For n in N,if I_n = int frac{sin nx}{sin x} d | vedclass

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(B) Let $I_n = \int_0^\pi \frac{\sin nx}{\sin x} dx$. Using the given recurrence relation: $I_n - I_{n-2} = \int_0^\pi \frac{\sin nx - \sin(n-2)x}{\si

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$1 - (-1)^n$

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(B) Let $I_n = \int_0^\pi \frac{\sin nx}{\sin x} dx$. Using the given recurrence relation: $I_n - I_{n-2} = \int_0^\pi \frac{\sin nx - \sin(n-2)x}{\sin x} dx = \int_0^\pi \frac{2 \cos(n-1)x \sin x}{\sin x} dx = \int_0^\pi 2 \cos(n-1)x dx$. For $n > 1$,$I_n - I_{n-2} = \left[ \frac{2 \sin(n-1)x}{n-1} \right]_0^\pi = 0$. Thus,$I_n = I_{n-2}$ for all $n > 1$. For $n=1$,$I_1 = \int_0^\pi \frac{\sin x}{\sin x} dx = \int_0^\pi 1 dx = \pi$. For $n=2$,$I_2 = \int_0^\pi \frac{\sin 2x}{\sin x} dx = \int_0^\pi 2 \cos x dx = [2 \sin x]_0^\pi = 0$. If $n$ is odd,$I_n = I_1 = \pi = 2 \cdot \frac{\pi}{2} = (1 - (-1)^n) \frac{\pi}{2}$. If $n$ is even,$I_n = I_2 = 0 = (1 - (-1)^n) \frac{\pi}{2}$. Therefore,$k = 1 - (-1)^n$.

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 $k =$

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