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?
 A: 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
A: Use the addition formulas for $\sin$ and $\cos$ to show that
$$\frac{\sin ((k+2)x)}{\sin(x)} = \frac{\sin (k x)}{\sin(x)} +  2\cos ((k+1)x)$$
From this formula the result you are after follows directly by using induction and the fact that $\int_0^\pi \cos(nx){\rm d}x = 0$ for all integers $n>0$. It also follows that the integral is $0$ for even $k$ although this can be shown more easily by using the fact that the integrand is odd about $x = \pi/2$.
A: 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
A: 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 $k\in\mathbb{Z}$.
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
&\bbox[10px,#ffd]{\left.\int_{0}^{\pi}{\sin\pars{kx} \over \sin\pars{x}}\,\dd x\,\right\vert_{\ k\ =\ \pm 1,\ \pm 3,\
\pm 5,\ldots}} =
2\,\mrm{sgn}\pars{k}\Im\int_{0}^{\pi/2}{\expo{\ic\verts{k}x} - 1 \over \sin\pars{x}}\,\dd x
\\[5mm] = &\
\left.2\,\mrm{sgn}\pars{k}\Im\int_{x\ =\ 0}^{x\ =\ \pi/2}{z^\verts{k} - 1 \over \pars{1 - z^{2}}\ic/\pars{2z}}\,{\dd z \over \ic z}\,\right\vert_{\ z\ =\ \exp\pars{\ic x}}
\\[5mm] = &\
\left.4\,\mrm{sgn}\pars{k}\Im\int_{x\ =\ 0}^{x\ =\ \pi/2}
{1 - z^\verts{k} \over 1 - z^{2}}\,\dd z\,\right\vert_{\ z\ =\ \exp\pars{\ic x}}
\\[5mm] =&
-4\,\mrm{sgn}\pars{k}\Im\int_{1}^{0}
{1 - y^\verts{k}\expo{\ic\verts{k}\pi/2} \over 1 + y^{2}}\,\ic\,\dd y =
4\,\mrm{sgn}\pars{k}\int_{0}^{1}{\dd y \over 1 + y^{2}}
\\[5mm] = &
\bbx{\pi\,\mrm{sgn}\pars{k}\,\qquad k = \pm 1,\ \pm 3,\ \pm 5\ldots}
\end{align}

Note that
  $\ds{\left.\cos\pars{k\,{\pi \over 2}}
\,\right\vert_{\ k\ \mrm{odd}} = \color{red}{\large 0}}$.

