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:

The definite integral \( \int_0^{\pi} \frac{\sin nx}{\sin x} dx \) has a known value:
\(\pi\) if n is an odd positive integer.
0 if n is an even positive integer.
This can be expressed as \( \frac{1 - (-1)^n}{2} \pi \).
This integral is related to Dirichlet kernel in Fourier series.

Answer 1

The Correct Option is B

Let \(J_n = \int_0^{\pi} \frac{\sin nx}{\sin x} dx\). We are given a recurrence relation for the indefinite integral \(I_n\). This recurrence is not standard for the definite integral \(J_n\). A known result for the definite integral \(J_n = \int_0^{\pi} \frac{\sin nx}{\sin x} dx\) is: \(J_n = \pi\) if n is odd. \(J_n = 0\) if n is even. This can be combined into a single expression. We can write this as \(J_n = \frac{1 - (-1)^n}{2} \pi\). If n is odd, \( (-1)^n = -1 \), so \( J_n = \frac{1 - (-1)}{2} \pi = \frac{2}{2} \pi = \pi \). If n is even, \( (-1)^n = 1 \), so \( J_n = \frac{1 - 1}{2} \pi = 0 \). This matches the known result. The problem states \(J_n = \int_0^{\pi} \frac{\sin nx}{\sin x} dx = \frac{k\pi}{2}\). So we have \( \frac{k\pi}{2} = \frac{1 - (-1)^n}{2} \pi \). Dividing by \(\pi/2\) (assuming \(\pi \neq 0\)): \( k = 1 - (-1)^n \). This matches option (b). The provided recurrence relation for \(I_n\) seems to be a distractor or for a different problem, as it's for the indefinite integral and doesn't directly help find the value of the definite integral \(J_n\) without more context or integration of the constant term if \(I_{n-2}\) becomes a constant. The direct evaluation of \(J_n\) is standard. \[ \boxed{1 - (-1)^n} \]