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

Answer: (C)

$\displaystyle \sum^{10}_{ m = 1 } \, I _{ 2 m } = 10$

Answer 2

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).