Answer
$\frac{1}{2}tan^{2}x+C$
Work Step by Step
Given:
$\int \frac{csc^{2}x}{cot^{3}x}dx$
Rewrite:
$\int (\frac{1}{sin^{2}x})\div (\frac{cos^{3}x}{sin^{3}x})dx$
$\int (\frac{1}{sin^{2}x})(\frac{sin^{3}x}{cos^{3}x})dx$
$\int \frac{sinx}{cos^{3}x}dx$
$\int tanxsec^{2}xdx$
Let $u=tanx$
$u=tanx$
$\frac{du}{dx}=sec^{2}x$
$dx=\frac{1}{sec^{2}x}du$
$=\int udu$
$=\frac{1}{2}u^{2}+C$
$=\frac{1}{2}tan^{2}x+C$
