For any $n \in N, \int\limits_0^\pi \frac{\sin (2 n+1) x}{\sin x} d x$ is equal to

$\pi$

We have,

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

$\Rightarrow I_{n+1}=\int\limits_0^\pi \frac{\sin (2 n+3) x}{\sin x} d x$

∴  $I_{n+1}-I_n=\int\limits_0^\pi \frac{\sin (2 n+3) x-\sin (2 n+1) x}{\sin x}$

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

$\Rightarrow I_{n+1}-I_n=\frac{1}{n+1}[\sin (2 n+2) x]_0^\pi=0$

$\Rightarrow I_{n+1}=I_n=I_{n-1}=...=I_2=I_1$

But, $ I_1=\int\limits_0^\pi \frac{\sin 3 x}{\sin x} d x=\int\limits_0^\pi\left(3-4 \sin ^2 x\right)=3 \pi-4 \times \frac{\pi}{2}=\pi$

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