Integral $ \int \frac{\operatorname d\!x}{\sin^3 x} $ I need to calculate the following integral for my homework, but I dont know how. If someone show me step by step solution I would really appreciate it.
 $$\int \frac {1}{\sin^3(x)} dx$$

Substitute $u=\tan\frac{x}{2}$. Then $\frac{x}{2}=\arctan u\Rightarrow x=2\arctan u$. This means $$\frac{dx}{du}=\frac{2}{1+u^2}$$
In addition, $$\sin x=\frac{2u}{u^2+1}$$
Then,
$$\int\frac{1}{\sin^3x}dx=\frac{2}{8}\int\frac{(u^2+1)^2}{u^3}du=\frac{1}{4}\int u+\frac{2}{u}+\frac{1}{u^3}du$$
which I think you know how to solve.
What I used here is the classic but always useful Tangent half-angle formula 
