Prove that $\int_{0}^{\frac{\pi}{2}}{\frac{\sin(2n+1)x}{\sin(x)}dx}=\frac{\pi}{2}$ for $n\ge0$ Prove that
$$\int_{0}^{\frac{\pi}{2}}{\frac{\sin(2n+1)x}{\sin(x)}dx}=\frac{\pi}{2}$$ for $n\ge0$
I am not able to proceed with the integral. For the case $k+1$ please guide me through the problem. 
Thanks in advance!
 A: Hint:
$$
\frac{1}{2} + \cos y + \cdots +\cos (ny) = \frac{\sin (n+1/2)y}{2 \sin (y/2)}.
$$
This is a useful identity that can be used to prove the convergence of Fourier series. It is proved by induction on $n \in \mathbb{N}$.
A: $$\frac{\sin((2n+1)x)}{\sin(x)} = \frac{e^{(2n+1)ix}-e^{-(2n+1)ix}}{e^{ix}-e^{-ix}}=\sum_{k=-n}^{n}e^{2ikx} $$
and for every $k\in\mathbb{Z}\setminus\{0\}$, 
$$ \int_{0}^{\pi/2} e^{2ikx}\,dx \in i\mathbb{R},$$
but the integral $\int_{0}^{\pi/2}\frac{\sin((2n+1)x)}{\sin(x)}\,dx$ has to be real, hence it equals the contribute given by $k=0$ only.
A: Using Prosthaphaeresis Formula,
$$\sin(m+2)x-\sin mx=2\sin x\cdot\cos(m+1)x$$
So, if $\displaystyle I_m=\int_0^{\pi/2}\dfrac{\sin mx}{\sin x}dx,$
$$I_{m+2}-I_m=2\int_0^{\pi/2}\cos(m+1)x\ dx=\dfrac2{m+1}\cdot\left\{\sin(m+1)\dfrac\pi2-\sin0\right\}$$
If $\displaystyle m+1$ is even $\displaystyle=2r$(say), $$I_{2r+1}=I_{2r-1}$$
What is $I_1?$
A: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
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${\large\mbox{With}\ m = 1, 3, 5, \ldots}$:
\begin{align}
&\color{#f00}{\int_{0}^{\pi/2}{\sin\pars{mx} \over \sin\pars{x}}\,\dd x} =
\Im\int_{0}^{\pi/2}{\expo{\ic mx} - 1 \over
\pars{\expo{\ic x} - \expo{-\ic x}}/\pars{2\ic}}\,\dd x =
\Im\int_{z\ =\ \exp\pars{\ic x} \atop {\vphantom{\Large A}0\ <\ x\ <\ \pi/2}}
\quad{z^{m} - 1 \over \pars{z^{2} - 1}/\pars{2\ic z}}\,{\dd z \over \ic z}
\\[3mm] = &\
2\,\Im\int_{z\ =\ \exp\pars{\ic x} \atop {\vphantom{\Large A}0\ <\ x\ <\ \pi/2}}
\quad{1 - z^{m} \over 1 - z^{2}}\,\dd z =
2\,\Im\pars{-\int_{1}^{0}{1 - y^{m}\expo{\ic m\pi/2} \over 1 + y^{2}}\,\ic\,\dd y} =
2\int_{0}^{1}{1 - y^{m}\ \overbrace{\cos\pars{m\pi/2}}^{\ds{=\ 0}\ } \over
1 + y^{2}}\,\dd y
\\[3mm] = &\
2\int_{0}^{1}{\dd y \over 1 + y^{2}} = \color{#f00}{{\pi \over 2}}
\end{align}
