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_{m=1}^{10} I_{2 m+1}=10 \pi$
• C. $\displaystyle\sum_{m=1}^{10} I_{2 m}=0$
• D. $I_n=I_{n+1}$

Answer 1

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

(1) $I_n=\int\limits_{-\pi}^\pi \frac{\sin n x}{\left(1+\pi^x\right) \sin x} d x$
$I_n=\int\limits_{-\pi}^\pi \frac{\pi^x \sin n x}{\left(1+\pi^x\right) \sin x} d x$
(by property $\int\limits_a^b f(x) d x=\int\limits_a^b f(a+b-x) d x$ )
$2 I_n=\int\limits_{-\pi}^\pi \frac{\sin n x}{\sin x} d x$
$2 I_n=2 \int\limits_0^\pi \frac{\sin n x}{\sin x} d x$
$I_n=\int\limits_0^\pi \frac{\sin n x}{\sin x} d x$
$I_{n+2}-I_n=\int\limits_0^\pi \frac{\sin (n+2) x-\sin n x}{\sin x} d x$
$=\int\limits_0^b \frac{2 \cos (n+1) x \sin x}{\sin x} d x$
$=2\left[\frac{\sin (n+1) x}{(n+1)}\right]_0^\pi=0$
$\Rightarrow I_{n+2}=I_n$
(2) $I_3=I_5=\ldots . .=I_{21}$
$ \therefore \displaystyle\sum_{m=1}^{10} I_{2 m+1}=10 I_3=10 \int\limits_0^\pi \frac{\sin 3 x}{\sin x} d x=10 \int\limits_0^\pi\left(3-4 \sin ^2 x\right) d x $
$ =10[3 x-2 x+2 \sin 2 x]_0^\pi=10 \pi$
(3) $I _2= I _4=\ldots \ldots . .= I _{20}$
$\displaystyle\sum_{m=1}^{10} I_{2 m}=10 \int\limits_0^\pi \frac{\sin 2 x}{\sin x} d x=20[\sin x]_0^\pi=0$