Solve integral $\int_0^{2\pi}\frac{\sin^2\frac{(N+1)x}{2}}{2\pi(N+1)\sin^2\frac x2} dx $ My question is to solve
$$\int_0^{2\pi}\frac{\sin^2\left(\frac{N+1}{2}x\right)}{2\pi(N+1)\sin^2(x/2)} dx  $$
Up to $1000$ or more, Wolfram-Alpha solves this integral and gives the value $1$ for every integer. But, I couldn't prove that this is the case for any $N\in\mathbb{N}$.
I know if we didn't have the $N+1$ part and squares, this integral should be $1$. We also have
$$ \frac{\sin^2\left(\frac{N+1}{2}x\right)}{\sin^2(x/2)}=\frac{\cos((N+1)x)-1}{\cos(x)-1}=\sum_{n=0}^N\left(\sum_{k=-n}^n e^{inx} \right)$$
and
$$ \frac{\sin\left((N+\frac{1}{2})x\right)}{\sin(x/2)}=\sum_{n=-N}^Ne^{inx}= \sum_{k=1}^N2\cos(kx)+1$$

\begin{align}
\frac{1}{2\pi(N+1)}\int_0^{2\pi}\frac{\sin^2\left(\frac{N+1}{2}x\right)}{\sin^2\left(\frac{x}{2}\right)} dx &\overset{\color{blue}{u = \frac{x}{2}}, \, \color{green}{n=N+1}}{=}\frac{1}{2\pi n}\color{blue}{2}\int_0^{\color{blue}{\pi}}\frac{\sin^2(nu)}{\sin^2(u)} du\\
& \overset{(\color{purple}{\text{Even})}}{=} \frac{1}{2\pi n}\int_{\color{purple}{-\pi}}^{\pi}\frac{\sin^2(nu)}{\sin^2(u)} du\\
& \overset{\color{blue}{s = u +\pi}}{=} \frac{1}{2\pi n}\int_{\color{blue}{0}}^{\color{blue}{2\pi}}\frac{(-1)^{2n}\sin^2(ns)}{(-1)^{2n}\sin^2(s)} ds\\
& = \frac{1}{2\pi n}2n\pi =1
\end{align}
where on the last step you use $\int_{0}^{2\pi} \frac{\sin^2(nx)}{\sin^2(x)} dx=2n\pi$ with $n \in \mathbb{N}$.
