In this section we will learn to evaluate trigonometric integrals of the form:
\[ \int \sin^m x \cos^n x\;dx,\; m,n\geq 0 \]
\[ \int \tan^m x \sec^n x\;dx,\; m,n\geq 0 \]
There are three cases for \(\displaystyle\int \sin^m x \cos^n x\;dx\):
- The power of sine is odd, i.e., \(m=2k+1\) for some integer \(k\geq 0\).
Steps:
\begin{align*}
\int \sin^{2k+1} x \cos^n x\;dx &= \int (\sin^2 x)^k \cos^n x \sin x\;dx \\
&= \int (1-\cos^2 x)^k \cos^n x \sin x\;dx \\
&= \int (1-u^2)^k u^n \;(-du) \,\, \left(\text{Let } u=\cos x.\; \text{Then } du=-\sin x \,dx \right)
\end{align*}
Example.
Evaluate \(\displaystyle \int \sin^3 x \cos x \,dx\).
Solution.
\[\begin{align*}
\int \sin^3 x \cos x\;dx &= \int \sin^2 x \cos x \sin x\;dx \\
&= \int (1-\cos^2 x) \cos x \sin x\;dx \\
&= \int (1-u^2) u \;(-du) \,\, \left(\text{Let } u=\cos x.\; \text{Then } du=-\sin x \,dx \right)\\
&= \int (-u+u^3) \;du\\
&= -\frac{u^2}{2}+\frac{u^4}{4} +C\\
&= -\frac{1}{2}\cos^2 x+\frac{1}{4}\cos^4 x +C.
\end{align*}\]
- The power of cosine is odd, i.e., \(n=2k+1\) for some integer \(k\geq 0\).
Steps:
\[\begin{align*}
\int \sin^m x \cos^{2k+1} x \;dx &= \int \sin^m x (\cos^2 x)^k \cos x\;dx \\
&= \int \sin^m x (1-\sin^2 x)^k \cos x\;dx \\
&= \int u^m (1-u^2)^k \;du \,\, \left(\text{Let } u=\sin x.\; \text{Then } du=\cos x \,dx \right)
\end{align*}\]
Example.
Evaluate \(\displaystyle \int \sin^2 x \cos^5 x \,dx\).
Solution.
\[\begin{align*}
\int \sin^2 x \cos^5 x \;dx &= \int \sin^2 x (\cos^2 x)^2 \cos x\;dx \\
&= \int \sin^2 x (1-\sin^2 x)^2 \cos x\;dx \\
&= \int u^2 (1-u^2)^2 \;du \,\, \left(\text{Let } u=\sin x.\; \text{Then } du=\cos x \,dx \right)\\
&= \int u^2 (1-2u^2+u^4) \;du\\
&= \int (u^2-2u^4+u^6) \;du\\
&= \frac{u^3}{3}-2\frac{u^5}{5}+\frac{u^7}{7}+C\\
&= \frac{1}{3}\sin^3 x-\frac{2}{5}\sin^5 x+\frac{1}{7}\sin^7 x+C.
\end{align*}\]
- The powers of sine and cosine are both even.
Steps: Write the integrand as an expression of cosines using the following trigonometric identities.
\[\begin{align*}
\cos^2 x &= \frac{1}{2}\left( 1+\cos (2x) \right) \\
\sin^2 x &= \frac{1}{2}\left( 1-\cos (2x) \right) \\
%\sin x \cos x &= \frac{1}{2} \sin(2x) \\
\cos x \cos y &= \frac{1}{2}\left[ \cos (x+y)+\cos(x-y) \right]
\end{align*}\]
Example.
Evaluate \(\displaystyle \int \sin^2 x \cos^2 x \,dx\).
Solution.
\[\begin{align*}
\sin^2 x \cos^2 x &= \frac{1}{2}\left( 1-\cos (2x) \right) \frac{1}{2}\left( 1+\cos (2x) \right) \\
&= \frac{1}{4} \left( 1-\cos^2 (2x) \right)\\
&= \frac{1}{4} \left[ 1-\frac{1}{2}\left( 1+\cos (4x) \right) \right]\\
&=\frac{1}{8} \left[ 1-\cos (4x) \right].\\
&\\
\int \sin^2 x \cos^2 x \,dx &= \int \frac{1}{8} \left[ 1-\cos (4x) \right]\;dx \\
&= \frac{1}{8} \int \left[ 1-\cos (4x) \right]\;dx \\
&= \frac{1}{8} \left[ x-\frac{\sin (4x)}{4} \right]+C\\
&= \frac{x}{8}-\frac{1}{32}\sin (4x)+C.
\end{align*}\]
For an alternative solution, we use the identity \(\sin(2x)=2\sin x \cos x\) which implies
\[ \sin^2 x \cos^2 x = (\sin x \cos x)^2 =\frac{\sin^2(2x)}{4}=\frac{1}{8} \left[ 1-\cos (4x) \right]. \]
Note that trigonometric (product-to-sum) identities can also be used to evaluate integrals of the form:
\[ \int \sin(mx) \cos(nx) \;dx,\; \int \sin(mx) \sin(nx) \;dx,\; \int \cos(mx) \cos(nx) \;dx. \]
Example.
Evaluate \(\displaystyle \int \cos(3x) \cos(5x) \,dx\).
Solution.
\[\begin{align*}
\cos(3x) \cos(5x) &= \frac{1}{2}\left[ \cos (3x+5x)+\cos(3x-5x) \right] \\
&=\frac{1}{2}\left[ \cos (8x)+\cos(-2x) \right] \\
&=\frac{1}{2}\left[ \cos (8x)+\cos(2x) \right].\\
&\\
\int \cos(3x) \cos(5x) \,dx &= \int \frac{1}{2}\left[ \cos (8x)+\cos(2x) \right]\;dx \\
&= \frac{1}{2} \int \left[ \cos (8x)+\cos(2x) \right]\;dx \\
&= \frac{1}{2} \left[ \frac{\sin (8x)}{8}+\frac{\sin (2x)}{2} \right]+C\\
&= \frac{1}{16}\sin (8x)+\frac{1}{4}\sin (2x) +C.
\end{align*}\]
To evaluate \(\displaystyle\int \tan^m x \sec^n x\;dx\), first recall the formulas:
\[\begin{align*}
\int \tan x \;dx &= \ln|\sec x| +C\\
\int \sec x \;dx &= \ln|\sec x +\tan x| +C
\end{align*}\]
- \(m=2k+1\) for some integer \(k\geq 0\) and \(n\geq 1\).
Steps:
\[\begin{align*}
\int \tan^{2k+1} x \sec^n x\;dx &= \int (\tan^2 x)^k \tan x \sec^{n-1} x \sec x\;dx \\
&= \int (\sec^2 x-1)^k \sec^{n-1} x \sec x \tan x \;dx \\
&= \int (u^2-1)^k u^{n-1} \;du \,\, \left(\text{Let } u=\sec x.\; \text{Then } du=\sec x \tan x \,dx \right)
\end{align*}\]
Example.
Evaluate \(\displaystyle\int \tan^3 x \sec^2 x\;dx\).
Solution.
\[\begin{align*}
\int \tan^3 x \sec^2 x\;dx &= \int \tan^2 x \tan x \sec x \sec x\;dx \\
&= \int (\sec^2 x-1) \sec x \sec x \tan x \;dx \\
&= \int (u^2-1) u \;du \,\, \left(\text{Let } u=\sec x.\; \text{Then } du=\sec x \tan x \,dx \right)\\
&= \int (u^3-u) \;du\\
&= \frac{u^4}{4}-\frac{u^2}{2}+C\\
&= \frac{1}{4}\sec^4 x-\frac{1}{2}\sec^2 x+C.
\end{align*}\]
- \(n=2k\) for some integer \(k\geq 1\).
Steps:
\[\begin{align*}
\int \tan^m x \sec^{2k} x\;dx &= \int \tan^m x (\sec^{2} x)^{k-1} \sec^2 x\;dx \\
&= \int \tan^m x (1+\tan^{2} x)^{k-1} \sec^2 x\;dx \\
&= \int u^m (1+u^2)^{k-1} \;du \,\, \left(\text{Let } u=\tan x.\; \text{Then } du=\sec^2 x \,dx \right)
\end{align*}\]
Example.
Evaluate \(\displaystyle\int \tan^2 x \sec^4 x\;dx\).
Solution.
\[\begin{align*}
\int \tan^2 x \sec^4 x\;dx &= \int \tan^2 x \sec^{2} x \sec^2 x\;dx \\
&= \int \tan^2 x (1+\tan^{2} x) \sec^2 x\;dx \\
&= \int u^2 (1+u^2) \;du \,\, \left(\text{Let } u=\tan x.\; \text{Then } du=\sec^2 x \,dx \right)\\
&= \int (u^2+u^4) \;du\\
&= \frac{u^3}{3}+\frac{u^5}{5}+C\\
&= \frac{1}{3}\tan^3 x+\frac{1}{5}\tan^5 x+C.
\end{align*}\]
- \(m\) is even and \(n\) is odd.
Steps: Write the integrand as an expression of powers of secants by using the identity \(\tan^2 x=\sec^2 x-1\)
and then apply the following reduction formula:
\[ \int \sec^n x \,dx=\frac{1}{n-1} \tan x\sec^{n-2}x +\frac{n-2}{n-1}\int \sec^{n-2}x \, dx,\; n\geq 2. \]
Example.
Evaluate \(\displaystyle\int \tan^2 x \sec^3 x\;dx\).
Solution.
\[\begin{align*}
\int \tan^2 x \sec^3 x\,dx &= \int (\sec^2 x-1) \sec^3 x\,dx \\
&= \int (\sec^5 x-\sec^3 x) \,dx \\
&= \int \sec^5 x \;dx -\int \sec^3 x \,dx \\
&=\frac{1}{4} \tan x\sec^{3}x +\frac{3}{4}\int \sec^3 x \,dx -\int \sec^3 x \,dx\\
&=\frac{1}{4} \tan x\sec^{3}x -\frac{1}{4}\int \sec^3 x \,dx \\
&=\frac{1}{4} \tan x\sec^{3}x -\frac{1}{4} \left( \frac{1}{2} \tan x\sec x +\frac{1}{2}\int \sec x \, dx \right) \\
&= \frac{1}{4} \tan x\sec^{3}x-\frac{1}{8} \tan x \sec x-\frac{1}{8} \ln|\sec x+\tan x| +C.
\end{align*}\]
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