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.

write your integrand as $$\frac{1}{\sin(x)}+\frac{\cos(x)^2}{\sin(x)^3}$$ and the first as $$\sin(x)+\frac{\cos(x)^2}{\sin(x)}$$
now you can set $$t=\sin(x)$$
