The integral $\int \frac { \sec ^ { 2 } x } { \tan ^ { 3 } x } d x$ is

A) $- \frac { \cot ^ { 2 } x } { 2 } + C$
B) $\csc x + C$
C) $\frac { \csc ^ { 2 } x } { 2 } + C$
D) $\frac { \sec ^ { 2 } x } { 2 } + C$
E) $\frac { \tan ^ { 2 } x } { 2 } + C$

Question

The integral $\int \frac { \sec ^ { 2 } x } { \tan ^ { 3 } x } d x$ is

A) $- \frac { \cot ^ { 2 } x } { 2 } + C$
B) $\csc x + C$
C) $\frac { \csc ^ { 2 } x } { 2 } + C$
D) $\frac { \sec ^ { 2 } x } { 2 } + C$
E) $\frac { \tan ^ { 2 } x } { 2 } + C$

Quizplus · Accepted Answer

The integral can be simplified by substituting $u = \tan x$, which leads to $du = \sec^2 x dx$. The integral becomes $\int \frac{1}{u^3} du$, which simplifies to $-\frac{1}{2u^2} + C$, or $-\frac{\cot^2 x}{2} + C$.

The Integral ∫sec⁡2xtan⁡3xdx\int \frac { \sec ^ { 2 } x } { \tan ^ { 3 } x } d x∫tan3xsec2x​dx

The Integral $\int \frac { \sec ^ { 2 } x } { \tan ^ { 3 } x } d x$