Find the indefinite integral. $\int \tan ^ { 3 } x \sec ^ { 2 } x d x$
A) $- \frac { 1 } { 6 } \tan ^ { 3 } x \sec ^ { 2 } x + C$
B) $\frac { 1 } { 12 } \tan ^ { 4 } x \sec ^ { 3 } x + C$
C) $\frac { 1 } { 4 } \tan ^ { 4 } x + C$
D) $\frac { 1 } { 4 } \tan ^ { 4 } x \sec ^ { 4 } x + C$
E) $\frac { 1 } { 4 } \sec ^ { 4 } x + C$

Question

Quizplus · Accepted Answer

To solve the integral $\int \tan^3 x \sec^2 x dx$, use the substitution $u = \tan x$, which implies $du = \sec^2 x dx$. The integral becomes $\int u^3 du$, which evaluates to $\frac{1}{4}u^4 + C$ or $\frac{1}{4}\tan^4 x + C$.

Find the Indefinite Integral ∫tan⁡3xsec⁡2xdx\int \tan ^ { 3 } x \sec ^ { 2 } x d x∫tan3xsec2xdx

Find the Indefinite Integral $\int \tan ^ { 3 } x \sec ^ { 2 } x d x$