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?