Problem with Indefinite Integral $\int\frac {\cos^4x}{\sin^3x} dx$ I'm stuck with this integral
$\int\frac {\cos^4x}{\sin^3x} dx$  
which I rewrote as
$\int \csc^3x \cos^4xdx$
then after using the half angle formula twice for $\cos^4x$ I got this
$\frac 14\int \csc^3x (1+\cos(2x))(1+\cos(2x))dx$ 
then after solving those products I got these integrals
$\frac 14 \{\int \csc^3xdx+2\int \csc^3x \cos(2x)dx + \int \csc^3x \cos^2(2x)dx\}$
I do know how to solve the $\int \csc^3xdx$ one but I'm totally lost on the other ones, any tips/help/advice would be highly appreciated! thanks in advance guys!   

Integrate by parts instead
\begin{align}
\int\frac {\cos^4x}{\sin^3x} dx
=&-\frac12\int \cos^3x\ d\bigg(\frac1{\sin^2x}\bigg)\\
\overset{ibp}= &-\frac{\cos^3x}{2\sin^2x}+\frac32\int \left(\sin x-\frac{1}{\sin x}\right)dx\\
 = &-\frac{\cos^3x}{2\sin^2x}-\frac32\cos x+\frac32\tanh^{-1}(\cos x)
\end{align}
