Question

The value of the integral $\int ^{\frac{\pi}{2}}_{-\frac{\pi}{2}} \bigg ( x^2 + \log \frac{\pi - x }{ \pi + x } \bigg ) \, \cos \, x \, dx$

• A. $0$
• B. $\frac{\pi^2 }{2} - 4$
• C. $\frac{\pi^2 }{2} + 4$
• D. $\frac{ \pi ^2 }{ 2 }$

Answer 1

Answer: (B)

$\frac{\pi^2 }{2} - 4$

Answer 2

$I = \int ^{\frac{\pi}{2}}_{-\frac{\pi}{2}} \bigg [ x^2 + \log \frac{\pi - x }{ \pi + x } \bigg ] \, \cos \, x \, dx$
$As, \int^a_{-a} f ( x ) \, dx = 0 , when \, f ( -x) = -f ( x )$
$\therefore I = I = \int ^{\frac{\pi}{2}}_{-\frac{\pi}{2}} x^2 \, \cos \, x \, dx + 0 = 2 \int^{\frac{\pi}{2} }_0 ( x^2 \cos x ) \, dx$
\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, = 2 \bigg \\{ ( x^2 \, \sin \, x )^{\frac{\pi}{2}}_0 - \int^{\frac{\pi}{2} }_0 \, 2 \, x.\sin\, x \, dx \bigg \\}
\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, = 2 \bigg [ \frac{\pi^2 }{4} -2 \bigg \\{ ( - x .\cos \, x )^{\frac{\pi}{2}} _0 - \int^{\frac{\pi}{2}}_0 1 . ( - \cos \, x ) \, dx \bigg \\} \bigg ]
$2 \bigg [ \frac{ \pi^2}{4} - 2 ( \sin \, x )^{\frac{\pi}{2}}_0\bigg ] = 2 \bigg [ \frac{\pi^2}{4} - 2 \bigg ] = \bigg ( \frac{ \pi^2 }{ 2} - 4\bigg )$