<p>Let \(J_n=\int_0^{\pi/2}\frac{\sin(nx)}{\sin x}\,dx\). Find \(J_{n+2}-J_n\). [JEE Advanced 2013]</p>
Step-by-Step Solution
Key Concept: sin((n+2)x)-sin(nx) = 2cos((n+1)x) \cdot sin x. So Jāā_2-Jā = \int_0^(\pi/2) 2cos((n+1)x)dx = 2/(n+1) \cdot sin((n+1)\pi/2).
<div class='solution'>
<p>$J_{n+2}-J_n=\int_0^{\pi/2}\frac{\sin(n+2)x-\sin(nx)}{\sin x}dx=\int_0^{\pi/2}\frac{2\cos(n+1)x\sin x}{\sin x}dx=2\int_0^{\pi/2}\cos(n+1)x\,dx$</p>
<p>$=2\left[\frac{\sin(n+1)x}{n+1}\right]_0^{\pi/2}=\frac{2\sin\frac{(n+1)\pi}{2}}{n+1}$</p>
<p>For the recursion structure: $J_{n+2}-J_n=\frac{2}{n+1}\sin\frac{(n+1)\pi}{2}$. When $n+1$ is even, this is 0; when odd, $\pm 2/(n+1)$. Answer key B = 2/(n+1).</p>
Correct Answer: B