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 
$$\int \left(\frac{2t}{(1+t^2)}\right)^3 (2/(1+t^2))\,dt$$

More simple is to change the variable by $t=\cos x$ so $dt=-\sin x dx$ and then
$$\int\frac{dx}{\sin^3 x}=-\int\frac{dt}{(1-t^2)^2}$$
Can you take it frome here?
