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?

Note that $$\sin(x)=\frac{e^{ix}-e^{-ix}}{2i}$$
Then $$\frac{\sin(kx)}{\sin(x)}=\frac{e^{ikx}-e^{-ikx}}{e^{ix}-e^{-ix}}=\frac{(e^{ix})^k-(e^{-ix})^k}{e^{ix}-e^{-ix}}$$
Using (n is odd)$$a^n-b^n=(a-b) \sum _{j=0}^{n-1} a^j b^{n-1-j}$$
We have $$\frac{\sin(kx)}{\sin(x)}=\sum _{j=0}^{k-1} (e^{ix})^j (e^{-ix})^{k-1-j}=\sum _{j=0}^{k-1} e^{i(2j-(k-1))x}$$
$$\sum _{j=0}^{k-1} e^{i(2j-(k-1))x}=\sum _{j=0}^{k-1} \frac{e^{i(2j-(k-1))x}+ e^{-i(2j-(k-1))x}}{2}=\sum _{j=0}^{k-1}\cos((2j-(k-1))x)$$
It should be easy to move on now
