Conjecture:$\int_0^{\pi/2} \left(\frac{\sin(2n+1)x}{\sin x}\right)^\beta dx$ is integer multiple of $\pi/2$,for integer $\beta>2$ I was solving the integral $$I_n=\int_0^{\frac {\pi}{2}} \left(\frac {\sin ((2n+1)x)}{\sin x}\right)^2 dx$$
With $n\ge 0$ And $n\in \mathbb{N}$
On solving, I got $$I_n =\frac {(2n+1)\pi}{2}$$
But, due to curiosity, I started investigating the family of integrals as

$$I_n(\beta) =\int_0^{\frac {\pi}{2}} \left(\frac {\sin (2n+1)x}{\sin x}\right)^{\beta} dx$$
On trying various values of $\beta\gt 2$ and $\beta\in \mathbb{N}$, I conjectured that
$$I_n(\beta) =c_{\beta} \frac{\pi}{2}$$
where $c_{\beta}$ denotes "Number of arrays of $\beta$ integers in $-n$ to $n$ with sum $0$"

But, on trying a lot, I couldn't prove this statement. Also, I suppose that the statement could be proved with help of Dirichlet kernel, but I couldn't get the way out through it.
Any help and hints to prove/disprove the conjecture are greatly appreciated.
 A: $\def\b{\beta}$\begin{align*}
\newcommand\cmt[1]{{\small\textrm{#1}}}
I_n(\b) &= \int_0^{\pi/2}
\left(\frac{\sin (2n+1)x}{\sin x}\right)^\b dx \\
&= \frac 1 4 \int_0^{2\pi}
\left(\frac{\sin (2n+1)x}{\sin x}\right)^\b dx 
    & \cmt{begin similar to user630708} \\
&= \frac{1}{4i} \oint_\gamma
\left(\frac{z^{4n+2}-1}{z^2-1}\right)^\b
\frac{dz}{z^{2n\b+1}}
 & \cmt{let $z=e^{ix}$} \\
&= \frac{1}{4i} \oint_\gamma
\left(\sum_{k=0}^{2n}z^{2k}\right)^\b
\frac{dz}{z^{2n\b+1}}
 & \cmt{partial sum of geometric series} \\
&= \left.\frac{1}{4i} \frac{2\pi i}{(2n\b)!}
\left(\frac{d}{dz}\right)^{2n\b}
\left(\sum_{k=0}^{2n}z^{2k}\right)^\b \right|_{z=0}
 & \cmt{Cauchy integral formula} \\
&= \left.\frac{\pi}{2} \frac{1}{(2n\b)!}
\left(\frac{d}{dz}\right)^{2n\b}
\sum_{\sum x_k=\b} \frac{\b!}{\prod x_k!}
\prod (z^{2k})^{x_k}
\right|_{z=0}
 & \cmt{multinomial expansion, $k=0,1,\ldots,2n$} \\
&= \left.\frac{\pi}{2} \frac{1}{(2n\b)!}
\left(\frac{d}{dz}\right)^{2n\b}
\sum_{\sum x_k=\b} \frac{\b!}{\prod x_k!}
z^{2\sum k x_k}
\right|_{z=0} \\
&= \frac{\pi}{2}
\sum_{\sum x_k=\b \atop \sum k x_k = n\b}
\frac{\b!}{\prod x_k!} 
 & \cmt{only surviving terms have $\sum k x_k = n\b$} \\
&= \frac{\pi}{2}
\sum_{\sum x_k=\b \atop \sum (n-k) x_k = 0}
\frac{\b!}{\prod x_k!} 
\end{align*}
In the last line note that 
$\sum_{k=0}^{2n} n x_k=n\b$ and so
$\sum_{k=0}^{2n} (n-k)x_k = 0$.
By inspection one can see that
$$\sum_{\sum_{k=0}^{2n} x_k=\b \atop \sum_{k=0}^{2n} (n-k) x_k = 0}
\frac{\b!}{\prod x_k!}
= \textrm{number of arrays of $\b$ integers in $-n,\ldots,n$ with sum equal to 0,}$$
i.e., 
$$I_n(\b) = \frac{\pi}{2} T(\b,n),$$
where $T(\b,n)$ is OEIS A201552, as pointed out by James Arathoon in the comments.
(On that page we also find an integral form of $T(\b,n)$ which, after a simple substitution, gives $I_n(\b) = \frac{\pi}{2} T(\b,n)$.)
A: This is easy using Residue Theory:
Note that by symmetry $\int_{0}^{\pi/2}...dx=1/4\int_{-\pi}^{\pi}…dx$ (use parity and a sub $y=\pi-x$ to Show that).
employing $z=e^{ix}$ we get
$$
4 I_{n,\beta}=\oint_C \left[\frac{z^{4n+2}-1}{z^2-1}\right]^{\beta}\frac{dz}{i z^{2\beta n+1}}
$$
where $C$ denotes the unit circle in the comlex plane. By the residue theorem (there is one pole inside the contour at $z=0$, using f.e. the geometric series you can Show that the Points $z=\pm i$ are removable singularities). We have
$$
4 I_{n,\beta}=2\pi  \text{Res}(\left[\frac{z^{4n+2}-1}{z^2-1}\right]^{\beta}\frac{1}{z^{2\beta n+1}}
,z=0)
$$
Using $\beta!(z^2-1)^{-\beta}=((2z)^{-1}\partial_z)^{\beta-1}(z^2-1)^{-1}$ we get
$$
(1-z^2)^{-\beta}=\frac{1}{2^{\beta-1}}\sum_{m\geq0}\binom{m+\beta-1}{\beta-1}z^{2m}\\
(1-z^{4n+2})^{\beta}=z^{2\beta}\sum_{k\geq0}(-1)^k\binom{\beta}{k}z^{4k}
$$
which  means that we have the condition $4k+2(m+\beta)-2\beta n-1=-1$ (since we are interested in $a_{-1}$ coefficent of the Laurent expansion) which essenitally kills one of the sums, and we end up with
$$
I_{n,\beta}=\frac{\pi}{2^{\beta}}\sum_{m\geq0}(-1)^{\beta n /2-(\beta+m)/2}\binom{m+\beta-1}{\beta-1}\binom{\beta}{\beta n /2-(\beta+m)/2}
$$
which is a finite sum, since the second binomial becomes zero when $m$ is large enough ($m> \beta (n-1)$)
