Integration of $1/\sin^3 x$ 
I need a explanation of this problem: 
     $$  \int \frac{1}{\sin^3 x}\,dx $$ 
Change the variable $$ t = \tan (x/2) $$
  With use of $\tan$, $\cos$, $\sin$ and $\cot$, only.

So how do I think of this and what are the steps? 
Progress
Well I have $\sin x = 2t/(1+t^2)$ and $t = \tan x/2$ gives 
$$2 \arctan t = x,\quad  \frac{2}{(1+t^2)} dt = dx$$
 This all gave me 
$$\int \left(\frac{2t}{(1+t^2)}\right)^3 (2/(1+t^2))\,dt$$

$$\int \left(\frac{2t}{(1+t^2)}\right)^3 (2/(1+t^2))\,= \int \frac{16t^3}{(1+t^2)^2}\,dt= \int \frac{16t(1+t^2-1)}{(1+t^2)^2}dt$$
Now, the substitution $u=t^2+1$ solves the problem.
