JEE Advanced 2009 Paper 2Question 1
If $$I_n=\int_{-\pi}^{\pi}\frac{\sin nx}{(1+\pi^x)\sin x}\,dx,\quad n=0,1,2,\ldots,$$ then
Definite Integration
Solution
Concept. Replace $x$ by $-x$ and add: the $\pi^x$ factor disappears and $I_n=\int_0^\pi\frac{\sin nx}{\sin x}dx$. Then compare $I_{n+2}$ with $I_n$. Formulas used. (1) $\frac{1}{1+\pi^x}+\frac{1}{1+\pi^{-x}}=1$(2) $\sin(n+2)x-\sin nx=2\cos(n+1)x\,\sin x$Step 1 - Remove $\pi^x$. $\frac{\sin nx}{\sin x}$ is even. With $x\to-x$ and (1), $$2I_n=\int_{-\pi}^{\pi}\frac{\sin nx}{\sin x}dx,\qquad I_n=\int_0^{\pi}\frac{\sin nx}{\sin x}dx.$$ Step 2 - Recurrence. By (2), $$I_{n+2}-I_n=\int_0^\pi2\cos(n+1)x\,dx=\Big[\frac{2\sin(n+1)x}{n+1}\Big]_0^\pi=0.$$ So $I_n=I_{n+2}$: (A) is true. Step 3 - Values. $I_0=0$ and $I_1=\int_0^\pi1\,dx=\pi$. So $I_n=\pi$ for odd $n$, $I_n=0$ for even $n$. Step 4 - Options. (B) $\sum_{m=1}^{10}I_{2m+1}=10\pi$: true. (C) $\sum_{m=1}^{10}I_{2m}=0$: true. (D) $I_0=0\ne I_1=\pi$: false. Answer: (A), (B), (C)