For any $n \in N$, the value of the integral $\int\limits_0^\pi \frac{\sin 2 n x}{\sin x} d x$, is

none of these

Let $I_n=\int\limits_0^\pi \frac{\sin 2 n x}{\sin x} d x$. Then,

$I_{n+1}=\int\limits_0^\pi \frac{\sin 2(n+1) x}{\sin x} d x$

∴  $I_{n+1}-I_n=\int\limits_0^\pi \frac{\sin (2 n+2) x-\sin 2 n x}{\sin x} d x$

$\Rightarrow I_{n+1}-I_n=2 \int\limits_0^\pi \frac{2 \sin x \cos (2 n+1) x}{\sin x} d x$

$\Rightarrow I_{n+1}-I_n=2 \int\limits_0^\pi \cos (2 n+1) x d x$

$\Rightarrow I_{n+1}-I_n=\frac{2}{2 n+1}[\sin (2 n+1) x]_0^\pi$

$\Rightarrow I_{n+1}-I_n=\frac{2}{2 n+1}[\sin (2 n+1) \pi-\sin 0]=0$

$\Rightarrow I_{n+1}=I_n$ for $n=1,2,3, ....$

$\Rightarrow I_n=I_{n-1}=...=I_1$

But, $I_1=\int\limits_0^\pi \frac{\sin 2 x}{\sin x} d x=2 \int\limits_0^\pi \cos x d x=0$

Hence, $I_n=0$ for all $n \in N$.