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. $$\int \frac {1}{\sin^3(x)} dx$$

There is a standard first-year calculus response. Rewrite the integrand as $\dfrac{\sin x}{\sin^4 x}=\dfrac{\sin x}{(1-\cos^2 x)^2}$. The substitution $u=\cos x$ leaves us integrating $-\dfrac{1}{(1-u^2)^2}$. Now partial fractions.

There are in many cases more efficient procedures, but one can in principle handle in this way all $$\int \sin^m x\cos^n x\,dx,$$ where $m$ and $n$ are integers and at least one of $m$ and $n$ is odd.

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