Show that $$ \int_0^{\pi\over 2}\frac{\sin(2nx)}{\sin^{2n+2}(x)}\frac{1}{e^{2\pi \cot x}-1}dx =(-1)^{n-1}\frac{2n-1}{4(2n+1)}
$$
My attempt
Lemma-1
\begin{align*}
\frac{\sin(2nx)}{\sin^{2n}(x)}&=\sum_{r=1}^{n}\binom{2n}{2r-1} (-1)^{r-1}\cot^{2n-2r+1}(x) \\
\end{align*}
proof
\begin{align*}
\frac{\sin(2nx)}{\sin^{2n}(x)} &= \text{Im}\frac{e^{2inx}}{\sin^{2n}(x)} \\
&= \text{Im} \;\left( \frac{e^{ix}}{\sin x}\right)^{2n} \\
&= \text{Im} \; \left( \frac{\cos x+i\sin x}{\sin x}\right)^{2n} \\
&= \text{Im} \; (\cot x+i)^{2n} \\
&= \text{Im} \sum_{r=0}^{2n}\binom{2n}{r}i^r \cot^{2n-r}(x) \\
&= \sum_{r=0}^{2n}\binom{2n}{r}\sin\left(\frac{\pi r}{2} \right) \cot^{2n-r}(x) \\
&= \sum_{r=1}^n \binom{2n}{2r-1} (-1)^{r-1}\cot^{2n-2r+1}(x)
\end{align*}
Let $I$ denote the integral. Now using the identity above I mentioned
$$I = \int_0^{\pi\over 2}\frac{\sin(2nx)}{\sin^{2n+2}(x)}\frac{1}{e^{2\pi \cot x}-1}dx
$$
$$
= \int_0^{\pi\over 2}\left( \sum_{r=1}^{n}\binom{2n}{2r-1} (-1)^{r-1}\cot^{2n-2r+1}(x)\right)\frac{\csc^2(x)}{e^{2\pi \cot x}-1}dx
$$
$$
= \sum_{r=1}^{n}\binom{2n}{2r-1} (-1)^{r-1} \int_0^{\pi \over 2}\cot^{2n-2r+1}(x)\frac{\csc^2(x)}{e^{2\pi \cot x}-1}dx
$$
$$
= \sum_{r=1}^{n}\binom{2n}{2r-1} (-1)^{r-1} \int_0^\infty \frac{t^{2n-2r+1}}{e^{2\pi t}-1}dt \quad \color{blue}{(t=\cot x)}
$$
$$
= \sum_{r=1}^{n}\binom{2n}{2r-1} (-1)^{r-1}\left\{ \frac{(2n-2r+1)!\zeta(2n-2r+2)}{(2\pi)^{2n-2r+2}}\right\}
$$
Best Answer
(Assuming $n$ is a positive integer.) As you have noted $$\frac{\sin 2nx}{\sin^{2n}x}=\Im(\cot x+i)^{2n}=\frac1{2i}\left((\cot x+i)^{2n}-(\cot x-i)^{2n}\right),$$ the given integral, after the substitution $t=\cot x$, equals $$I=\frac1{2i}\int_0^\infty\frac{(t+i)^{2n}-(t-i)^{2n}}{e^{2\pi t}-1}\,dt.$$
Now, for $0<r<1$, consider the contour $\lambda_r$ that is the boundary of $$\{z\in\mathbb{C} : \Re z>0 \land |\Im z|<1\}\setminus\{z\in\mathbb{C} : |z|<r\}$$ (goes along $+\infty-i\to-i\to i\to+\infty+i$ with a notch around $z=0$). Then \begin{align} 0=\int_{\lambda_r}f(z)\,dz &=\int_0^\infty\big(f(x+i)-f(x-i)\big)\,dx\\ &+i\int_r^1\big(f(ix)+f(-ix)\big)\,dx\\ &+\int_{-\pi/2}^{\pi/2}f(re^{it})\,rie^{it}\,dt, \end{align} where we set $f(z)=\displaystyle\frac{z^{2n}-i^{2n}}{e^{2\pi z}-1}$.
The first of the three terms is $2iI$; as $r\to 0$, the second one tends to $i^{2n+1}$ times $$\int_0^1\left(\frac{x^{2n}-1}{e^{2i\pi x}-1}+\frac{x^{2n}-1}{e^{-2i\pi x}-1}\right)dx=\int_0^1(1-x^{2n})\,dx=\frac{2n}{2n+1},$$ and the last one tends to $i\pi\displaystyle\operatorname*{Res}_{z=0}f(z)=-i^{2n+1}/2$. The expected result follows.