I would like to compute:
$ \int_{0}^{2\pi} \frac{\sin(nx)^2}{\sin(x)^2} \mathrm dx $
I think that :
$ \int_{0}^{2\pi} \frac{\sin(nx)^2}{\sin(x)^2} \mathrm dx=2n\pi $
I tried to use induction:
$ \sin((n+1)x)^2=\sin(nx)^2(1-\sin(x)^2)+\sin(x)^2(1-\sin(nx)^2)+2\sin(x)\cos(x)\sin(nx)\cos(nx)$
$ \frac{\sin((n+1)x)^2}{\sin(nx)}=\frac{\sin(nx)^2}{\sin(x)^2}-\sin(nx)^2+1-\sin(nx)^2+2\frac{\cos(x)}{\sin(x)}\sin(nx)\cos(nx)$
$ \int_{0}^{2\pi}\frac{\sin((n+1)x)^2}{\sin(nx)} \mathrm dx=2(n+1)\pi+2\int_{0}^{2\pi}\sin(nx)(\frac{\cos(x)}{\sin(x)}\cos(nx)-\sin(nx)) \mathrm dx $
So does $ \int_{0}^{2\pi}\sin(nx)(\frac{\cos(x)}{\sin(x)}\cos(nx)-\sin(nx)) \mathrm dx=0 $ hold?
But this is probably not the best method.
Could you help me?