For $n \in N$, if $I_n=\int \frac{\sin n x}{\sin x} d x=\frac{2}{n-1} \sin (n-1) x+I_{n-2}$ and $\int_0^\pi…
For $n \in N$, if $I_n=\int \frac{\sin n x}{\sin x} d x=\frac{2}{n-1} \sin (n-1) x+I_{n-2}$ and $\int_0^\pi \frac{\sin n x}{\sin x} d x=\frac{k \pi}{2}$, then $k=$
$(-1)^n-1$
$1-(-1)^n$
$(-1)^n$
$(-1)^{n+1}$
Solution
$\begin{aligned} & \text { Consider } I_n=\int_0^\pi \frac{\sin n x}{\sin x} d x, n \in N \\ & I_n=\left[\frac{2}{n-1} \sin (n-1) x\right]_0^n+I_{n-2}=I_{n-2}\end{aligned}$
$\begin{aligned} & \Rightarrow \quad I_1=I_3=I_5=I_7 \ldots . \\ & \text { and } I_2+I_4=I_6=I_8 \\ & \text { Now, } I_1=\int_0^{\pi \sin x} \frac{\sin x}{\sin } d x=\pi \\ & \text { and } I_2=\int_0^\pi 2 \cos x d x=0 \\ & \therefore \quad I_1=I_3=I_5 \ldots \ldots . \\ & =\pi=2 .\left(\frac{\pi}{2}\right)=\left(1-(-1)^n\right) \frac{\pi}{2}, n \in \text { odd } \\ & \text { and } I_2=I_4=I_6 \ldots . .=0 \\ & =\left(1-(-1)^n\right) \frac{\pi}{2}, n \text { is even. } \\ & \end{aligned}$