dx

Answer

\ds e^x+\ln(e^x+1)+C

26. \ds \int \theta\cdot\sin(\theta^2)\cdot\cos(\theta^2) d\theta

27. \ds \int e^{\theta}\cos(2e^{\theta})}\sqrt{\sin(2e^{\theta})+1} d\theta

Answer

\ds \frac{1}{3}\big(\sin(2e^{\theta}+1)\big)^{3\text{/}2}+C

In the following exercises, use a change of variables to evaluate the definite integral.

28. \ds \int\limits_{0}^{1} \sqrt{3t+1}dt

29. \ds \int\limits_{-1}^{0} 2^{1-t}dt

Answer

\ds \frac2{\ln(2)}

30. \ds \int\limits_{0}^{1}x(1-{x}^{2})^{20}\,dx

31. \ds \int\limits_{0}^{1}\frac{x}{\sqrt{1+{x}^{2}}}\,dx

Answer

\ds \sqrt{2}-1

32. \ds \int\limits_{1}^{e^{\pi}}\frac{\sin(\ln(t))}{t}dt

33. \ds \int\limits_{0}^{1}\frac{t^4+2}{t^5+10t+3}dt

Answer

\ds \frac15\ln\left(\frac{14}3\right)

34. \ds \int\limits_{-\pi \text{/}6}^0{\text{sec}}^{2}(\theta) \phantom{\rule{0.1em}{0ex}}\text{tan}\phantom{\rule{0.1em}{0ex}}(\theta) d\theta

35. \ds \int\limits_{0}^{\pi \text{/}4}\frac{\text{sin}\phantom{\rule{0.1em}{0ex}}(\theta) }{{\text{cos}}^{4}(\theta) }d\theta

Answer

\ds \frac{1}{3}\left(2\sqrt{2}-1\right)

36. \ds \int\limits_{0}^{\pi}\sin(\theta)e^{\cos(\theta)}d\theta

37. \ds \int\limits_{\pi \text{/}3}^{\pi \text{/}2}\cot(\theta)d\theta

(Hint: Write \cot(\theta) in terms of \sin(\theta) and \cos(\theta) and make a substitution u=\sin(\theta))

Answer

\ds -\ln\left(\frac{\sqrt3}2\right)

In the following exercises, evaluate the indefinite integral \ds \int f\left(x\right)\,dx with constant \ds C=0 using a substitution. Then, graph the function and the antiderivative over the indicated interval. If possible, estimate a value of C that would need to be added to the antiderivative to make it equal to the definite integral \ds F\left(x\right)=\int\limits_{a}^{x}f\left(t\right)dt, with a being the left endpoint of the given interval.

38. [T]\ds \int \left(2x+1\right){e}^{{x}^{2}+x-6}\,dx over \ds \left[-3,2\right]

39. [T]\ds \int \frac{\text{cos}\left(\text{ln}\left(2x\right)\right)}{x}\,dx over \ds \left[0,2\right]

Answer

 

Two graphs. The first shows the function f(x) = cos(ln(2x)) / x, which increases sharply over the approximate interval (0,.25) and then decreases gradually to the x axis. The second shows the function f(x) = sin(ln(2x)), which decreases sharply on the approximate interval (0, .25) and then increases in a gently curve into the first quadrant.


The antiderivative is \ds y=\text{sin}\left(\text{ln}\left(2x\right)\right). Since the antiderivative is not continuous at \ds x=0, one cannot find a value of C that would make \ds y=\text{sin}\left(\text{ln}\left(2x\right)\right)-C work as a definite integral.

40. [T]\ds \int \frac{3{x}^{2}+2x+1}{\sqrt{{x}^{3}+{x}^{2}+x+4}}\,dx over \ds \left[-1,2\right]

41. [T]\ds \int \frac{\text{sin}\phantom{\rule{0.1em}{0ex}}x}{{\text{cos}}^{3}x}\,dx over \ds \left[-\frac{\pi }{3},\frac{\pi }{3}\right]

Answer

 

Two graphs. The first is the function f(x) = sin(x) / cos(x)^3 over [-5pi/16, 5pi/16]. It is an increasing concave down function for values less than zero and an increasing concave up function for values greater than zero. The second is the fuction f(x) = ½ sec(x)^2 over the same interval. It is a wide, concave up curve which decreases for values less than zero and increases for values greater than zero.


The antiderivative is \ds y=\frac{1}{2}\phantom{\rule{0.05em}{0ex}}{\text{sec}}^{2}x. You should take \ds C=-2 so that \ds F\left(-\frac{\pi }{3}\right)=0.

42. [T]\ds \int \left(x+2\right){e}^{ - {x}^{2}-4x+3}\,dx over \ds \left[-5,1\right]

43. [T]\ds \int 3{x}^{2}\sqrt{2{x}^{3}+1}\,dx over \ds \left[0,1\right]

Answer

 

Two graphs. The first shows the function f(x) = 3x^2 * sqrt(2x^3 + 1). It is an increasing concave up curve starting at the origin. The second shows the function f(x) = 1/3 * (2x^3 + 1)^(1/3). It is an increasing concave up curve starting at about 0.3.


The antiderivative is \ds y=\frac{1}{3}{\left(2{x}^{3}+1\right)}^{3\text{/}2}. One should take \ds C=-\frac{1}{3}.

44. If \ds h\left(a\right)=h\left(b\right) in \ds \int\limits_{a}^{b}g ' \left(h\left(x\right)\right)h\left(x\right)\,dx , what can you say about the value of the integral?

45. Is the substitution \ds u=1-{x}^{2} in the definite integral \ds \int\limits_{0}^{2}\frac{x}{1-{x}^{2}}\,dx okay? If not, why not?

Answer

No, because the integrand is discontinuous at \ds x=1.

In the following exercises, use a change of variable to show that each definite integral is equal to zero.

46. \ds \int\limits_{0}^{\pi }{\text{cos}}^{2}\left(2\theta \right)\text{sin}\left(2\theta \right)d\theta

47. \ds \int\limits_{0}^{\sqrt{\pi }}t\phantom{\rule{0.1em}{0ex}}\text{cos}\left({t}^{2}\right)\text{sin}\left({t}^{2}\right)dt

Answer

\ds u=\text{sin}\left({t}^{2}\right); the integral becomes \ds \frac{1}{2}\int\limits_{0}^{0}udu.

48. \ds \int\limits_{0}^{1}\left(1-2t\right)dt

49. \ds \int\limits_{0}^{1}\frac{1-2t}{1+{\left(t-\frac{1}{2}\right)}^{2}}dt

Answer

\ds u=1+{\left(t-\frac{1}{2}\right)}^{2}; the integral becomes \ds - \int\limits_{5\text{/}4}^{5\text{/}4}\frac{1}{u}du.

50. \ds \int\limits_{0}^{\pi }\text{sin}\left({\left(t-\frac{\pi }{2}\right)}^{3}\right)\text{cos}\left(t-\frac{\pi }{2}\right)dt

51. \ds \int\limits_{0}^{2}\left(1-t\right)\text{cos}\left(\pi t\right)dt

Answer

\ds u=1-t; the integral becomes

\ds \begin{array}{ll}\ds\int\limits_{1}^{-1}u\phantom{\rule{0.1em}{0ex}}\text{cos}\left(\pi \left(1-u\right)\right)(-1)du\hfill&\ds =\int\limits_{1}^{-1}u(-\cos(u))(-1)du\hfill \\[5mm] &\ds=\int\limits_{1}^{-1}u\phantom{\rule{0.1em}{0ex}}\text{cos}\phantom{\rule{0.1em}{0ex}}(u)du\hfill=-\int\limits_{-1}^{1}u\phantom{\rule{0.1em}{0ex}}\text{cos}\phantom{\rule{0.1em}{0ex}}(u)\,du=0\hfill \end{array}
since the integrand is odd.

52. \ds \int\limits_{\pi \text{/}4}^{3\pi \text{/}4}{\text{sin}}^{2}(t)\phantom{\rule{0.1em}{0ex}}\text{cos}\phantom{\rule{0.1em}{0ex}}(t)dt

53. Show that the average value of \ds f\left(x\right) over an interval \ds \left[a,b\right] is the same as the average value of \ds f\left(cx\right) over the interval \ds \left[\frac{a}{c},\frac{b}{c}\right] for \ds c\symbol{"3E}0.

Answer

Setting \ds u=cx and \ds du=c\,dx we obtain

\ds \frac{1}{\frac{b}{c}-\frac{a}{c}}\int\limits_{a\text{/}c}^{b\text{/}c}f\left(cx\right)\,dx =\frac{c}{b-a}\phantom{\rule{0.2em}{0ex}}\int\limits_{u=a}^{u=b}f\left(u\right)\frac{du}{c}=\frac{1}{b-a}\int\limits_{a}^{b}f\left(u\right)du.

54. Find the area under the graph of \ds f\left(t\right)=\frac{t}{{\left(1+{t}^{2}\right)}^{a}} between \ds t=0 and \ds t=x where \ds a\symbol{"3E}0, \ds a\ne 1 is fixed, and evaluate the limit as \ds x\to \infty .

55. Find the area under the graph of \ds g\left(t\right)=\frac{t}{{\left(1-{t}^{2}\right)}^{a}} between \ds t=0 and \ds t=x, where \ds 0<x<1 and \ds a\symbol{"3E}0 is fixed. Evaluate the limit as \ds x\to 1.

Answer

\ds \int\limits_{0}^{x}g\left(t\right)dt=\frac{1}{2}\int\limits_{u=1-{x}^{2}}^{1}\frac{du}{{u}^{a}}=\frac{1}{2\left(1-a\right)}{u}^{1-a}{}\Big|_{u=1-{x}^{2}}^{1}=\frac{1}{2\left(1-a\right)}\left(1-{\left(1-{x}^{2}\right)}^{1-a}\right).

As \ds x\to 1 the limit is \ds \frac{1}{2\left(1-a\right)} if \ds a<1, and the limit does not exist and has a trend of \infty if \ds a\symbol{"3E}1.

56. The area of a semicircle of radius 1 can be expressed as \ds \int\limits_{-1}^{1}\sqrt{1-{x}^{2}}\,dx . Use the substitution \ds x=\text{cos}\phantom{\rule{0.1em}{0ex}}t to express the area of a semicircle as the integral of a trigonometric function. You do not need to compute the integral.

57. The area of the top half of an ellipse with a major axis that is the x-axis from \ds x=-1 to a and with a minor axis that is the y-axis from \ds y= - b to b can be written as \ds \int\limits_{ - a}^{a}b\sqrt{1-\frac{{x}^{2}}{{a}^{2}}}\,dx . Use the substitution \ds x=a\phantom{\rule{0.1em}{0ex}}\text{cos}\phantom{\rule{0.1em}{0ex}}t to express this area in terms of an integral of a trigonometric function. You do not need to compute the integral.

Answer

\ds \int\limits_{\pi }^{0}b\sqrt{1-{\text{cos}}^{2}t}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\left( - a\phantom{\rule{0.1em}{0ex}}\text{sin}\phantom{\rule{0.1em}{0ex}}t\right)dt=\int\limits_{0}^{\pi }ab{\text{sin}}^{2}tdt

58. [T] The following graph is of a function of the form \ds f\left(t\right)=a\phantom{\rule{0.1em}{0ex}}\text{sin}\left(nt\right)+b\phantom{\rule{0.1em}{0ex}}\text{sin}\left(mt\right). Estimate the coefficients a and b, and the frequency parameters n and m. Use these estimates to approximate \ds \int\limits_{0}^{\pi }f\left(t\right)dt.

A graph of a function of the given form over [0, 2pi], which has six turning points. They are located at just before pi/4, just after pi/2, between 3pi/4 and pi, between pi and 5pi/4, just before 3pi/2, and just after 7pi/4 at about 3, -2, 1, -1, 2, and -3. It begins at the origin and ends at (2pi, 0). It crosses the x axis between pi/4 and pi/2, just before 3pi/4, pi, just after 5pi/4, and between 3pi/2 and 4pi/4.

59. [T] The following graph is of a function of the form \ds f\left(x\right)=a\phantom{\rule{0.1em}{0ex}}\text{cos}\left(nt\right)+b\phantom{\rule{0.1em}{0ex}}\text{cos}\left(mt\right). Estimate the coefficients a and b and the frequency parameters n and m. Use these estimates to approximate \ds \int\limits_{0}^{\pi }f\left(t\right)dt.

The graph of a function of the given form over [0, 2pi]. It begins at (0,1) and ends at (2pi, 1). It has five turning points, located just after pi/4, between pi/2 and 3pi/4, pi, between 5pi/4 and 3pi/2, and just before 7pi/4 at about -1.5, 2.5, -3, 2.5, and -1. It crosses the x axis between 0 and pi/4, just before pi/2, just after 3pi/4, just before 5pi/4, just after 3pi/2, and between 7pi/4 and 2pi.

Answer

\ds f\left(t\right)=2\phantom{\rule{0.1em}{0ex}}\text{cos}\left(3t\right)-\text{cos}\left(2t\right);\int\limits_{0}^{\pi \text{/}2}\left(2\phantom{\rule{0.1em}{0ex}}\text{cos}\left(3t\right)-\text{cos}\left(2t\right)\right)=-\frac{2}{3}

Glossary

change of variables
the substitution of a variable 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

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Calculus: Volume 2 (Second University of Manitoba Edition) by Gilbert Strang and Edward 'Jed' Herman, modified by Varvara Shepelska is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License, except where otherwise noted.

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