What is $\int\frac{dx}{\sin x}$? I'm looking for the antiderivatives of $1/\sin x$. Is there even a closed form of the antiderivatives? Thanks in advance. 

Hint:   Write this as $$\int \frac{\sin (x)}{\sin^2 (x)} dx=\int \frac{\sin (x)}{1-\cos^2(x)} dx.$$  Now let $u=\cos(x)$, and use the fact that $$\frac{1}{1-u^2}=\frac{1}{2(1+u)}+\frac{1}{2(1-u)}.$$
Added: I want to give credit to my friend Fernando who taught me this approach.
