How to take the integral $\int\frac{dx}{\sin^3x}$ There's
$$\int\frac{\mathrm dx}{\sin^3x}.$$
I tried to write it like
$$\int\frac{(\sin^2x+\cos^2x)}{\sin^3x}\,\mathrm dx,$$
and then made partial fractions from it, but it didn't help much, the answer is still incorrect.

Hint:
Bioche's rules lead to to make the substitution $u=\cos x$, $\;\mathrm d u=\sin x \,\mathrm d x$. You obtain
$$\int\frac{\mathrm d x}{\sin^3 x}=\int\frac{-\mathrm d u}{\sin^4 x}=\int\frac{-\mathrm d u}{(1-u^2)^2}.$$
There remains to compute the partial fraction decomposition of $\;\dfrac{-1}{(1-u^2)^2}$.
