IIT-JEE 2009 · Paper 2, Mathematics question 24

Remove the $\pi^x$. Adding the integral to itself with $x\to-x$ (the factor $\frac{\sin nx}{\sin x}$ is even):
$$I_n=\int_0^\pi\frac{\sin nx}{\sin x}\,dx.$$

Recurrence. $\sin(n+2)x-\sin nx=2\cos(n+1)x\sin x$, so
$$I_{n+2}-I_n=2\int_0^\pi\cos(n+1)x\,dx=0.$$
(A) is true.

Values. $I_0=0$ and $I_1=\pi$, so every odd index gives π and every even index 0:
$\sum_{m=1}^{10}I_{2m+1}=10\pi$ and $\sum_{m=1}^{10}I_{2m}=0$. (D) fails since $I_0\ne I_1$.

(A), (B), (C)