Elegant way to evaluate $\int_0^\pi\frac{\sin\left(M+\frac{1}{2}\right)\theta\;\sin\left(N+\frac{1}{2}\right)\theta}{\sin^2(\theta/2)}d\theta$? Solve in terms of $M, N$
$$I(M. N) = \int_0^\pi\frac{\sin\left[\left(M + \frac{1}{2}\right)\theta\right]\sin\left[\left(N + \frac{1}{2}\right)\theta\right]}{\sin^2\left(\frac{\theta}{2}\right)}d\theta$$
where $M, N$ are non-negative integers.
I've tried solving it by using trigonometric identities and brute force, but it gets extremely annoying. Is there a simpler way?
 A: Hint
$${\displaystyle {\begin{aligned}1+2\sum _{n=1}^{N}\cos(n\theta )&={\frac {\sin \left(\left(N+{\frac {1}{2}}\right)\theta \right)}{\sin \left({\frac {\theta }{2}}\right)}}\end{aligned}}}$$
and use the following fact
$$\int_{0}^{\pi}\cos(n \theta)\cos(m \theta)d \theta=0$$
when $$m \ne n$$
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
&\bbox[10px,#ffd]{\left.\vphantom{\Large A}\mrm{I}\pars{M,N}\,
\right\vert_{\ M, N\ \in\ \mathbb{N}_{\large\ \geq 0}}} \equiv
\int_{0}^{\pi}{\sin\pars{\bracks{M + 1/2}\theta}
\sin\pars{\bracks{N + 1/2}\theta} \over \sin^{2}\pars{\theta/2}}\,\dd\theta
\\[5mm] = &\
{1 \over 2}\int_{0}^{\pi}{\cos\pars{\bracks{M - N}\theta}
-\cos\pars{\bracks{M + N + 1}\theta} \over \sin^{2}\pars{\theta/2}}\,\dd\theta
\\[5mm] = &\
{1 \over 2}\int_{0}^{\pi}{1 - \cos\pars{\bracks{M + N + 1}\theta} \over \sin^{2}\pars{\theta/2}}\,\dd\theta -
{1 \over 2}\int_{0}^{\pi}{1 - \cos\pars{\verts{M - N}\theta} \over \sin^{2}\pars{\theta/2}}\,\dd\theta
\\[5mm] = &\
\bbox[10px,#ffd]{\mc{J}\pars{M + N + 1} - \mc{J}\pars{\verts{M - N}}}
\label{1}\tag{1}
\\[5mm] &\ \mbox{where}\quad
\left\{\begin{array}{rcl}
\ds{\left.\vphantom{\Large A}\mc{J}\pars{a}
\,\right\vert_{\ a\ \in\ \mathbb{N}_{\ \geq\ 0}}} & \ds{\equiv} & \ds{\int_{0}^{\pi/2}
{1 - \cos\pars{2a\theta} \over \sin^{2}\pars{\theta}}\,\dd\theta}
\\[2mm]
& \ds{=} & \ds{\Re\int_{0}^{\pi/2}
{1 + 2\ic a\theta - \expo{2\ic a\theta} \over \sin^{2}\pars{\theta}}\,\dd\theta}
\end{array}\right.
\end{align}

Lets evaluate $\ds{\mrm{J}\pars{a}}$:

\begin{align}
\mc{J}\pars{a} & =
\Re\int_{0}^{\pi/2}
{1 + 2\ic a\theta - \expo{2\ic a\theta} \over \sin^{2}\pars{\theta}}\,\dd\theta
\\[5mm] & =
\left.\Re\int_{\theta\ =\ 0}^{\theta\ =\ \pi/2}
{1 + 2\ic a\bracks{-\ic\ln\pars{z}} - z^{2a} \over
-\pars{1 - z^{2}}^{2}/\pars{4z^{2}}}\,{\dd z \over \ic z}
\,\right\vert_{\ z\ =\ \exp\pars{\ic\theta}}
\\[5mm] & =
\left.4\,\Im\int_{\theta\ =\ 0}^{\theta\ =\ \pi/2}
{z^{2a} - 2a\ln\pars{z} - 1 \over
\pars{1 - z^{2}}^{2}}\,z\,\dd z
\,\right\vert_{\ z\ =\ \exp\pars{\ic\theta}}
\\[5mm] & =
-4\,\Im\int_{1}^{0}
{y^{2a}\expo{\ic\pars{2a}\pi/2} - 2a\bracks{\ln\pars{y} + \ic\pi/2} - 1 \over
\pars{1 + y^{2}}^{2}}\pars{\ic y}\,\ic\,\dd y
\\[5mm] & =
-4\int_{0}^{1}
{y^{2a}\
\overbrace{\sin\pars{\pi a}}^{\ds{\color{red}{=\ 0}}}\ -\
\pi a \over
\pars{1 + y^{2}}^{2}}\,y\,\dd y
\\[5mm] & \stackrel{y^{2}\ \mapsto\ y}{=}\,\,\,
2\pi a\int_{0}^{1}{\dd y \over\pars{1 + y}^{2}} = \bbx{\pi a}
\label{2}\tag{2}
\end{align}

With \eqref{1} and \eqref{2}:

$$
\bbx{\mrm{I}\pars{M,N} =
\pi\pars{M + N + 1 - \verts{M - N}}}
$$
