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

If we let
$$I_n=\int \frac{\sin{((2n+1)x)}}{\sin{(x)}} dx$$
One can quite easily show that
$$I_n=I_{n-1}+\frac{\sin{(2nx)}}{n}$$
Solving this recurrence relation gives
$$I_n=\sum_{k=1}^n \frac{\sin{(2kx)}}{k}+x+c$$
So, for the integral in question,
$$\int_0^\pi \frac{\sin{((2n+1)x)}}{\sin{(x)}}dx=[\sum_{k=1}^n \frac{\sin{(2kx)}}{k}+x]_0^\pi = \pi$$
As $\sin{(2k\pi)}=0$ for all $k\in\mathbb{Z}$.
