Question

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

The correct options are
A $I_n=I_{n+2}$
B $\sum _{m=1}^{10}{I}_{2m+1}=10\pi$
C $\sum _{m=1}^{10}{I}_{2m}=0$

$I_n={\int }_{-\pi }^{\pi }\frac{\sin nx}{(1+{\pi }^x)\sin x}dx$ .....(1)
$I_n={\int }_{-\pi }^{\pi }\frac{{\pi }^x\sin nx}{(1+{\pi }^x)\sin x}dx$ ($\because {\int }_a^{b}f\left(x\right)dx={\int }_a^{b}f(a+b-x)$) .....(2)
Adding (1) and (2), we get
$2I_n={\int }_{-\pi }^{\pi }\frac{(1+{\pi }^x)\sin nx}{(1+{\pi }^x)\sin x}dx$
$2I_n={\int }_{-\pi }^{\pi }\frac{\sin nx}{\sin x}dx$
Here, $f(-x)=f\left(x\right)$ i.e. even function.
$2I_n=2{\int }_0^{\pi }\frac{\sin nx}{\sin x}dx$
$\Rightarrow I_n={\int }_0^{\pi }\frac{\sin nx}{\sin x}dx$
$I_{n+2}={\int }_0^{\pi }\frac{\sin (n+2)x}{\sin x}dx$
$\Rightarrow I_{n+2}-I_n={\int }_0^{\pi }\frac{\sin (n+2)x-\sin nx}{\sin x}dx$
$I_{n+2}-I_n={\int }_0^{\pi }\frac{2\cos (n+1)x\sin x}{\sin x}dx$
$I_{n+2}-I_n=2{\int }_0^{\pi }\cos (n+1)xdx$
$=2{\left[\frac{\sin (n+1)x}{(n+1)}\right]}_0^{\pi }=0$
$\Rightarrow I_{n+2}=I_n$
On substituting the values of n=0,1....
$I_2=I_0$
$I_3=I_1$
$I_4=I_2$.....
Here, $I_1=\pi$
$I_2=2{\int }_0^{\pi }\cos xdx=0$
Hence, ${\sum }_{m=1}^{10}{I}_{2m}=0$
${\sum }_{m=1}^{10}{I}_{2m+1}=I_3+I_5+.....I_{21}=10\pi$