TITLE: Computing $ \int_{0}^{2\pi} \frac{\sin(nx)^2}{\sin(x)^2} \mathrm dx $
QUESTION [15 upvotes]: 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?

REPLY [12 votes]: We have $$\frac{\sin(nx)}{\sin(x)} = \frac{e^{inx} - e^{-inx}}{e^{ix} - e^{-ix}} = e^{-i(n-1)x} \frac{\left(e^{2ix}\right)^n-1}{e^{2ix} - 1} = e^{-i(n-1)x} \sum_{k=0}^{n-1} e^{2ikx} $$
This shows that the Fourier series $$\sum_{k=-\infty}^\infty a_k e^{ikx}=\frac{\sin(nx)}{\sin(x)} = e^{-i(n-1)x} \sum_{k=0}^{n-1} e^{2ikx}$$ has exactly $n$ coefficients which are $=1$ and all others vanish. Therefore by Parseval's identity, we get
$$\frac 1{2\pi}\int_0^{2\pi} \frac{\sin^2(nx)}{\sin^2(x)}\, dx = \sum_{k=-\infty}^\infty |a_k|^2 = n$$
i.e. $$\int_0^{2\pi} \frac{\sin^2(nx)}{\sin^2(x)} \, dx = 2n\pi$$