OBJECTIVES

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

1. Find integrals of the form $\int \hbox{sin}^{n}x {dx}$ or $\int \hbox{cos} ^{ n} x\,{dx}$, $n \ge$ 2 an integer (p. 480)
2. Find integrals of the form $\int \hbox{sin}^{m} x \hbox{cos}^{n} x {dx}$ (p. 483)
3. Find integrals of the form $\int {tan}^{m}x {sec}^{n}x {dx}$ or $\int {cot}^{m} x {csc}^{n}x {dx}$ (p. 483)
4. Find integrals of the form $\int \hbox{sin}{(ax)} \hbox{sin}{(bx)}{dx}$, $\int \hbox{sin}{(ax)} \hbox{cos}{(bx)}{dx}$, or $\int \hbox{cos}{(ax)} \hbox{cos}{(bx)} {dx}$ (p. 485)

Find $\int \sin ^{5}x\,dx$.

Solution Since the exponent 5 is odd, we write $\int \sin ^{5}x\,dx=\int \sin ^{4}x \sin x\,dx$, and use the identity $\sin ^{2}x=1-\cos ^{2}x$. $$\begin{eqnarray*} \int \sin ^{5}x\,dx&=&\int \sin ^{4}x\sin x\,dx=\int (\sin ^{2}x) ^{2}\sin x\,dx=\int (1-\cos ^{2}x)^{2}\sin x\,dx\\[4pt] &=&\int (1-2\cos ^{2}x+\cos ^{4}x)\sin x\,dx \end{eqnarray*}$$

Now we use the substitution $u=\cos x$. Then $du=-\sin x\,dx$, and $$\begin{eqnarray*} \int \sin ^{5}x\,dx&=&-\int (1-2u^{2}+u^{4})~du=-u+\dfrac{2}{3}u^{3}-\dfrac{1}{5}u^{5}+C\\[3pt] &=&-\cos x+\dfrac{2}{3}\cos ^{3}x-\dfrac{1}{5}\cos ^{5}x+C \end{eqnarray*}$$

NOW WORK

Problem 3.

NEED TO REVIEW?

Trigonometric identities are discussed in Appendix A.4, pp. A-32 to A-35.

Find $\int \sin ^{2}x\,dx$.

Solution Since the exponent of $\sin x$ is an even integer, we use the identity $$\sin ^{2}x=\dfrac{1-\cos (2x) }{2}$$

Then $$\begin{eqnarray*} \int \sin ^{2}x\,dx&=&\dfrac{1}{2}\int \,[ 1-\cos (2x) ] \,dx =\dfrac{1}{2}\int\, dx-\dfrac{1}{2}\int \cos (2x) \,dx \\[4pt] &=& \dfrac{1}{2} x + C_1 - \dfrac{1}{2} \int \cos u \dfrac{du}{2} \qquad {\color{#0066A7}{\hbox{\(u=2x, du=2\, dx\).}}} \\[4pt] &=& \dfrac{1}{2} x + C_1 - \dfrac{1}{4} \sin (2x) + C_2 \end{eqnarray*}$$

Since $C_1$ and $C_2$ are constants, we write the solution as $$\int \sin^2 x \, dx =\dfrac{1}{2}x -\dfrac{1}{4} \sin (2x) +C$$

where $C=C_1+C_2$.

NOTE

Usually we will just add the constant of integration at the end of the integration to avoid letting $C=C_1+C_2$.

NOW WORK

Problem 5.

Find the average value $\bar{y}$ of the function $f( x) =\cos ^{4}x$ over the closed interval $[0,\pi]$.

Solution The average value $\bar{y}$ of a function $f$ over $[a,b]$ is $\bar{y}=\dfrac{1}{b-a}\int_{a}^{b}f(x)\,dx.$

For $f( x) =\cos ^{4}x$ on $[0,\pi]$, we have

Now

To find $\int_{0}^{\pi }\cos ^{2}(2x)\, dx,$ we use the identity $\cos ^{2}\theta =\dfrac{1+\cos (2\theta )}{2}$ again to write $\cos ^{2}(2x) =\dfrac{1+\cos (4x) }{2}$. Then $$\begin{eqnarray*} \int_{0}^{\pi }\cos ^{2}(2x) \,dx &=&\int_{0}^{\pi }\dfrac{ 1+\cos (4x) }{2}\,dx=\dfrac{1}{2}\left[ \int_{0}^{\pi }dx+\int_{0}^{\pi }\cos ( 4x) \,dx\right]\\[7pt] &=& \dfrac{1}{2}\left[ \pi +\int_{0}^{4\pi }\cos u\,\dfrac{du}{4}\right] \qquad {\color{#0066A7}{\hbox{\(u=4x\); \(du=4\,dx\)}}} \end{eqnarray*}$$

So, from (1), $$\bar{y}=\dfrac{1}{4\pi }\left[ \pi +0+\dfrac{\pi }{2}\right] =\dfrac{3}{8}$$

NOW WORK

Problem 57.

Find $\int \sin ^{5} x \sqrt{\cos x} dx = \int \sin^{5}x\,\cos ^{1/2}x\,dx$.

Solution The exponent of $\sin x$ is $5,$ a positive, odd integer. We factor $\sin x$ from $\sin ^{5}x$ and write $$\begin{align*} \int \sin ^{5} x\cos ^{1/2} x\,dx &= \int \sin ^{4} x\,\cos ^{1/2} x\sin x\,dx=\int(\sin ^{2}x) ^{2}\,\cos ^{1/2} x\sin x\,dx\\[4pt] &=\int ( 1-\cos ^{2}x) ^{2}\cos ^{1/2} x\sin xdx \end{align*}$$

Now we use the substitution $u=\cos x.$ $$\begin{eqnarray*} \int \sin ^{5}x\,\cos ^{1/2}x\,dx &=&\int (1-\cos ^{2}x) ^{2}\,\cos ^{1/2}x\,\sin x\,dx \nonumber \\ &\underset{\underset{\underset{\color{#0066A7}{\hbox{\(du=-\sin x~dx\)}}}{\color{#0066A7}{\hbox{\(u=\cos x\)}}}}{\color{#0066A7}{\uparrow }}}{=}&\int (1-u^{2})^{2}u^{1/2}(-du) \nonumber \\ &=&-\int (u^{1/2}-2u^{5/2}+u^{9/2})\,du=-\dfrac{2}{3}u^{3/2}+\dfrac{4}{7}u^{7/2}-\dfrac{2}{11}u^{11/2}+C \nonumber \\ &=&u^{3/2}\left[ -\dfrac{2}{3}+\dfrac{4}{7}u^{2}-\dfrac{2}{11}u^{4}\right] +C \nonumber \\ &\underset{\underset{\color{#0066A7}{\hbox{\(u=\cos x\)}}}{\color{#0066A7}{\uparrow }}}{=}&(\cos x) ^{3/2}\left[ -\dfrac{2}{3}+\dfrac{4}{7}\cos ^{2}x-\dfrac{2}{11}\cos ^{4}x\right] +C \end{eqnarray*}$$

NOW WORK

Problem 11.

Find $\int \tan ^{3} x\sec ^{4} x\,dx$.

Solution Here, $\tan x$ is raised to the odd power $3.$ We factor $\tan x$ from $\tan ^{3}x$ and use the identity $\tan ^{2} x=\sec^{2}x-1.$ $$\begin{eqnarray*} \begin{array}{@rcl@l} \int \tan ^{3}x\sec ^{4}x\,dx &=& \int \tan ^{2}x\,\tan x\,\sec ^{4}x\,dx & {\color{#0066A7}{\hbox{Factor tan \(x\) from \(\tan^{3} x\).}}}\\[6pt] &=& \int (\sec ^{2}x-1) \tan x\sec ^{4}x\,dx & {\color{#0066A7}{\hbox{\(\tan^{2} x = \sec^{2} x - 1\)}}}\\[6pt] &=& \int ( \sec ^{2}x-1) \sec^{3}x\sec x\tan x\,dx & {\color{#0066A7}{\hbox{Factor sec \(x\) from \(\sec^{4} x\).}}}\\[6pt] &=& \int (u^{2}-1) u^{3}du & {\color{#0066A7}{\hbox{Substitute \(u=\sec x\);}}}\\[-4pt] &&& {\color{#0066A7}{\hbox{\(du = \sec x\) \(\tan x\,dx\).}}}\\[6pt] &=& \int (u^{5}-u^{3}) \,du =\dfrac{u^{6}}{6}-\dfrac{u^{4}}{4}+C \underset{\underset{\color{#0066A7}{\hbox{\(u=\sec x\)}}}{\color{#0066A7}{\uparrow}}}{=} \dfrac{\sec ^{6}x}{6}-\dfrac{\sec ^{4}x}{4}+C &\\[-13pt] \end{array} \end{eqnarray*}$$

NOW WORK

Problem 19.

Find $\int \tan ^{2}x\sec ^{4}x\,dx.$

Solution Here, $\sec x$ is raised to a positive even power. We factor $\sec ^{2}x$ from $\sec ^{4}x$ and use the identity $\sec ^{2}x=1+\tan ^{2}x.$ Then $$\begin{eqnarray*} \begin{array}{rcl@l} \int \tan ^{2}x\sec ^{4}x\,dx &=&\int \tan ^{2}x\sec ^{2}x\cdot \sec ^{2}x\,dx & {\color{#0066A7}{\hbox{Factor \(\sec^2 x\) from \(\sec^4 x\).}}} \\[5pt] &=&\int \tan ^{2}x(1+\tan ^{2}x)\sec ^{2}x\,dx & {\color{#0066A7}{\hbox{\(\sec ^{2}{x=1+}\tan ^{2}{x}\)}}} \nonumber \\[5pt] &=&\int u^{2}(1+u^{2})\,du & {\color{#0066A7}{\hbox{Substitute \(u=\tan x\);}}} \nonumber \\[0pt] &&&{\color{#0066A7}{\hbox{\(du=\sec ^{2} x\,dx\).}}}\nonumber \\[5pt] &=&\int (u^{2}+u^{4})\,du = \dfrac{u^{3}}{3}+\frac{u^{5}}{5}+C \underset{\underset{\color{#0066A7}{\hbox{\(u=\tan x\)}}}{\color{#0066A7}{\uparrow}}}{=}\dfrac{\tan ^{3}x}{3}+\dfrac{\tan ^{5}x}{5}+C\\[-8pt] \end{array} \end{eqnarray*}$$

NOW WORK

Problem 21.

Find $\int \tan ^{2}x\sec x\,dx$.

Solution Here, tan $x$ is raised to an even power and sec $x$ to an odd power. We use the identity $\tan ^{2}x=\sec ^{2}x-1$ to write $$\begin{eqnarray*} \int \tan ^{2}x\sec x\,dx &=&\int (\sec ^{2}x-1)\sec x\,dx=\int ( \sec ^{3}x-\sec x)~dx\nonumber\\[4pt] &=&\int \sec ^{3}x\,dx-\int \sec x\,dx\tag{2} \end{eqnarray*}$$

Next we integrate $\int \sec ^{3}x\,dx$ by parts. Choose $$\begin{array}{rcl@{\qquad}c@{\qquad}rcl} u &=&\sec x &\hbox{and}& dv &=&\sec ^{2}x\,dx \\[3pt] du &=&\sec x\tan x\,dx &\hbox{}& v &=&\int \sec ^{2}x\,dx=\tan x \end{array}$$

Then $$\begin{array}{@rcl@{\quad}l} \int \sec ^{3}x\,dx &=&\sec x\,\tan x-\int \tan ^{2}x\,\sec x\,dx & {\color{#0066A7}{\int udv = uv-\int vdu}}\\ &=&\sec x\,\tan x-\int ( \sec ^{2}x-1) \,\sec x\,dx & {\color{#0066A7}{\hbox{\(\tan ^{2}{x=}\sec ^{2}{x-1}\)}}} \\ &=&\sec x\,\tan x-\int \sec ^{3}x\,dx+\int \sec x\,dx & {\color{#0066A7}{\hbox{Write the integral as the}}} \\ &&& {\color{#0066A7}{\hbox{sum of two integrals.}}} \\ 2\int \sec ^{3}x\,dx &=&\sec x\,\tan x+\int \sec x\,dx & {\color{#0066A7}{\hbox{Add \(\int \sec ^{3}x~dx\) to both sides.}}} \\ \int \sec ^{3}x\,dx &=&\dfrac{1}{2}[ \sec x\,\tan x+\ln \vert \sec x+\tan x\vert ] & {\color{#0066A7}{\hbox{Solve for \(\int \sec ^{3}x~dx\);} }}\\ &&&{\color{#0066A7}{\hbox{\(\int \sec x~dx=\ln \left\vert \sec {x+}\tan {x}\right\vert\).}}} \end{array}$$

Now we substitute this result in (2). $$\begin{eqnarray*} \int \tan ^{2}x\sec x\,dx &=&\int \sec ^{3}x\,dx-\int \sec x\,dx\\[4pt] &=& \dfrac{1}{2}\left[ \sec x\,\tan x+\ln \vert \sec x+\tan x\vert \right] -\ln \vert \sec x+\tan x\vert +C \nonumber \\[4pt] &=&\dfrac{1}{2}\left[ \sec x\tan x-\ln \,\vert \sec x+\tan x\vert \right] +C \end{eqnarray*}$$

NOTE

Substituting $n = 3$ into the reduction formula for $\int \sec ^{n} x\,dx$ (derived in Example 8 of Section 7.1) could also have been used to find $\int \sec ^{3} x\,dx.$

NOW WORK

Problem 23.

• $2\sin A\sin B=\cos (A-B)-\cos (A+B)$
• $2\sin A\cos B=\sin (A+B)+\sin (A-B)$
• $2\cos A\cos B=\cos (A-B)+\cos (A+B)$

Find $\int \sin (3x) \sin (2x) \,dx$.

Solution We use the product-to-sum identity $2\sin A\sin B=\cos(A-B)-\cos (A+B).$

Then $$\begin{eqnarray*} 2\sin (3x) \sin (2x) &=&\cos (3x-2x) -\cos (3x+2x) \\[4pt] \sin (3x) \sin (2x) &=&\dfrac{1}{2}[ \cos x-\cos (5x) ] \end{eqnarray*}$$

Then $$\begin{eqnarray*} \int \sin (3x) \sin (2x) \,dx&=&\dfrac{1}{2}\int [\cos x-\cos (5x) ] \,dx=\dfrac{1}{2}\int \cos x~dx-\dfrac{1}{2}\int \cos (5x) \,dx\\[5pt] &&\underset{\underset{\underset{\color{#0066A7}{\hbox{\(du=5~dx\)}}}{\color{#0066A7}{\hbox{\(u=5x\)}}}}{\color{#0066A7}{\uparrow }}}{=} \dfrac{1}{2}\sin x-\dfrac{1}{2}\int \cos u\,\dfrac{du}{5}=\dfrac{1}{2}\sin x-\dfrac{1}{10}\sin (5x) +C \end{eqnarray*}$$

NOW WORK

Problem 27.