analysis - Integral $int_0^pi cot(x/2)sin(nx),dx$

It seems that $$\int_0^\pi \cot(x/2)\sin(nx)\,dx=\pi$$ for all positive integers $n$.

But I have trouble proving it. Anyone?

Answer

Use this famous sum:

$$1+2\cos x+2\cos 2x+\cdots+2\cos nx=\frac{\sin (n+\frac{1}{2})x}{\sin \frac{x}{2}}=\sin nx\cot\left(\frac{x}{2}\right)+\cos nx$$

Hence

$$\int_0^{\pi}\cot \left(\frac{x}{2}\right)\sin n x\,dx=\int_0^{\pi}1+2\cos x+2\cos 2x+\cdots +\cos nx\,dx$$

All cosine terms obviously evaluate to zero.