Partial Integration $ \int \frac{x\cos x}{\sin^3x}dx $ The problem: $$ \int \frac{x\cos x}{\sin^3x}dx $$ Can someone give me a hint on how to solve this without using cosecant? The solution provided is: $$ -\frac{1}{2}\left(\frac{x}{\sin^2x}+\cot x\right)\:+\:C $$

HINT:

$$\int x\cdot\frac{\cos x}{\sin^3x}dx=x\int\frac{\cos x}{\sin^3x}dx-\int\left(\frac{dx}{dx}\int\frac{\cos x}{\sin^3x}dx\right)dx$$

For $\int\dfrac{\cos x}{\sin^3x}dx$ write $\sin x=u$