Answer
$$\frac{-1}{2}\sin^{-2}x -2\ln \sin x +\frac{1}{2}\sin^2 x+c$$
Work Step by Step
\begin{align*}
\int \frac{\cos ^{5} x}{\sin ^{3} x} d x&= \int \sin^{-3}x\cos^4x\cos xdx\\
&= \int \sin^{-3}x(1-\sin^2x)^2\cos xdx\\
&=\int \sin^{-3}x(1-2\sin^2 x+\sin^4x)\cos xdx\\
&= \int \left( \sin^{-3}x -2\frac{1}{\sin x} + \sin x\right)\cos xdx\\
&= \frac{-1}{2}\sin^{-2}x -2\ln \sin x +\frac{1}{2}\sin^2 x+c
\end{align*}
