Question

If $I_{n}=\int\limits_{-\pi}^{\pi} \frac{\sin n x}{\left(1+\pi^{x}\right) \sin x} d x, n=0,1,2, \ldots$, then

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

Answer 1

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

Given $ I_{n}=\int\limits_{-\pi}^{\pi} \frac{\sin n x}{\left(1+\pi^{x}\right) \sin x} d x ....$ (i)
Using $ \int\limits_{a}^{b} f(x) d x=\int\limits_{a}^{b} f(b+a-x) d x$, we get
$I_{n}=\int\limits_{-\pi}^{\pi} \frac{\pi^{x} \sin n x}{\left(1+\pi^{x}\right) \sin x} d x .....$(ii)
On adding Eqs. (i) and (ii), we have
$2 I_{n}=\int\limits_{-\pi}^{\pi} \frac{\sin n x}{\sin x} d x=2 \int\limits_{0}^{\pi} \frac{ \sin n x}{\sin x}$
${\left[\because f(x)=\frac{\sin n \sin x}{\sin } d x\right.}$ is an even function]
$\Rightarrow I_n = \int\limits_{0}^{\pi} \frac{\sin n x}{\sin x} d x $
Now, $ I_{n+2}-I_{n}=\int\limits_{0}^{\pi} \frac{\sin (n+2) x-\sin n x}{\sin x} d x$
$ = \int\limits_{0}^{\pi} \frac{2\cos (n + 1) x \cdot \sin x}{\sin x} dx$
$=2 \int\limits_{0}^{\pi} \cos (n+1) x d x=2\left[\frac{\sin (n+1) x}{(n+1)}\right]_{0}^{\pi}=0$
$\therefore I_{n+2}=I_{n} .....$(iii)
Since, $ I_{n}=\int\limits_{0}^{\pi} \frac{\sin n x}{\sin x} d x$
$\Rightarrow I_{1}=\pi$ and $I_{2}=0$
From Eq. (iii) $ I_{1}=I_{3}=I_{5}=\ldots=\pi $
and $ I_{2}=I_{4}=I_{6}=\ldots=0$
$\Rightarrow \displaystyle\sum_{m=1}^{10} I_{2 m+1}=10 \pi$
and $\displaystyle\sum_{m=1}^{10} I_{2 m}=0$
$\therefore$ Correct options are (a), (b), (c).