I need to calculate the following integral:
$\int \frac{dx}{\sin^3x}$
I noticed that $\int \frac{dx}{\sin^3x}=\int \frac{dx}{\sin x \sin^2x}=-\int \frac{dx}{-\sin x (1-\cos^2x)} (A)$
Let $v=\cos u \Leftrightarrow dv=-\sin u du$
Therefore: $(A)= -\int \frac{dv}{(1-v^2)^2} $ Is that correct ?
How do I calculate then the final integral.
Thank you in advance