# 8.3 | Trigonometric Integrals

For trigonometric integrals of the form
$$\int \sin ^m x \cdot \cos ^n x \; dx,$$
if

• $$m$$ is odd, then retain a factor of $$\sin x$$ and use the identity $$\sin^2 x = 1- \cos^2 x$$ to rewrite the remaining power of $$\sin x$$ in terms of $$\cos x$$
• $$n$$ is odd, then retain a factor of $$\cos x$$ and use the identity $$\cos^2 = 1 – \sin^2 x$$ to rewrite the remaining power of $$\cos x$$ in terms of $$\sin x$$
• $$m$$ and $$n$$ are both even, then apply the identities
$$\sin^2 x = \frac{1- \cos(2x)}{2} \text{ and } \cos^2 x = \frac{1+ \cos(2x)}{2}$$
$$\sin^2 x = \frac{1- \cos(2x)}{2}$$
and
$$\cos^2 x = \frac{1+ \cos(2x)}{2}$$
$$\sin^2 x = \frac{1- \cos(2x)}{2}$$
and
$$\cos^2 x = \frac{1+ \cos(2x)}{2}$$

to reduce the powers.

Then apply $$u$$-substitution.

For trigonometric integrals of the form
$$\int \sec^m x \cdot \tan^n x \; dx,$$
if

• $$n$$ is odd, then retain a factor of $$\sec x \tan x$$ and use the identity $$\tan^2 x = \sec^2x -1$$ to rewrite the remaining power of $$\tan x$$ in terms of $$\sec x$$.
• $$m$$ is even, then retain a factor of $$\sec^2 x$$ and use the identity $$\sec^2x = \tan^2 x +1$$ to rewrite the remaining power of $$\sec x$$ in terms of $$\tan x$$.

Then apply $$u$$-substitution.

Trigonometric Identities
\begin{aligned} \sin m x \sin nx & = \frac{1}{2}[ \cos(m-n)x – \cos(m+n)x] \\ \sin m x \cos n x & = \frac{1}{2}[ \sin(m-n)x + \sin(m+n)x] \\ \cos mx \cos nx & = \frac{1}{2} [ \cos(m-n)x + \cos(m+n)x] \end{aligned}
Find
$$\int \sin^7 x \cos^4 x \; dx.$$

\begin{aligned} \int \sin^7 x \cos^4 x \; dx & = \int \sin x (\sin^6 x) \cos ^4 x \; dx \\ & = \int \sin x (\sin^2 x )^3 \cos^4 x \; dx \\ & = \int \sin x (1- \cos^2 x)^3 \cos^4 x \; dx \\ & = \int \sin x (1- u^2)^3 u^4 \frac{du}{- \sin x} \\ & = -\int (1- u^2)^3 u^4 \; du \\ & = -\int (1-3u^2+3u^4 +u^6) u^4 \; du \\ & = -\int u^4-3u^6+3u^8 +u^{10} \; du \\ & = -\frac{u^5}{5}+\frac{3u^7}{7}-\frac{u^9}{3} -\frac{u^{11}}{11}+C \\ & = -\frac{\cos^5x}{5}+\frac{3\cos^7x}{7}-\frac{\cos^9x}{3} -\frac{\cos^{11}x}{11}+C \\ \end{aligned}
\begin{aligned} \int & \sin^7 x \cos^4 x \; dx \\ & = \int \sin x (\sin^6 x) \cos ^4 x \; dx \\ & = \int \sin x (\sin^2 x )^3 \cos^4 x \; dx \\ & = \int \sin x (1- \cos^2 x)^3 \cos^4 x \; dx \\ & = \int \sin x (1- u^2)^3 u^4 \frac{du}{- \sin x} \\ & = -\int (1- u^2)^3 u^4 \; du \\ & = -\int (1-3u^2+3u^4 +u^6) u^4 \; du \\ & = -\int u^4-3u^6+3u^8 +u^{10} \; du \\ & = -\frac{u^5}{5}+\frac{3u^7}{7}-\frac{u^9}{3} -\frac{u^{11}}{11}+C \\ & = -\frac{\cos^5x}{5}+\frac{3\cos^7x}{7}-\frac{\cos^9x}{3} -\frac{\cos^{11}x}{11}+C \end{aligned}
\begin{aligned} & \int \sin^7 x \cos^4 x \; dx \\ & = \int \sin x (\sin^6 x) \cos ^4 x \; dx \\ & = \int \sin x (\sin^2 x )^3 \cos^4 x \; dx \\ & = \int \sin x (1- \cos^2 x)^3 \cos^4 x \; dx \\ & = \int \sin x (1- u^2)^3 u^4 \frac{du}{- \sin x} \\ & = -\int (1- u^2)^3 u^4 \; du \\ & = -\int (1-3u^2+3u^4 +u^6) u^4 \; du \\ & = -\int u^4-3u^6+3u^8 +u^{10} \; du \\ & = -\frac{u^5}{5}+\frac{3u^7}{7}-\frac{u^9}{3} -\frac{u^{11}}{11}+C \\ & = -\frac{\cos^5x}{5}+\frac{3\cos^7x}{7}-\frac{\cos^9x}{3} \\ & \qquad -\frac{\cos^{11}x}{11}+C \\ \end{aligned}