Evaluate the integral.
-$$\int \tan ^ { 5 } 3 x d x$$
A) $\frac { 1 } { 4 } \tan ^ { 4 } 3 x - \frac { 1 } { 2 } \tan ^ { 2 } 3 x - \ln | \cos x | + C$
B) $\frac { 1 } { 12 } \tan ^ { 4 } 3 x - \frac { 1 } { 6 } \tan ^ { 2 } 3 x - \frac { 1 } { 3 } \ln | \cos x | + C$
C) $- \frac { 1 } { 12 } \tan ^ { 4 } 3 x + \frac { 1 } { 6 } \tan ^ { 2 } 3 x - \frac { 1 } { 3 } \ln | \cos x | + C$
D) $\frac { 1 } { 12 } \tan ^ { 4 } 3 x - \frac { 1 } { 6 } \tan ^ { 2 } 3 x + \frac { 1 } { 3 } \ln | \cos x | + C$

Question

Quizplus · Accepted Answer

Using the substitution $u = \tan 3x$, $du = 3 \sec^2 3x dx$, we get$$\int \tan^5 3x dx = \frac{1}{3} \int u^5 \frac{du}{\sec^2 3x} = \frac{1}{3} \int u^5 \cos^2 3x dx$$$$= \frac{1}{3} \int u^5 (1 - \sin^2 3x) dx = \frac{1}{3} \int (u^5 - u^5 \sin^2 3x) dx$$$$= \frac{1}{3} \left(\frac{1}{6} u^6 - \frac{1}{2} u^4 + C\right) = \frac{1}{18} u^6 - \frac{1}{6} u^4 + C$$Substituting back $u = \tan 3x$, we get$$\int \tan^5 3x dx = \frac{1}{18} \tan^6 3x - \frac{1}{6} \tan^4 3x + C$$

Evaluate the Integral ∫tan⁡53xdx\int \tan ^ { 5 } 3 x d x∫tan53xdx

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