Question

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

(1) $I_n=\underset{-\pi }{\overset{\pi }{\int }}\frac{\sin nx}{\left(1+{\pi }^x\right)\sin x}dx$
$I_n=\underset{-\pi }{\overset{\pi }{\int }}\frac{{\pi }^x\sin nx}{\left(1+{\pi }^x\right)\sin x}dx$
(by property $\underset{a}{\overset{b}{\int }}f(x)dx=\underset{a}{\overset{b}{\int }}f(a+b-x)dx$ )
$2I_n=\underset{-\pi }{\overset{\pi }{\int }}\frac{\sin nx}{\sin x}dx$
$2I_n=2\underset{0}{\overset{\pi }{\int }}\frac{\sin nx}{\sin x}dx$
$I_n=\underset{0}{\overset{\pi }{\int }}\frac{\sin nx}{\sin x}dx$
$I_{n+2}-I_n=\underset{0}{\overset{\pi }{\int }}\frac{\sin (n+2)x-\sin nx}{\sin x}dx$
$=\underset{0}{\overset{b}{\int }}\frac{2\cos (n+1)x\sin x}{\sin x}dx$
$=2{\left[\frac{\sin (n+1)x}{(n+1)}\right]}_0^{\pi }=0$
$\Rightarrow I_{n+2}=I_n$
(2) $I_3=I_5=\dots ..=I_{21}$
$\therefore \sum _{m=1}^{10}{I}_{2m+1}=10I_3=10\underset{0}{\overset{\pi }{\int }}\frac{\sin 3x}{\sin x}dx=10\underset{0}{\overset{\pi }{\int }}\left(3-4{\sin }^{2}x\right)dx$
$=10[3x-2x+2\sin 2x{]}_0^{\pi }=10\pi$
(3) $I_2=I_4=\dots \dots ..=I_{20}$
$\sum _{m=1}^{10}{I}_{2m}=10\underset{0}{\overset{\pi }{\int }}\frac{\sin 2x}{\sin x}dx=20[\sin x{]}_0^{\pi }=0$