Question

$ \, if \, I_ n = \int^{\pi}_{ -\pi} \frac { \sin \, n \, x }{ ( 1 + \pi ^x ) \sin \, x } dx , \, n = 0 , 1 , 2 , .............., 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

The Correct Option is C

Given \(\, if \, I_ n = \int^{\pi}_{ -\pi} \frac { sin \, n \, x }{ ( 1 + \pi ^x ) \, sin \, x } dx ,...............(1)\) 
Using \(\int^b_a \, f ( x ) \, dx = \int^b _a \, f ( b + a - x ) \, dx , we \, get\)
\(\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, I_n = \int^{\pi}_ {-\pi } \frac { \pi^{x } \, sin \, n x}{ ( 1 + \pi^x ) sin \, x } \, dx ................(2)\) 
On adding Eqs. (i) and (ii), we have 
\(\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, 2I_n = \int^{\pi}_ {-\pi } \frac { sin \, n x}{ sin \, x } \, dx = 2 \int^{\pi}_ 0 \frac { sin \, n x}{ sin \, x } \, dx\)
\(\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, [ \because f ( x ) = \frac { sin \, n x}{ sin \, x } \, is \, an \, even \, function ]\)
\(\Rightarrow \, \, \, \, \, \, \, \, \,\) \(I_n = \int^\pi _ 0 \frac { sin \, nx }{ simn \, x } \, dx\)
\(Now , I_{ n + 2 } - I_n = \int^{\pi}_0 \frac { sin ( n - 2 ) \, x - sin \, nx }{ sin \, x } \, dx\)
\(\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, = \int^\pi_0 \frac { 2 \, cos ( n + 1 ) \, x . sin \, x }{ sin \, x } \, dx\)
\(\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, = 2 \int^\pi_0 cos ( n + 1 ) \, x \, dx = 2 \bigg [ \frac { sin ( n + 1 ) \, x }{ ( n + 1 ) } \bigg ]^{\pi}_0 = 0\)
\(\therefore \, I_{ n + 2 } = I_n \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, ...................(3)\)
\(Since , \, \, \, \, \, \, \, \, \, \, \, \, \, \, I_n = \int^{\pi}_0 \frac {sin \, nx }{ sin \, x } \, dx\)
\(\Rightarrow\) \(\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, I_1 = \pi \, and \, I_2 = 0\) 
From E (iii) \(I_1 = I_3 = I_5 = .................= \pi\) 
and \(\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \,\) \(I_2 = I_4 = I_6 = .................= 0\)
\(\Rightarrow\) \(\displaystyle \sum^{10} I_{2m + 1 } = 10 \pi \, and \, \displaystyle \sum^{10} I_{2m } = 0\)
\(\therefore\) Correct options are (A), (B), (C).