Why is this integral equal to $-\frac{2n}{\pi}\cdot\frac{\cos({1\over2}\pi n)}{n^2-1}$

Question

could someone provide some indication about why the following equality is true (for $n\geq 2) $ please$${1\over\pi}\int\limits_{-\pi}^\pi\sin(x)\sin(nx)dx=-\frac{2n}{\pi}\cdot\frac{\cos({1\over2}\pi n)}{n^2-1}$$

Answer 1

Hint: $$\sin(\alpha)\sin(\beta)=\frac{\cos(\alpha-\beta)-\cos(\alpha+\beta)}{2}$$