TITLE: Showing that $ \int_0^\pi \frac{\sin{(kx)}}{\sin(x)} d x = \pi $ for odd integer $k$
QUESTION [8 upvotes]: I am trying to show that 

$$ \int_0^\pi \frac{\sin{(kx)}}{\sin(x)}  d x = \pi $$
  for odd integer $k$.

It seems like this could be done using multiple angle formulae, but I'm stuck.
I can get to
$$ \int_0^\pi \frac{\sin{((2N+1)x)}}{\sin(x)}  d x  $$
$$ = \sum_{\ell = 0} ^N  (-1)^\ell \frac{(2N+1)!}{(2\ell+1)!(2(N-\ell))!} \int_0 ^\pi (\cos(x))^{2(N-\ell)} (\sin(x))^{2\ell} d x, $$ 
using a multiple angle formula due to Viete (from the trig formulae Wikipedia page), but I don't know what to do from here. Is there a simpler way?

REPLY [4 votes]: Let $k=2n+1$ then 
$$I=\int_0^{\pi} \frac {\sin ((2n+1)x)}{\sin x} dx=\int_0^{\pi} \frac {\sin \left(\left(n+\frac 12\right)2x\right)}{\sin x} dx$$
Now substituting $x=\frac t2$ we get $$I=\frac 12\int_0^{2\pi} \frac {\sin \left(\left(n+\frac 12\right)t\right)}{\sin \left(\frac t2\right)} dt= 
\frac 12\int_0^{2\pi}  D_n(t) dt$$
Where $D_n(t)$ represents the Dirichlet Kernel. And using its property that $$D_n(t)=1+2\sum_{k=0}^n \cos(kt)$$
We get $$I=\frac 12\int_0^{2\pi} \left(1+2\sum_{k=0}^n \cos(kt)\right)dt=\pi+ \sum_{k=0}^n \int_0^{2\pi} \cos (kt) dt $$
From where you notice that the last integral is $0$ for all integer $k$ from $0$ to $n$
Q.E.D