Family of Integrals $\int_0^{\frac{\pi}{2}} {\left(\frac{\sin{(ax)}}{\sin{(bx)}}\right)}^{2n} \; dx$ Background:
I want to preface this by saying that I'm not sure if these generalized integrals have been brought up before, but I haven't seen anything on them.   I'm creating this post because I am interested to find out more intriguing information about these integrals, to see if there are any mistakes in my following observations, and maybe even to see if anyone has derivations of these observations ( I used Wolfram Alpha to calculate most of the integrals ).
Note, for all of the following integrals assume that
$\displaystyle\underline{a > b\ \mbox{and}\ a, b, n, k \in \mathbb{Z}^{+}}$:
Powers of 2:
First, for the generalized integrals with a power of $2$:
$$\int_0^{\frac{\pi}{2}} {\left(\frac{\sin{(ax)}}{\sin{(bx)}}\right)}^2 \; dx = \begin{cases}
\frac{a \pi}{2b} & \text{if} \; a \bmod b=0 &\\
\text{Diverges} & \text{if} \; a \bmod b \neq 0
\end{cases}$$

And making the upper bound dependent on $b$ and just substituting $u=bx$ yields:
$$\int_0^{\frac{\pi}{b}} {\left(\frac{\sin{(ax)}}{\sin{(bx)}}\right)}^2 \; dx = \begin{cases}
\frac{a \pi}{b^2} & \text{if} \; a \bmod b=0 \\
\text{Diverges} & \text{if} \; a \bmod b \neq 0
\end{cases}$$

Even powers:
Then even powers, although this has been a bit challenging.
For $a=2b$, I found that:
$$\int_0^{\frac{\pi}{2}} {\left(\frac{\sin{(2bx)}}{\sin{(bx)}}\right)}^{2n} \; dx = \frac{\pi}{2}  \cdot \frac{(2n)!}{{\left(n!\right)}^2}$$
I evaluated the integrals at varying powers of $n$ using Wolfram, and created a sequence using the coefficients of the result of the integrals.  Then, I used OEIS to recognize the sequence, which is the "central binomial coefficients", or sum of squares of entries in the $n^{\text{th}}$ row of the triangle of binomial coefficients.

For $a=3b$, I found that:
$$\int_0^{\frac{\pi}{2}} {\left(\frac{\sin{(3bx)}}{\sin{(bx)}}\right)}^{2n} \; dx =\frac{\pi}{2} \displaystyle\sum_{k=0}^n {2k \choose k}{2n \choose k}$$
According to OEIS, the sequence is equivalent to the sum of squares of entries in the $n^{\text{th}}$ row of the triangle of trinomial coefficients.  Note that the sequence is every other central trinomial coefficient.

For $a=4b$, I found that:
$$\int_0^{\frac{\pi}{2}} {\left(\frac{\sin{(4bx)}}{\sin{(bx)}}\right)}^{2n} \; dx =\frac{ \pi}{2} \displaystyle\sum_{k=0}^{ \lfloor{3n/4} \rfloor} {(-1)}^k {2n \choose k} {5n-4k-1 \choose 3n-4k}$$
According to OEIS, the sequence is equivalent to the "central quadrinomial coefficients".

For $a=5b$ and indeed it appears to follow this sequence, but I couldn't find a closed form for sum of squares of entries in the $n^{\text{th}}$ row of the triangle of 5-nomial (I'm not sure what it is called) coefficients

Conjecture:
From these observations, I conjecture the following with the aforementioned conditions:
$$\int_0^{\frac{\pi}{2}} {\left(\frac{\sin{(kbx)}}{\sin{(bx)}}\right)}^{2n} \; dx =\frac{ \pi}{2} \rho$$
where $\rho$ is the sum of squares of entries in the $n^{\text{th}}$ row of the triangle of $k^{\text{th}}$ multinomial coefficients.  I believe this is equivalent to the central $k^{\text{th}}$ multinomial coefficients for even valued $k$, but is the alternating central coefficients for odd valued $k$.
Is there a closed form expression for this (the sum of squares of entries in the $n^{\text{th}}$ row of the triangle of $k^{\text{th}}$ multinomial coefficients) and are my observations correct?

 A: In this answer I'll provide an intuition as to why multinomial coefficients occur. Let $$I=\int_0^{\pi/2}\left(\frac{\sin(kbx)}{\sin(bx)}\right)^{2n}\,dx=\frac14\int_0^{2\pi}\left(\frac{\sin(kbx)}{\sin(bx)}\right)^{2n}\,dx$$ and perform the substitution $z:=e^{ix}$. Then \begin{align}I&=\frac14\oint_{|z|=1}\left(\frac{z^{kb}-z^{-kb}}{z^b-z^{-b}}\right)^{2n}\,\frac{dz}{iz}\\&=\frac1{4i}\oint_{|z|=1}z^{-1-2nb(k-1)}\left(1+z^{2b}+z^{4b}+\cdots+z^{2b(k-1)}\right)^{2n}\,dz\end{align} since $(z^{2kb}-1)/(z^{2b}-1)$ has removable singularities at the roots of unity. Hence the residue theorem gives $$I=\frac\pi2\cdot\frac1{(2nb(k-1))!}\lim_{z\to0}\frac{d^{2nb(k-1)}}{dz^{2nb(k-1)}}\left(1+z^{2b}+z^{4b}+\cdots+z^{2b(k-1)}\right)^{2n}$$ and the only nonzero term will come from the coefficient of $z^{b(k-1)}$ in the polynomial.
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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$\ds{\Large\underline{\mbox{A}\ Finite\ Sum}:}$
\begin{align}
&\bbox[#ffd,10px]{\left.\int_{0}^{\pi/2}\bracks{\sin\pars{kbx} \over \sin\pars{bx}}^{2n}\,\dd x
\,\right\vert_{\ b, k, n\ \in\ \mathbb{N}_{\large\ \geq\ 2}}}
\\[5mm] \stackrel{\Large\color{red}{r\ =\ 1^{-}}}{=}\,\,\, &
\Re\int_{0}^{\pi/2}\pars{\expo{\ic kbx}\,
{{1 - r\expo{-2\ic kbx}} \over 2\ic}}^{2n}
\pars{\expo{\ic bx}\,{{1 - r\expo{-2\ic bx}} \over 2\ic}}^{-2n}\,\dd x
\\[5mm] = &\
\Re\int_{0}^{\pi/2}\bracks{\expo{2nkbx\ic}
\sum_{\ell = 0}^{2n}{2n \choose \ell}\pars{-r\expo{-2kbx\ic}}^{\ell}}
\\[2mm] &\ \phantom{\Re\int_{0}^{\pi/2}\!\!\!\!\!}\times \bracks{\expo{-2nbx\ic}
\sum_{m = 0}^{\infty}{-2n \choose m}\pars{-r\expo{-2\ic bx}}^{m}}
\dd x
\\[5mm] = &\
\Re\sum_{\ell = 0}^{\infty}\sum_{m = 0}^{\infty}{2n \choose \ell}
{-2n \choose m}\pars{-r}^{\ell + m}
\\[2mm] & \times \int_{0}^{\pi/2}
\exp\bracks{\pars{2nkb - 2\ell kb - 2nb - 2mb}\ic x}\,\dd x
\\[5mm] = &\
{\pi \over 2}\sum_{\ell = 0}^{\infty}\sum_{m = 0}^{\infty}{2n \choose \ell}{-2n \choose m}
\pars{-r}^{\ell + m}\,\,\delta_{\large m,nk - \ell k - n}
\\[5mm] = &\
{\pi \over 2}\sum_{\ell = 0}^{\infty}{2n \choose \ell}
{-2n \choose nk - \ell k - n}\pars{-r}^{\pars{n - \ell}\pars{k - 1}}\
\bracks{nk - \ell k - n \geq 0}
\\[5mm] = &\
\bbx{{\pi \over 2}\sum_{\ell = 0}^{\left\lfloor\pars{1 - 1/k}n\right\rfloor}{2n \choose \ell}
{-2n \choose nk - \ell k - n}\pars{-1}^{\pars{n - \ell}\pars{k - 1}}\,,
\quad \color{red}{\large r \to 1^{-}}}
\\ &\ \mbox{}
\end{align}
