Question

For \(n\in N\), if \[ I_n=\int \frac{\sin nx}{\sin x}\,dx = \frac{2}{n-1}\sin(n-1)x+I_{n-2} \] and \[ \int_0^\pi \frac{\sin nx}{\sin x}\,dx=\frac{k\pi}{2}, \] then \(k=\)

• A. \((-1)^n-1\)
• B. \(1-(-1)^n\)
• C. \((-1)^n\)
• D. \((-1)^{n+1}\)

Hint:

For the integral \[ \int_0^\pi \frac{\sin nx}{\sin x}\,dx, \] the value is \[ \pi \] when \(n\) is odd, and \[ 0 \] when \(n\) is even.

Answer 1

The Correct Option is B

Step 1: Use the given recurrence relation.
We are given \[ I_n=\frac{2}{n-1}\sin(n-1)x+I_{n-2}. \] Now apply limits from \(0\) to \(\pi\).
So, \[ \int_0^\pi \frac{\sin nx}{\sin x}\,dx = \left[\frac{2}{n-1}\sin(n-1)x\right]_0^\pi + \int_0^\pi \frac{\sin(n-2)x}{\sin x}\,dx. \]

Step 2: Evaluate the sine term.
At \(x=\pi\), \[ \sin(n-1)\pi=0. \] At \(x=0\), \[ \sin 0=0. \] Therefore, \[ \left[\frac{2}{n-1}\sin(n-1)x\right]_0^\pi=0. \] Hence, \[ \int_0^\pi \frac{\sin nx}{\sin x}\,dx = \int_0^\pi \frac{\sin(n-2)x}{\sin x}\,dx. \]

Step 3: Reduce the integral repeatedly.
By repeated reduction, \[ \int_0^\pi \frac{\sin nx}{\sin x}\,dx \] reduces to either \[ \int_0^\pi 1\,dx \] or \[ \int_0^\pi 2\cos x\,dx. \] If \(n\) is odd, it reduces to \[ \int_0^\pi 1\,dx=\pi. \] If \(n\) is even, it reduces to \[ \int_0^\pi 2\cos x\,dx=0. \] Thus, \[ \int_0^\pi \frac{\sin nx}{\sin x}\,dx = \begin{cases} \pi, & n \text{ is odd},\\ 0, & n \text{ is even}. \end{cases} \]

Step 4: Compare with the given form.
Given, \[ \int_0^\pi \frac{\sin nx}{\sin x}\,dx=\frac{k\pi}{2}. \] Now, \[ 1-(-1)^n= \begin{cases} 2, & n \text{ is odd},\\ 0, & n \text{ is even}. \end{cases} \] Therefore, \[ \frac{\{1-(-1)^n\}\pi}{2} = \begin{cases} \pi, & n \text{ is odd},\\ 0, & n \text{ is even}. \end{cases} \] Hence, \[ k=1-(-1)^n. \]

Step 5: Final conclusion.
Therefore, \[ \boxed{1-(-1)^n} \]