Find the integral. Select the correct answer.
$$\int \tan ^ { 3 } x \sec ^ { 5 } x d x$$
A) $\frac { 1 } { 5 } \sec ^ { 5 } x - \frac { 1 } { 3 } \sec ^ { 3 } x + C$
B) $\frac { 1 } { 5 } \sec ^ { 5 } x + \frac { 1 } { 3 } \sec ^ { 3 } x + C$
C) $\frac { 1 } { 7 } \sec ^ { 7 } x - \frac { 1 } { 5 } \sec ^ { 5 } x + C$
D) $\frac { 1 } { 7 } \sec ^ { 7 } x + \frac { 1 } { 5 } \sec ^ { 5 } x + C$

Question

Quizplus · Accepted Answer

We can rearrange the integrand as $\sec^2x\cdot	an x\cdot\sec^4xdx$. Let $u=\sec x,\ du=\sec x 	an x dx$. Then we can rewrite the integral as $\int u^4du - \int u^6du = \frac { 1 } { 5 } \sec ^ { 5 } x - \frac { 1 } { 7 } \sec ^ { 7 } x + C$. Thus, the answer is $\boxed{	extbf{(C)}}$.

Find the Integral ∫tan⁡3xsec⁡5xdx\int \tan ^ { 3 } x \sec ^ { 5 } x d x∫tan3xsec5xdx

Find the Integral $\int \tan ^ { 3 } x \sec ^ { 5 } x d x$