Computing $ \int_{0}^{2\pi}\frac{\sin(nx)}{\sin(x)} \mathrm dx $ I would like to compute:
$$ \int_{0}^{2\pi}\frac{\sin(nx)}{\sin(x)} \mathrm dx $$
We have:
$$ \int_{0}^{2\pi}\frac{\sin(nx)}{\sin(x)}\mathrm dx=2\int_{0}^{\pi}\frac{\sin(nx)}{\sin(x)}\mathrm dx$$
So: $ n \geq 2$ $$ \int_{0}^{\pi}\frac{\sin((n+1)x)}{\sin(x)}\mathrm dx= \ln(\sin(x))\sin(nx)\vert_0^{\pi}-n\int_{0}^{\pi} \ln(\sin(x))\cos(nx)\mathrm dx
$$
$$ =-n\int_{0}^{\pi} \ln(\sin(x))\cos(nx)\mathrm dx$$
...
 A: Use the identity 
$$\sin((n+1)x)-\sin((n-1) x)=2\cos(nx)\sin(x).$$
So it is an induction like the one you attempted. But instead of incrementing exponents by $1$, which creates a bit of a mess,  we increment by $2$. Start with the cases $n=1$ and $n=2$, and treat the even and odd cases separately.  Dividing the right-hand side by $\sin x$ leaves something very pleasant, whose integral is even more pleasant!
Remark: Note that for $n=2$, we are integrating $\frac{\sin 2x}{x}$. Since $\sin 2x=2\sin x\cos x$, our integral is $0$. So the odd and even cases yield quite different answers. If you ask a graphing program to graph $\frac{\sin nx}{x}$ for a few even $n$, you will see why. The cancellation is also visualizable without electronic aids. And you can prove directly that the integral is $0$, by breaking up the interval of integration into $0$ to $\pi$, and $\pi$  to $2\pi$. For even $n$, a suitable change of variable transforms the second integral into the negative of the first.
A: $$\sin(nx) =  \sin x \cos(n-1)x + \cos x \sin(n-1)x$$
$\\$
  $$\int\frac{\sin x \cos((n-1)x) + \cos x \sin((n-1)x)}{\sin x} \text{d}x $$
or:
$$ \int (\cos((n-1)x) + \frac{\cos x \sin((n-1)x)}{\sin x})\text{d}x $$
Which equals to:
$$\frac{\sin((n-1)x)}{n-1}+\int\frac{\cos x \sin((n-1)x)}{\sin x}\text{d}x$$
So, the problem is evaluating, 
$$\int\frac{\cos x \sin((n-1)x)}{\sin x})\text{d}x$$
From some trig identities, we will get that,
$$\cos x \sin((n-1)x) = \frac{1}{2}(\sin(nx) + \sin((n-2)x)$$
Hence our "problematic" integral becomes to:
$$ \int\frac{\frac{1}{2}(\sin(nx) + \sin((n-2)x)}{\sin x} $$
Or:
$$ \frac{1}{2}\int\frac{\sin(nx)}{\sin x}+\frac{1}{2}\int\frac{\sin(n-2)x}{\sin x}$$
And if we go back and say note our original integral as:
$$ I_n=\int\frac{\sin(nx)}{\sin x}\text{d}x $$
by our above observation, we can conclude:
$$ I_n=\frac{\sin((n-1)x}{n-1}+\frac{1}{2}I_n+\frac{1}{2}I_{n-2}$$
or:
$$ I_n=2 \frac{\sin((n-1)x)}{n-1}+I_{n-2}$$
Could you proceed? 
A: $$ \sin((n+2)x)-\sin(nx)=2\sin(x)\cos((n+1)x)$$
$$ \int_{0}^{2\pi}\frac{\sin((n+2)x)}{\sin(x)}\mathrm dx-\int_{0}^{2\pi}\frac{\sin(nx)}{\sin(x)}\mathrm dx=2\int_{0}^{2\pi} \cos((n+1)x)\mathrm dx=0$$
$$ \int_{0}^{2\pi}\frac{\sin((n+2)x)}{\sin(x)}\mathrm dx=\int_{0}^{2\pi}\frac{\sin(nx)}{\sin(x)}\mathrm dx $$
If n is odd:
$$ \int_{0}^{2\pi}\frac{\sin(nx)}{\sin(x)}\mathrm dx=\int_{0}^{2\pi}\frac{\sin(x)}{\sin(x)}\mathrm dx=2\pi $$
If n is even:
$$ \int_{0}^{2\pi}\frac{\sin(nx)}{\sin(x)}\mathrm dx=\int_0^{2\pi}0 \; \mathrm dx=0 $$
A: Note: We are going to prove that for odd $n$ the expression indeed equals $2\pi$. For even $n$ we want to prove it equals $0$.
We want to use induction. Prove it for $n=1,2,3,4$ and assume $n\geq 4$.
Apply the addition theorem $\sin(nx)=\sin((n-1)x)\cos(x)+\cos((n-1)x)\sin(x)$ twice to obtain
$$\begin{align}
\frac{\sin(nx)}{\sin(x)} &=\frac{\sin((n-2)x)}{\sin(x)}-\sin((n-2)x)\sin(x)+\cos((n-2)x)\cos(x)+\cos((n-1)x) \\
&=\frac{\sin((n-2)x)}{\sin(x)}+2\cos((n-1)x)
\end{align}$$
If you integrate this from $0$ to $2\pi$ the second term will give $0$. Now use the induction hypothesis!
Sorry I realised too late that this is actually exactly what André is suggesting.
