Question

If $I_n={\int }_{-\pi }^{\pi }\frac{\sin nx}{(1+{\pi }^x)\cdot \sin x}dx,n=0,1,2,...,$then

• A. $I_n=I_{n+2}$
• B. $\sum _{m=1}^{10}{I}_{2m+1}=10\pi$
• C. $\sum _{m=1}^{10}{I}_{2m}=0$
• D. $I_n=I_{n+1}$

Answer 1

Answer: (A), (B), (C)

$I_n=I_{n+2}$
$\sum _{m=1}^{10}{I}_{2m+1}=10\pi$
$\sum _{m=1}^{10}{I}_{2m}=0$

Given $I_n={\int }_{-\pi }^{\pi }\frac{\sin nx}{(1+{\pi }^x)\sin x}dx$ ...(1)
using ${\int }_a^{b}f\left(x\right)dx={\int }_a^{b}f(b+a-x)dx$
we get $I_n={\int }_{-\pi }^{\pi }\frac{{\pi }^x\sin nx}{(1+{\pi }^x)\sin x}dx$ ...(2)
Adding equation (1) and (2), we have
$2I_n={\int }_{-\pi }^{\pi }\frac{\sin nx}{\sin x}dx=2{\int }_0^{\pi }\frac{\sin nx}{\sin x}dx$
($\because f\left(x\right)=\frac{\sin nx}{\sin x}$ is an even function)
$\Rightarrow I_n={\int }_0^{\pi }\frac{\sin nx}{\sin x}dx$
Now, $I_{n+2}-I_n={\int }_0^{\pi }\frac{\sin (n+2)x-\sin nx}{\sin x}dx={\int }_0^{\pi }\frac{2\cos (n+1)x.\sin x}{\sin x}dx$
$=2{\int }_0^{\pi }\cos (n+1)xdx=2{\left[\frac{\sin (n+1)x}{(n+1)}\right]}_0^{\pi }=0$
$\therefore I_{n+2}=I_n$ ...(3)
Since $I_n={\int }_0^{\pi }\frac{\sin nx}{\sin x}dx\Rightarrow I_1=\pi$ and $I_2=0$
From equation (3) $I_1=I_3=I_5=.....=\pi$ and $I_2=I_4=I_6=...=0$
$\Rightarrow {\sum }_{m=1}^{10}{I}_{2m+1}=10\pi$ and ${\sum }_{m=1}^{10}{I}_{2m}=0$
Therefore option (a),(b),(c) are correct