Use any method to evaluate the integral.
-$$\int \frac { 6 \csc ^ { 3 } x } { \tan x } d x$$
A) $- 2 \csc ^ { 3 } x + C$
B) $- 2 \csc ^ { 4 } x + C$
C) $\frac { 3 } { 2 } \csc ^ { 4 } x \cot x + C$
D) $- 2 \cot ^ { 3 } x + C$

Question

Quizplus · Accepted Answer

The integral can be simplified using trigonometric identities. $\csc x = \frac{1}{\sin x}$ and $\tan x = \frac{\sin x}{\cos x}$, so $\frac{\csc^3 x}{\tan x} = \frac{1}{\sin^3 x} \cdot \frac{\cos x}{\sin x} = \frac{\cos x}{\sin^4 x} = \cos x \csc^4 x$. The integral becomes $\int 6 \cos x \csc^4 x dx$. Recognizing that the derivative of $\csc x$ is $-\csc x \cot x$, and thus the derivative of $\csc^3 x$ involves $\cos x$, we can directly integrate to get $-2 \csc^3 x + C$.

Use Any Method to Evaluate the Integral ∫6csc⁡3xtan⁡xdx\int \frac { 6 \csc ^ { 3 } x } { \tan x } d x∫tanx6csc3x​dx

Use Any Method to Evaluate the Integral $\int \frac { 6 \csc ^ { 3 } x } { \tan x } d x$