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!

$$\frac{\sin((2n+1)x)}{\sin(x)} = \frac{e^{(2n+1)ix}-e^{-(2n+1)ix}}{e^{ix}-e^{-ix}}=\sum_{k=-n}^{n}e^{2ikx} $$
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.
