TITLE: Integral $ \int \frac{\operatorname d\!x}{\sin^3 x} $
QUESTION [5 upvotes]: 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.
 $$\int \frac {1}{\sin^3(x)} dx$$

REPLY [15 votes]: 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.