OBJECTIVES

When you finish this section, you should be able to:

1. Find an indefinite integral using substitution (p. 387)
2. Find a definite integral using substitution (p. 391)
3. Integrate even and odd functions (p. 393)
4. Solve differential equations: Newton's Law of Cooling (p. 394)

NEED TO REVIEW?

Differentials are discussed in Section 3.4, pp. 230-232.

NOTE

When using the method of substitution, once the substitution $u$ is chosen, we write the original integral in terms of $u$ and $du$.

NEED TO REVIEW?

The Chain Rule is discussed in Section 3.1, pp. 198-200.

Find:

Solution (a) Since we know $\int \sin x\,dx,$ we let $u = 3x + 2$. Then $du = 3\,dx$ so $dx = \dfrac{du}{3}$. $$\begin{eqnarray*} \int \sin (\underset{\color{#0066A7}{\hbox{\(u\)}}}{\underbrace{(3x+2)}}\enspace\underset{\color{#0066A7}{\tfrac{du}{2}}}{\underbrace{dx}}&=& \int {\sin }\,u\dfrac{du}{3}=\dfrac{1}{3}\int {\sin }\,u\,du\\ &=&\dfrac{1}{3}({-}\cos u)+C \underset{\underset{{\color{#0066A7}{\hbox{\(u=3 x+2\)}}}}{\color{#0066A7}{\uparrow }}}{=} -\dfrac{1}{3}{\cos }(3x+2)+C \end{eqnarray*}$$

(b) We let $u = x^{2} + 1$. Then $du = 2x\,dx$, so $x\,dx = \dfrac{du}{2}$. $$\begin{eqnarray*} \int x\sqrt{x^{2}+1}\,dx &=& \int \underset{\color{#0066A7}{\hbox{\(u\)}}}{{\underbrace{\sqrt{x^{2}+1}}}\,}\underset{\color{#0066A7}{\hbox{\(\tfrac{du}{2}\)}}}{\underbrace{x\,dx}}=\int \sqrt{u}\,\dfrac{du}{2}=\dfrac{1}{2}\int u^{1/2}du=\dfrac{1}{2}\left(\dfrac{u^{3/2}}{\dfrac{3}{2}}\right) +C \\ &= & \dfrac{(x^{2}+1) ^{3/2}}{3} +C \end{eqnarray*}$$

(c) We let $u=\sqrt{x}=x^{1/2}$. Then $du=\dfrac{1}{2}x^{-1/2}dx=\dfrac{dx}{2\sqrt{x}}$, so $\dfrac{dx}{\sqrt{x}}$ $=2du$. $$\int \dfrac{e^{\sqrt{{ x}}}}{\sqrt{x}}dx=\int e^{\sqrt{x}}\cdot \dfrac{dx}{\sqrt{x}}=\int e^{u}\cdot 2du=2e^{u}+C=2e^{\sqrt{{ x}}}+C$$

NOW WORK

Problems 5 and 11.

Find:

Solution (a) Notice that the numerator equals the derivative of the denominator, except for a constant factor. So, we try substitution. Let $\ u=4x^{3}-1.$ Then $du=12x^{2}dx$ so $5x^{2}dx=\dfrac{5}{12}\,du$. $$\int \dfrac{5x^{2}dx}{4x^{3}-1}=\int \dfrac{\dfrac{5}{12}\,du}{u}=\dfrac{5}{12}\int \dfrac{du}{u}=\dfrac{5}{12}\ln \left\vert u\right\vert +C=\dfrac{5}{12}\ln \left\vert 4x^{3}-1\right\vert +C$$

(b) Here, the numerator equals the derivative of the denominator. So, we use the substitution $u=e^{x}+4$. Then $du=e^{x}dx$. $$\begin{eqnarray*} \int \dfrac{e^{x}}{e^{x}+4}dx=\int \dfrac{1}{e^{x}+4}\cdot e^{x}dx=\int \dfrac{1}{u}\,du=\ln \vert u\vert +C \underset{\underset{{\color{#0066A7}{\hbox{\(u=e^{x}+4>0\)}}}}{\color{#0066A7}{\uparrow}}}{=} \ln (e^{x}+4) +C \\ \end{eqnarray*}$$

NOW WORK

Problem 17.

Show that:

Solution (a) Since $\tan x=\dfrac{\sin x}{\cos x},$ we let $u=\cos x.$ Then $du=-\sin x\, dx$ and $$\begin{eqnarray*} \int \tan x dx = \int \dfrac{\sin x}{\cos x}dx=\int -\dfrac{du}{u}=-\ln \left\vert u\right\vert +C=-\ln \left\vert \cos x\right\vert +C \\ \underset{ \underset{ \color{#0066A7}{\hbox{\(r\ln x=\ln x^{r}\)}} } {\color{#0066A7}{\uparrow}} } {=} \ln \left\vert \cos x\right\vert^{-1}+C =\ln \left\vert \dfrac{1}{\cos x}\right\vert + C =\ln \left\vert \sec x\right\vert +C \end{eqnarray*}$$

(b) To find $\int \sec x dx$, we multiply the integrand by $\dfrac{\sec x+\tan x}{\sec x+\tan x}.$ $$\int \sec x dx=\int \sec x\cdot \dfrac{\sec x+\tan x}{\sec x+\tan x} dx=\int \dfrac{\sec ^{2}x+\sec x\tan x}{\sec x+\tan x} dx$$

Now the numerator equals the derivative of the denominator. So if $u=\sec x+\tan x,$ then $du=( \sec x\tan x+\sec ^{2}x) dx.$ $$\int \sec x dx=\int \dfrac{du}{u}=\ln \left\vert u\right\vert +C=\ln \left\vert \sec x+\tan x\right\vert +C$$

NOW WORK

Problem 27.

IN WORDS

If the numerator of the integrand equals the derivative of the denominator, then the integral equals a logarithmic function.

Find $\int {x\sqrt{4+x}}\,dx$.

Solution Substitution I  Let $u = 4 + x$. Then $du = dx.$ Since $u=4+x,$ $x=u-4.$ Substituting gives $$\begin{eqnarray*} \int {x\sqrt{4+x}}\,dx \underset{\underset{{\color{#0066A7}{\hbox{\(u=x+4\)}}}}{\color{#0066A7}{\uparrow }}} =&& \int \underset{\color{#0066A7}{\hbox{\(x\)}}}{\underbrace{( u-4) }} \underset{\underset{{\color{#0066A7}{\hbox{\(4+x\)}}}}{\color{#0066A7}{\uparrow }}}{{\sqrt{u}}}\,\,\underset{\color{#0066A7}{\hbox{\(dx\)}}}{\underbrace{du}}\, =\int {({u^{3/2}-4u^{1/2}})}\,du \\ &=& {\dfrac{{u^{5/2}}}{{{\dfrac{{5}}{{2}}}}}}-4\cdot {\dfrac{{u^{3/2}}}{{{\dfrac{{3}}{{2}}} }}}+C \\ &=&{\dfrac{{2({4+x})^{5/2}}}{{5}}}-{\dfrac{{8({4+x})^{3/2}}}{{3}}}+C \end{eqnarray*}$$

Substitution II  Let $u = \sqrt{4\,{+}\,x}$, so $u^{2} = 4 + x$ and $x = u^{2} - 4$. Then $dx=2u\,du$ and $$\begin{eqnarray*} \int {x\sqrt{4+x}}\,dx & = &\int {(}\underset{\color{#0066A7}{\hbox{\(x\)}}}{\underbrace{{u^{2}-4}}}{)(u)(}\underset{\color{#0066A7}{\hbox{\(dx\)}}}{\underbrace{{2u\,du}}}{)= 2 \int {(u^{4}-4u^{2})\,du}}=2\left[ {\dfrac{{u^{5}}}{{5}}}-{\dfrac{{4u^{3}}}{{3}}}\right] +C \\ &=&\dfrac{2}{5} \big(\sqrt{4+x}\,\big) ^{5} -\dfrac{8}{3} \big(\sqrt{4+x}\,\big) ^{3}+C= \dfrac{2(4+x)^{5/2}}{5} -\dfrac{8(4+x)^{3/2}}{3}+C   \end{eqnarray*}$$

NOW WORK

Problem 35.

Find:

Solution (a) $\int \dfrac{dx}{\sqrt{4-x^{2}}}$ resembles $\int \dfrac{1}{\sqrt{1-x^{2}}}\,dx=$ $\sin ^{-1}x+C.$ We begin by rewriting the integrand as $$\dfrac{1}{\sqrt{4-x^{2}}}=\dfrac{1}{\sqrt{4\left( 1-\dfrac{x^{2}}{4}\right) }}=\dfrac{1}{2\sqrt{1-\left( \dfrac{x}{2}\right) ^{2}}}$$

Now we let $u=\dfrac{x}{2}$. Then $du=\dfrac{dx}{2}$, so $dx=2\,du.$ $$\begin{eqnarray*} \int \dfrac{dx}{\sqrt{4-x^{2}}}&=&\int \dfrac{dx}{2\sqrt{1-\left( \dfrac{x}{2}\right) ^{2}}} \underset{\underset{{\underset{{\color{#0066A7}{\hbox{\(dx=2du\)}}}}{\color{#0066A7}{\hbox{\(u=\tfrac{x}{2}\)}}}}}{\color{#0066A7}{\uparrow}}} = \int \dfrac{2 du}{2\sqrt{1-u^{2}}}=\int \dfrac{du}{\sqrt{1-u^{2}}}=\sin ^{-1}u+C \\ &=&\sin ^{-1}\left( \dfrac{x}{2}\right) +C \end{eqnarray*}$$

(b) $\int \dfrac{dx}{9+4x^{2}}$ resembles $\int \dfrac{1}{1+x^{2}}dx=\tan ^{-1}x+C$. We rewrite the integrand as $$\dfrac{1}{9+4x^{2}}=\dfrac{1}{9\left( 1+\dfrac{4x^{2}}{9}\right) }=\dfrac{1}{9\left[ 1+\left( \dfrac{2x}{3}\right) ^{2}\right] }$$

Now let $u=\dfrac{2x}{3}.$ Then $du=\dfrac{2}{3}dx$, so $dx=\dfrac{3}{2} du$. $$\begin{eqnarray*} \int \dfrac{dx}{9+4x^{2}}&=&\int \dfrac{dx}{9\left[ 1+\left( \dfrac{2x}{3} \right) ^{2}\right] }=\int \dfrac{\dfrac{3}{2}du}{9( 1+u^{2}) }=\dfrac{1}{6}\int \dfrac{du}{1+u^{2}} \\ &=& \dfrac{1}{6}\tan ^{-1}u+C=\dfrac{1}{6}\tan ^{-1}\left( \dfrac{2x}{3}\right) +C \end{eqnarray*}$$

NOW WORK

Problem 39.

• Method 1: Find the related indefinite integral using substitution, and then use the Fundamental Theorem of Calculus.
• Method 2: Find the definite integral directly by making a substitution in the integrand and using the substitution to change the limits of integration.

Find $\int_{0}^{2}x\sqrt{4-x^{2}}\,dx$.

Solution

Method 1: Use the related indefinite integral and then apply the Fundamental Theorem of Calculus. The related indefinite integral $\int {x\sqrt{4-x^{2}}\,dx}$ can be found using the substitution $u = 4 - x^{2}$. Then $du\,=-2x\,dx$, so $x\,dx=-\dfrac{du}{2}.$ $$\begin{eqnarray*} \int {x\sqrt{4-x^{2}}\,dx}&=&\int {\sqrt{u}\left( {-}{\dfrac{{du}}{{2}}}\right) }={-}{\dfrac{{1}}{{2}}}\int u^{1/2}du=-\dfrac{1}{2}\cdot {{\dfrac{{u^{3/2}}}{{{ \dfrac{{3}}{{2}}}}}}}+C \\ &=&-\dfrac{1}{3}{{(4-x^{2})^{3/2}}}+C \end{eqnarray*}$$

Then by the Fundamental Theorem of Calculus, $$\int_{0}^{2} x\sqrt{4-x^{2}}\,dx = -\dfrac{1}{3} \Big[(4-x^{2})^{3/2}\Big]_{0}^{2} = -\dfrac{1}{3}\big[0-4^{3/2}\big] =\dfrac{8}{3}$$

Method 2: Find the definite integral directly by making a substitution in the integrand and changing the limits of integration. We let $u=4-x^{2}$; then $du=-2x\,dx.$ Now use the function $u=4-x^{2}$ to change the limits of integration.

• The lower limit of integration is $x=0$ so, in terms of $u,$ it becomes $u=4-0^{2}=4$.
• The upper limit of integration is $x=2$ so the upper limit becomes $u=4-2^{2}=0$.

Then

$$\begin{eqnarray*} {\int_{0}^{2}{x\sqrt{4-x^{2}}\,dx}} && \underset{\underset{{\underset{{\color{#0066A7}{\hbox{\(x dx=-\tfrac{1}{2}du\)}}}}{{\color{#0066A7}{\hbox{\(u=4-x^{2} \)}}}}}}{\color{#0066A7}{\uparrow }}}{=} \int_{4}^{0}\sqrt{u}\left( -\dfrac{du}{2}\right) =-\dfrac{1}{2}\int_{4}^{0} \sqrt{u} du=-\dfrac{1}{2}\cdot \left[ \dfrac{u^{3/2}}{\dfrac{3}{2}}\right] _{4}^{0} \\ {\;\;} &&= -{\dfrac{1}{3}}(0-8)={\dfrac{{8}}{{3}}} \end{eqnarray*}$$

NOW WORK

Problem 45.

Find $\int_{0}^{\pi /2}\dfrac{{1-\cos}(2\theta)}{2}\,d\theta$.

Solution We use properties of integrals to simplify before integrating. $$\begin{eqnarray*} \int_{0}^{\pi /2}\dfrac{1-\cos (2\theta ) }{2}\,d\theta &=&\dfrac{1}{2}\int_{0}^{\pi /2}\left[ 1-\cos (2\theta ) \right] \,d\theta \\ &=& \dfrac{1}{2}\left[ \int_{0}^{\pi /2}d\theta -\int_{0}^{\pi /2}\cos (2\theta ) \,d\theta \right]\\ & =& \dfrac{1}{2}\int_{0}^{\pi /2}d\theta -\dfrac{1}{2}\int_{0}^{\pi /2}\cos (2\theta ) \,d\theta \\ &=& \dfrac{1}{2} \Big[\theta\Big] _{0}^{\pi /2}-\dfrac{1}{2}\int_{0}^{\pi /2}\cos (2\theta ) \,d\theta \\ &=& \dfrac{\pi }{4}-\dfrac{1}{2}\int_{0}^{\pi /2}\cos (2\theta ) \,d\theta \end{eqnarray*}$$

In the integral on the right, we use the substitution $u = 2\theta$. Then $du = 2\,d\theta$ so $d\theta =\dfrac{du}{2}$. Now we change the limits of integration:

• when $\theta = 0,$   then $u=2(0) =0$
• when $\theta =\dfrac{\pi }{2},$   then $u=2\left( \dfrac{\pi }{2}\right) =\pi$

Now $$\int_{0}^{\pi /2}\cos (2\theta ) \,d\theta =\int_{0}^{\pi }{\cos u\,\dfrac{{du}}{{2}}}= \dfrac{1}{2}\Big[\sin u\Big] _{0}^{\pi }=\dfrac{1}{2}\left( \sin \pi -\sin 0\right) =0$$

Then, $$\int_{0}^{\pi /2}\dfrac{1-\cos (2\theta ) }{2}d\theta =\dfrac{\pi}{4}-\dfrac{1}{2}\int_{0}^{\pi /2}{\cos }(2\theta ) {\,d\theta }=\dfrac{\pi }{4}$$

NOW WORK

Problem 53.

NEED TO REVIEW?

Even and odd functions are discussed in Section P.1, pp. 9-10.

THEOREM The Integrals of Even and Odd Functions

Let a function $f$ be continuous on a closed interval $[-a,a]$, $a>\,0$.

• If $f$ is an even function, then $$\bbox[5px, border:1px solid black, #F9F7ED]{\bbox[5pt]{ \int_{-a}^{a}f(x)\,dx=2\int_{{0}}^{a}f(x)\,dx }}$$
• If $f$ is an odd function, then $$\bbox[5px, border:1px solid black, #F9F7ED]{\bbox[5pt]{ \int_{-a}^{a} f(x)\,dx=0 }}$$

Proof

$f$ is an even function:  Since $f$ is continuous on the closed interval $[ -a,\,a] ,$ $a>0,$ and $0$ is in the interval $\left[ -a,a\right]$, we have $$\begin{equation*} \int_{-a}^{a} f(x)\,dx=\int_{-a}^{0} f(x)\,dx+ \int_{0}^{a}f(x)\,dx=-\int_{0}^{-a}f(x)\,dx+\int_{0}^{a}f(x)\,dx \tag {2} \end{equation*}$$

In $- {\int_{0}^{-a}{f(x)\,dx}}$, we use the substitution $u = -x$. Then $du = -dx$. Also, if $x=0$, then $u=0$, and if $x=-a$, then $u = a$. Therefore, $$\begin{eqnarray*} -{\int_{0}^{-a}{f(x)\,dx={\int_{0}^{a}{f({-}u)\,du}}}} \underset{\underset{{\underset{{\color{#0066A7}{\hbox{\(f(-u) =f(u) \)}}}}{\color{#0066A7}{\hbox{\(f\) is even}}}}}{\color{#0066A7}{\uparrow }}}= \int_{0}^{a}f(u) du=\int_{0}^{a}f(x)\, dx \tag{3} \end{eqnarray*}$$

Combining (2) and (3), we obtain $${\int_{{-}a}^{a}{f(x)\,dx={\int_{0}^{a}{f(x)\,dx}}}}+{\int_{0}^{a}{f(x)\,dx}}=2{\int_{0}^{a}{f(x)\,dx}}$$

Find:

Solution (a) If $f(x)={{x^{7}-4x^{3}+x,}}$ then $f({-}x)=({-}x)^{7}-4({-}x)^{3}+({-}x)={-}(x^{7}-4x^{3}+x)={-}f(x).$ Since $f$ is an odd function, $$\int_{-3}^{3}{(x^{7}-4x^{3}+x)}\,dx=0$$

(b) If $g(x) =x^{4}-x^{2}+3,$ then $g(-x) =(-x) ^{4}-(-x) ^{2}+3=x^{4}-x^{2}+3=g(x).$ Since $g$ is an even function, $$\begin{eqnarray*} \int_{-2}^{2}( x^{4}-x^{2}+3) dx &=& 2\int_{0}^{2}(x^{4}-x^{2}+3)\, dx=2\left[ \dfrac{x^{5}}{5}-\dfrac{x^{3}}{3}+3x\right]_{0}^{2} \\ &=& 2\left[ \dfrac{32}{5}-\dfrac{8}{3}+6\right] =\dfrac{292}{15} \end{eqnarray*}$$

NOW WORK

Problems 63 and 67.

If $f$ is an even function and $\int_{0}^{2}f(x)\, dx=-6$ and $\int_{-5}^{0}f(x)\, dx=8,$ find $\int_{2}^{5}f(x)\, dx.$

Solution $\int_{2}^{5}f(x)\, dx=\int_{2}^{0}f(x)\, dx+\int_{0}^{5}f(x)\, dx$

Now $\int_{2}^{0}f(x)\, dx=-\int_{0}^{2}f(x)\, dx=6.$

Since $f$ is even, $\int_{0}^{5}f(x)\, dx=$ $\int_{-5}^{0}f(x)\, dx$ $\,=8.$ Then $$\int_{2}^{5}f(x)\, dx=\int_{2}^{0}f(x)\, dx+\int_{0}^{5}f(x)\, dx=6+8=14$$

An object is heated to $90{}^{\circ}{\rm C}$ and allowed to cool in a room with a constant ambient temperature of $20{}^{\circ}{\rm C}$. If after $10$ min the temperature of the object is $60{}^{\circ}{\rm C}$, what will its temperature be after $20$ min?

Solution When $t=0$, $u(0) =u_{0}=90{}^{\circ}{\rm C}$, and when $t=10$ min, $u\left( 10\right) =60{}^{\circ}{\rm C}$. Given that the ambient temperature $T$ is $20{}^{\circ}{\rm C}$, we substitute these values into equation (5). $$\begin{eqnarray*} \begin{array}{rcl@{\qquad}l@{\hspace*{-6pc}}} u(t) &=&( u_{0}-T) e^{kt}+T \\ 60 &=&( 90-20) e^{10k}+20 & {\color{#0066A7}{\hbox{\(u=60 \hbox{ when }t=10; \ T=20;\ u_{0} =90\)}}} \\ \dfrac{40}{70} &=&e^{10k} \\ k &=&\dfrac{1}{10}\ln \dfrac{4}{7} = 0.1 0 \ln \dfrac{4}{7} \end{array} \end{eqnarray*}$$

The temperature $u$ is $$u(t) =70 e^{[0.1 \ln (4/7)]t} +20$$

Then when $t=20$, the temperature $u$ of the object is $$u (20) =70e ^{[0.1 \ln (4/7)] (20)} +20=70e^{2\ln (4/7)}+20\approx 42.86{}^{\circ}{\rm C}$$

NOW WORK

Problem 109.