$\displaystyle \int{csc2x}dx$

1. I used the identity $\displaystyle csc2x=\frac{1}{2\sin{x}\cos{x}}$

I tried letting $\displaystyle u=\sin{x}$. So

$\displaystyle \frac{1}{2}\int{\frac{1}{u-u^3}}du$.

I need an explanation from top to bottom guys, please.

I'm purposefully trying not to use partial fraction decomp.