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!   

Let $t=\cos x$, then
$$\int \frac{\cos^4 x}{\sin^3 x}dx=-\int \frac{t^4}{(1-t^2)^2}dt.$$
Can you proceed?
