Answer
$$ -\frac{1}{2}\cot^2 x-\ln \sin x+c$$
Work Step by Step
\begin{align*}
\int \cot^3 xdx&= \int \cot x\cot^2 xdx\\
&= \int \cot x(\csc^2 x-1)dx\\
&= \int \cot x\csc^2 xdx-\int \cot x dx\\
&= -\frac{1}{2}\cot^2 x-\ln \sin x+c
\end{align*}