When you finish this section, you should be able to:
In this section, we develop techniques to find certain trigonometric integrals. When studying these techniques, concentrate on the strategies used in the examples rather than trying to memorize the results.
Although we could use integration by parts to obtain reduction formulas for integrals of the form \(\int \sin ^{n}x\,dx\) or \(\int \cos ^{n}x\,dx\), \(n \ge 2\) an integer, these integrals also can be found using other, often easier, techniques. We consider two cases:
Suppose we want to find \(\int \sin ^{n}x~dx\) when \(n \ge 3\) is an odd integer. We begin by writing the integral in the form \[\bbox[5px, border:1px solid black, #F9F7ED]{\int \sin^{n}x~dx = \int \sin^{n-1}x \sin x~dx} \]
Since \(n\) is odd, \((n-1)\) is even and we can use the identity \(\sin ^{2}x=1-\cos ^{2}x\).
Then the substitution \(u=\cos x\), \(du=-\sin x~dx\), leads to an integral involving integer powers of \(u\).
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Find \(\int \sin ^{5}x\,dx\).
The method of substitution is discussed in Section 5.6, pp. 387-393.
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 \[ \begin{eqnarray*} \int \sin ^{5}x\,dx&=&-\int (1-2u^{2}+u^{4})~du=-u+\dfrac{2}{3}u^{3}-\dfrac{1}{5}u^{5}+C\\[3pt] &=&-\cos x+\dfrac{2}{3}\cos ^{3}x-\dfrac{1}{5}\cos ^{5}x+C \end{eqnarray*} \]
A similar technique is used to find \(\int \cos ^{n}x\,dx,\) when \(n \geq 3\) is an odd integer. In this case, we write \[\bbox[5px, border:1px solid black, #F9F7ED]{ \int \cos ^{n}x\,dx=\int \cos ^{n-1}x\cos x\,dx } \]
and use the trigonometric identity \(\cos ^{2}x=1-\sin ^{2}x.\) Then we use the substitution \(u=\sin x\). For example, \[ \begin{eqnarray*} \int \cos ^{3}x~dx &=&\int \cos ^{2}x\,\cos x\,dx=\int ( 1-\sin ^{2}x) \cos x\,dx\\ &\underset{\underset{\underset{\color{#0066A7}{\hbox{\(du=\cos x\, dx\)}}}{\color{#0066A7}{\hbox{\(u=\sin x\)}}}}{\color{#0066A7}{\uparrow }}} {=}&\int (1-u^{2})~ du=u-\dfrac{u^{3}}{3}+C=\sin x- \dfrac{\sin ^{3}x}{3}+C\\ \end{eqnarray*} \]
Problem 3.
To find \(\int \sin ^{n}x\,dx\) or \(\int \cos ^{n}x\,dx\) when \(n \geq 2\) is an even integer, the preceding strategy does not work. (Try it for yourself.) Instead, we use one of the identities below: \[\bbox[5px, border:1px solid black, #F9F7ED]{ \sin ^{2}x=\dfrac{1-\cos (2x) }{2}\qquad \cos ^{2}x=\dfrac{1+\cos (2x)} {2}} \]
to obtain a simpler integrand.
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*} \]
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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 \(C=C_1+C_2\).
Usually we will just add the constant of integration at the end of the integration to avoid letting \(C=C_1+C_2\).
Problem 5.
Find the average value \(\bar{y}\) of the function \(f( x) =\cos ^{4}x\) over the closed interval \([0,\pi]\).
The average value of a function is discussed in Section 5.4, pp. 373-374.
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), \[ \bar{y}=\dfrac{1}{4\pi }\left[ \pi +0+\dfrac{\pi }{2}\right] =\dfrac{3}{8} \]
If \(f\) is nonnegative on an interval \([ a,b]\), the average value of \(f\) over the interval \([ a,b]\) represents the height of a rectangle with width \(b-a\) whose area equals the area under the graph of \(f\) from \(a\) to \(b.\) Figure 2 shows the graph of \(f\) from Example 3 and the rectangle of height \(\bar{y}=\dfrac{3}{8}\) and base \(\pi\) whose area is equal to the area under the graph of \(f\).
Problem 57.
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Integrals of the form \(\int \sin ^{m}x\,\cos ^{n}x\,dx\) are found using variations of previous techniques. We discuss two cases:
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*} \]
Problem 11.
If \(m\) and \(n\) are both positive even integers in \(\int \sin ^{m}x\,\cos ^{n}x\,dx\), we use the trigonometric identity \(\sin ^{2}x+\cos ^{2}x=1\) to obtain a sum of integrals, each integral involving only even powers of either \(\sin x\) or \(\cos x\). For example, \[ \begin{equation*} \int \sin ^{2}x\cos ^{4}x\,dx=\int (1-\cos ^{2}x)\cos ^{4}x\,dx=\int \cos ^{4}x\,dx-\int \cos ^{6}x\,dx \end{equation*} \]
The two integrals on the right are now of the form \(\int \cos ^{n}x\,dx\), \(n\) a positive even integer, and we can integrate them using the techniques discussed in Examples 2 and 3.
We consider three cases involving integrals of the form \(\int \tan ^{m}x\,\sec ^{n}x\,dx\):
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The idea is to express the integrand so that we can use either the substitution \(u=\tan x\) and \(du=\sec ^{2}x\,dx\) or the substitution \(u=\sec x\) and \(du=\sec x \tan x\,dx\), while leaving an even power of one of the functions. Then we use a Pythagorean identity to express the integrand in terms of only one trigonometric function.
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*} \]
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*} \]
Problem 21.
When the tangent function is raised to a positive even integer \(m\) and the secant function is raised to a positive odd integer \(n,\) the approach is slightly different. Rather than factoring, we begin by using the identity \(\tan ^{2}x=\sec ^{2}x-1.\)
485
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). \[ \begin{eqnarray*} \int \tan ^{2}x\sec x\,dx &=&\int \sec ^{3}x\,dx-\int \sec x\,dx\\[4pt] &=& \dfrac{1}{2}\left[ \sec x\,\tan x+\ln \vert \sec x+\tan x\vert \right] -\ln \vert \sec x+\tan x\vert +C \nonumber \\[4pt] &=&\dfrac{1}{2}\left[ \sec x\tan x-\ln \,\vert \sec x+\tan x\vert \right] +C \end{eqnarray*} \]
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 \(\int \sec ^{3} x\,dx.\)
Problem 23.
To find integrals of the form \(\int \cot ^{m}x\csc ^{n}x\,dx\), we use the same strategies, but with the identity \(\csc ^{2}x=1+\cot ^{2}x.\)
Trigonometric integrals of the form \[ \begin{equation*} {\int \sin (ax) \sin (bx) \,dx\qquad \int \sin (ax) \cos (bx) \,dx\qquad \int \cos (ax) \cos (bx) \,dx} \end{equation*} \]
are integrated using the product-to-sum identities:
These identities transform the integrand into a sum of sines and/or cosines.
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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*} \]
Problem 27.