Many trigonometric functions can be integrated by combining substitution and Integration by Parts with the appropriate trigonometric identities. First, consider \[ \int \sin^mx~\cos^nx\, dx \]
where \(m, n\) are whole numbers. The easier case is when at least one of \(m, n\) is odd.
Evaluate \(\displaystyle\int \sin^3x\, dx\).
Solution Because \(\sin^3x\) is an odd power, the identity \(\sin^2 x = 1 - \cos^2 x\) allows us to split off a factor of \(\sin x\,dx\): \[ \sin^3x\,dx = \sin^2x (\sin x\,dx) = (1 - \cos^2x)\sin x\,dx \]
and use the substitution \(u = \cos x\), \(du = -\sin x\, dx\): \[ \begin{align*} \int \sin^3x\, dx &= \int (1 - \cos^2x) \sin x\, dx = - \int (1 - u^2)\,du\\ &= \frac{u^3}{3} - u+ C = \frac{\cos^3 x}{3} - \cos x + C \end{align*} \]
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The strategy of the previous example works when \(\sin^mx\) appears with \(m\) odd. Similarly, if \(n\) is odd, write \(\cos^n x\) as a power of \((1 - \sin^2 x)\) times \(\cos x\).
Integrating \(\sin^m x~\cos^n x\) \[ \fbox{\(Case~1: m = 2k + 1~odd\)} \]
Write \(\sin^{2k+1} x\) as \((1 -\cos^2 x)^k \sin x\). Then \(\int\sin^{2k+1} x~\cos^n x\, dx\) becomes \[ \int\sin x(1 - \cos^2 x)^k \cos^n x\, dx \]
Substitute \(u=\cos x, -du = \sin x\, dx\). \[ \fbox{\(Case~2: n = 2k + 1~odd\)} \]
Write \(\cos^{2k +1} x\) as \((1 - \sin^2 x)^k \cos x\) Then \( \int \sin^m x~\cos^{2k +1} x\, dx\) becomes \[ \int \sin^m x (1-\sin^2 x)^k \cos x\, dx \]
Substitute \(u=\sin x, du = \cos x\, dx\). \[ \fbox{\(Case~3: m, n~both~even\)} \]
Use reduction formulas (1) or (2) as described below or use the method of Exercises 65–68.
Evaluate \(\displaystyle\int \sin^4x~\cos^5x\, dx\).
Solution We take advantage of the fact that \(\cos^5x\) is an odd power to write \[ \sin^4x~\cos^5x\,dx = \sin^4x\,\cos^4x (\cos x\,dx) = \sin^4x(1 - \sin^2x)^2 (\cos x\,dx)\\ \]
This allows us to use the substitution \(u = \sin x\), \(du=\cos x\,dx\): \[ \begin{align*} \int \sin^4x~\cos^5x\, dx &= \int (\sin^4x) (1 - \sin^2x)^2\cos x \, dx\\ &=\int u^4(1 - u^2)^2 \, du = \int (u^4 - 2u^6 + u^8)\, du\\ &= \frac{u^5}{5} - \frac{2u^7}{7} + \frac{u^9}{9}+ C = \frac{\sin^5 x}{5} - \frac{2\sin^7 x}{7} + \frac{\sin^9 x}{9} + C \end{align*} \]
The following reduction formulas can be used to integrate \(\sin^nx\) and \(\cos^nx\) for any exponent \(n\), even or odd (their proofs are left as exercises; see Exercise 64).
\[ \begin{eqnarray} &\boxed{\bbox[#FAF8ED,5pt]{\displaystyle\int \sin^n x\, dx = -\frac{1}{n}\sin^{n-1} x~\cos x + \frac{n-1}{n}\int \sin^{n-2} x\,dx}}\tag{1} \\ &\boxed{\bbox[#FAF8ED,5pt]{\displaystyle\int \cos^n x\, dx = \frac{1}{n}\cos^{n-1}x~\sin x + \frac{n- 1}{n}\int \cos^{n-2}x\, dx}}\tag{2} \end{eqnarray} \]
Evaluate \(\displaystyle\int \sin^4x\, dx\).
Solution Apply Eq. (1) with \(n = 4\), \[ \int \sin^4x\, dx = -\frac{1}{4}\sin^3x~\cos x + \frac{3}{4}\int \sin^2x\,dx\tag{3} \]
Then apply Eq. (1) again, with \(n=2\), to the integral on the right: \[ \int \sin^2x\,dx = -\frac{1}{2}\sin x~\cos x + \frac{1}{2}\int\,dx = -\frac{1}{2}\sin x~\cos x+\frac12\,x + C\tag{4} \]
Using Eq. (4) in Eq. (3), we obtain \[ \int \sin^4x\, dx = -\frac{1}{4}\sin^3x~\cos x -\frac38 \sin x~\cos x + \frac38 x + C \]
Trigonometric integrals can be expressed in many different ways because trigonometric functions satisfy a large number of identities. For example, a computer algebra system might evaluate the integral in the previous example as \[ \int\sin^4 x\,dx = \frac1{32}(x - 8\sin 2x+\sin 4x ) +C \]
You can check that this agrees with the result in Example 3 (Exercise 61).
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More work is required to integrate \(\sin^m x~\cos^nx\) when both \(m\) and \(n\) are even. First of all, we have the following formulas, which are verified using the identities recalled in the margin. \[ \begin{eqnarray*} &\displaystyle\boxed{\bbox[#FAF8ED,5pt]{\int \sin^2x\, dx = \frac{x}{2} - \frac{\sin 2x }{4} + C = \frac{x}{2} - \frac12\sin x~\cos x+ C}} \\ &\displaystyle\boxed{\bbox[#FAF8ED,5pt]{\int \cos^2x\, dx = \frac{x}{2} + \frac{\sin 2x }4 + C= \frac{x}{2} + \frac12\sin x~\cos x+ C}} \end{eqnarray*} \]
Useful Identities: \[ \begin{align*} \sin^2x &= \frac12(1-\cos 2x)\\ \cos^2x &= \frac12(1+\cos 2x)\\ \sin 2x &= 2\sin x~\cos x\\ \cos 2x &= \cos^2x - \sin^2x \end{align*} \]
Here is a method for integrating \(\sin^m x~\cos^nx\) when both \(m\) and \(n\) are even. Another method is used in Exercises 65–68.
Expand the integral on the right to obtain a sum of integrals of powers of \(\cos x\) and use reduction formula (2).
Expand the integral on the right to obtain a sum of integrals of powers of \(\sin x\), and again evaluate using reduction formula (1).
As we have noted, trigonometric integrals can be expressed in more than one way. According to Mathematica, \[ \begin{align*} \int &\sin^2x~\cos^4x\, dx \\ =&\tfrac1{16} x + \tfrac1{64}\sin 2x - \tfrac1{64}\sin 4x-\tfrac1{192}\sin 6x +C \end{align*} \]
Trigonometric identities show that this agrees with Eq. (6) in Example 4.
Evaluate \(\displaystyle\int \sin^2x~\cos^4x\, dx\).
Solution Here \(m=2\) and \(n=4\). Since \(m < n\), we replace \(\sin^2 x\) by \(1 - \cos^2 x\): \[ \begin{align} \int \sin^2x~\cos^4x\, dx &= \int (1 - \cos^2x)\cos^4x\, dx= \int \cos^4x\, dx - \int \cos^6x\, dx\tag{5} \end{align} \]
The reduction formula for \(n=6\) gives \[ \int \cos^6x\, dx = \frac{1}{6}\cos^{5}x~\sin x + \frac{5}{6}\int \cos^4x\,dx \]
Using this result in the right-hand side of Eq. (5), we obtain \[ \begin{align*} \int \sin^2x~\cos^4x\, dx &=\int \cos^4 x\, dx - \left( \frac16 \cos^5 x~\sin x + \frac56 \int \cos^4 x\, dx \right)\\ &= -\frac{1}{6}\cos^{5}x~\sin x + \frac16 \int \cos^4x\,dx \end{align*} \]
Next, we evaluate \({\int \cos^4x\,dx}\) using the reduction formulas for \(n=4\) and \(n=2\): \[ \begin{align*} \int \cos^4x\, dx &= \frac14\cos^{3}x~\sin x + \frac34\int \cos^{2}x\, dx \\ &= \frac14\cos^{3}x~\sin x + \frac34\left(\frac12 \cos x~\sin x+\frac12 x\right)+C\\ &= \frac{1}{4}\cos^{3}x~\sin x + \frac38 \cos x~\sin x + \frac38 x +C \end{align*} \]
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Altogether, \[ \begin{align} \int \sin^2x~\cos^4x\, dx &= -\frac{1}{6}\cos^{5}x~\sin x + \frac16 \left(\frac{1}{4}\cos^{3}x~\sin x + \frac38 \cos x~\sin x + \frac38 x\right)+C \notag\\ &= -\frac{1}{6}\cos^{5}x~\sin x + \frac{1}{24}\cos^{3}x \sin x + \frac1{16} \cos x~\sin x + \frac1{16} x +C\tag{6} \end{align} \]
We turn now to the integrals of the remaining trigonometric functions.
The integral \(\int\sec x\,dx\) was first computed numerically in the 1590s by the English mathematician Edward Wright, decades before the invention of calculus. Although he did not invent the concept of an integral, Wright realized that the sums that approximate the integral hold the key to understanding the Mercator map projection, of great importance in sea navigation because it enabled sailors to reach their destinations along lines of fixed compass direction. The formula for the integral was first proved by James Gregory in 1668.
Derive the formulas \[ \int \tan x\, dx = \ln |\sec x| + C,\qquad \int \sec x\, dx = \ln \big| {\sec x + \tan x}\big| +C \]
Solution To integrate \(\tan x\), use the substitution \(u = \cos x\), \(du = -\sin x\,dx\): \[ \begin{align*} \int \tan x\, dx &= \int \frac{\sin x}{\cos x}\, dx = -\int\frac{du}u = -\ln |u|+C = - \ln |{\cos x}|+C \\ &= \ln \frac{1}{|{\cos x}|} + C = \ln |{\sec x}| + C \end{align*} \]
To integrate \(\sec x\), we employ a clever substitution: \(u = \sec x + \tan x\). Then \[ du = (\sec x~\tan x + \sec^2 x)\, dx = (\sec x)\underbrace{(\tan x + \sec x)}_{u}\, dx = (\sec x)u\,dx \]
Thus \(du = (\sec x)u\,dx\), and dividing by \(u\) gives \({du}/{u} = \sec x\, dx\). We obtain \[ \int \sec x\, dx = \int \frac{du}{u} = \ln |u| + C = \ln \big|\sec x + \tan x\big| + C \]
The table of integrals at the end of this section (page 410) contains a list of additional trigonometric integrals and reduction formulas.
Evaluate \(\int_0^{\pi/4} \tan^3 x\, dx\).
Solution We use reduction formula (16) in the table with \(k=3\). \[ \begin{align*} \int_0^{\pi/4} \tan^3 x\, dx &= \frac{\tan^2x}{2}\bigg|_0^{\pi/4} - \int_0^{\pi/4} \tan x\, dx = \left(\frac12\tan^2x - \ln|\sec x|\right)\bigg|_0^{\pi/4} \\ &= \left(\frac12\tan^2\frac{\pi}4 - \ln\left\vert\sec \frac{\pi}4\right\vert\right) - \left(\frac12 \tan^20 - \ln|\sec 0|\right) \\ &= \left(\frac12(1)^2 - \ln \sqrt 2\right) - \left(\frac12 0^2 - \ln|1|\right) = \frac12-\ln\sqrt 2 \end{align*} \]
In the margin we describe a method for integrating \(\tan^m~x~\sec^nx\).
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Integrating \(\tan^m~x~\sec^nx\) \[ \fbox{\(Case~1: m=2k+1 odd~and~n\ge 1\)} \]
Use the identity \(\tan^2x= \sec^2x-1\) to write \(\tan^{2k+1}x~\sec^nx\) as \[ (\sec^2x-1)^k(\sec^{n-1}x)(\sec x~\tan x) \]
Then substitute \(u=\sec x\), \(du=\sec x~\tan x\,dx\) to obtain an integral involving only powers of \(u\). \[ \fbox{\(Case~2: n = 2k~even\)} \]
Use the identity \(\sec^2x = 1 + \tan^2x\) to write \(\tan^mx\,\sec^nx\) as \[ (\tan^mx)(1+\tan^2x)^{k-1}\sec^2x \]
Then substitute \(u=\tan x\), \(du=\sec^2x\,dx\) to obtain an integral involving only powers of \(u\). \[ \fbox{\(Case~3: m~even~and~n~odd\)} \]
Use the identity \(\tan^2~x = \sec^2x-1\) to write \(\tan^m~x~\sec^n~x\) as \[ (\sec^2x-1)^{m/2}\sec^nx \]
Expand to obtain an integral involving only powers of \(\sec x\) and use the reduction formula (20).
Evaluate \(\int \tan^2x\sec^3x\,dx\).
Solution Our integral is covered by Case 3 in the marginal note, because the integrand is \(\tan^m~x~\sec^n~x\), with \(m=2\) and \(n=3\).
The first step is to use the identity \(\tan^2x=\sec^2x-1\): \[ \int \tan^2x~\sec^3x\,dx = \int (\sec^2x-1)\sec^3x \,dx = \int \sec^5x\, dx - \int \sec^3x\, dx\tag{7} \]
Next, use the reduction formula (20) in the table on page 140 with \(m=5\): \[ \begin{align} \int \sec^5 x\, dx &= \frac{\tan x~\sec^3 x}{4} + \frac{3}{4}\int \sec^3 x\, dx\nonumber\\ \end{align} \]
Substitute this result in Eq. (7): \[ \begin{align} \int \tan^2x~\sec^3x\,dx &= \left(\frac{\tan x~\sec^3 x}{4} + \frac{3}{4}\int \sec^3 x\, dx\right) - \int \sec^3x\, dx\nonumber\\ &=\frac14\tan x~\sec^3 x - \frac14\int \sec^3 x\, dx\tag{8} \end{align} \]
and use the reduction formula (20) again with \(m = 3\) and formula (19): \[ \begin{align*} \int \sec^3 x\, dx &= \frac{\tan x~\sec x}{2} + \frac12\int \sec x\, dx\\ &=\frac12 \tan x~\sec x + \frac12\ln|{\sec x + \tan x}|+C \end{align*} \]
Then Eq. (8) becomes \[ \begin{align*} \int \tan^2x~\sec^3x\,dx & = \frac14\tan x~\sec^3 x - \frac14\left( \frac12 \tan x~\sec x + \frac12\ln|\sec x + \tan x|\right)+C\\ &= \frac14 \tan x~\sec^3 x - \frac18\tan x~\sec x - \frac18\ln|{\sec x + \tan x}|+C \end{align*} \]
Formulas (23)–(25) in the table describe the integrals of the products \(\sin mx\sin nx\), \(\cos mx~\cos nx\), and \(\sin mx~\cos nx\). These integrals appear in the theory of Fourier Series, which is a fundamental technique used extensively in engineering and physics.
Evaluate \(\displaystyle\int_0^{\pi}\sin 4x~\cos 3x\, dx\).
Solution Apply reduction formula (24), with \(m=4\) and \(n=3\): \[ \begin{align*} \int_0^{\pi} \sin 4x~\cos 3x\, dx &= \left(-\frac{\cos(4-3)x}{2(4-3)} - \frac{\cos(4+3)x}{2(4+ 3)}\right)\bigg|_0^{\pi} \\ &= \left(-\frac{\cos x}{2} - \frac{\cos 7 x}{14}\right)\bigg|_0^{\pi}\\ &=\left(\frac{1}2+\frac{1}{14}\right)-\left(-\frac12-\frac1{14}\right) = \frac87 \end{align*} \]
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| \[ \begin{eqnarray} &&\int \sin^2x\, dx = \frac{x}{2} - \frac{\sin 2x}4 + C = \frac{x}{2} - \frac{1}{2}\sin x~\cos x + C\tag{9} \\ &&\int \cos^2x\, dx = \frac{x}{2} + \frac{\sin 2x}4 + C = \frac{x}{2} + \frac{1}{2}\sin x~\cos x + C\tag{10}\\ &&\int \sin^n x\, dx = -\frac{\sin^{n-1}x~\cos x}n +\frac{n-1}n\int \sin^{n-2}x \, dx\tag{11} \\ &&\int \cos^n x\, dx = \frac{\cos^{n-1}x~\sin x}n +\frac{n-1}n\int \cos^{n-2}x \, dx\tag{12} \\ &&\int \sin^m x~\cos^nx \, dx = \frac{\sin^{m+1}x~\cos^{n-1} x}{m+n} +\frac{n-1}{m+n}\int \sin^mx~\cos^{n-2}x \, dx \label{8.trigintegrals.sinncosn1}\tag{13} \\ &&\int \sin^m x~\cos^nx \, dx = -\frac{\sin^{m-1}x~\cos^{n+1} x}{m+n} +\frac{m-1}{m+n}\int \sin^{m-2}x~\cos^nx \, dx\label{8.trigintegrals.sinncosn2}\tag{14}\\ &&\int \tan x\, dx = \ln |{\sec x}| +C = -\ln |{\cos x}| +C\tag{15}\\ &&\int \tan^m x\, dx = \frac{\tan^{m-1}x}{m-1} - \int \tan^{m-2}x \, dx\label{8.trigintegrals.tank}\tag{16} \\ &&\int \cot x\, dx = -\ln |{\csc x}| +C = \ln |{\sin x}| +C\tag{17}\\ &&\int \cot^m x\, dx = -\frac{\cot^{m-1}x}{m-1} - \int \cot^{m-2}x \, dx\tag{18}\\ &&\int \sec x\, dx = \ln \bigl|\sec x+\tan x \bigr| + C\label{8.trigintegrals.intofsec}\tag{19}\\ &&\int \sec^m x\, dx = \frac{\tan x~\sec^{m-2} x}{m-1} + \frac{m-2}{m-1}\int \sec^{m-2} x\, dx\label{8.trigintegrals.seck} \tag{20}\\ &&\int \csc x\, dx = \ln \bigl|{\csc x-\cot x}\bigr| + C\tag{21}\\ &&\int \csc^m x\, dx = - \frac{\cot x\csc^{m-2} x}{m-1} + \frac{m-2}{m-1}\int \csc^{m-2} x\,dx\tag{22}\\ &&\int \sin mx\sin nx\, dx = \frac{\sin(m-n)x}{2(m-n)} - \frac{\sin(m+n)x}{2(m + n)} +C \quad( m\ne \pm n) \label{8.trigintegrals.mnsin}\tag{23}\\ &&\int \sin mx~\cos nx\, dx = -\frac{\cos(m-n)x}{2(m-n)} - \frac{\cos(m+n)x}{2(m+n)} +C\quad( m\ne \pm n) \label{8.trigintegrals.mnsincos}\tag{24}\\ &&\int \cos mx~\cos nx\, dx = \frac{\sin(m-n)x}{2(m-n)} + \frac{\sin(m+n)x}{2(m+n)} +C\quad( m\ne \pm n)\tag{25} \end{eqnarray} \] |
Then use the substitution \(u = \cos x\), \(du = -\sin x\, dx\).
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Then use the substitution \(u = \sin x\), \(du = \cos x\, dx\).
Expand the right-hand side to obtain a sum of powers of \(\cos x\) or powers of \(\sin x\). Then use the reduction formulas \[ \begin{align*} \int \sin^n x\, dx &= -\frac{1}{n}\sin^{n-1} x~\cos x + \frac{n-1}{n}\int \sin^{n-2} x\,dx \\ \int \cos^n x\, dx &= \frac{1}{n}\cos^{n-1}x~\sin x + \frac{n- 1}{n}\int \cos^{n-2}x\, dx \end{align*} \]