QJEEAdv16Ma5fb0: JEE Advanced 2016 Mathematics - Class 12 JEEAdv16Ma5fb0 Solution

Let $I = \int_{ - \pi /2}^{\pi /2} {{{{x^2}\cos x} \over {1 + {e^x}}}dx}$ ...... (i) [ $\because$ $\int_a^b {f(x)dx = \int_a^b {f(a + b - x)dx} }$ ]

$\Rightarrow I = \int_{ - \pi /2}^{\pi /2} {{{{x^2}\cos ( - x)} \over {1 + {e^{ - x}}}}dx}$ ...... (ii)

On adding Eqs. (i) and (ii), we get

$2I = \int_{ - \pi /2}^{\pi /2} {{x^2}\cos x\left[ {{1 \over {1 + {e^x}}} + {1 \over {1 + {e^{ - x}}}}} \right]dx}$

$= \int_{ - \pi /2}^{\pi /2} {{x^2}\cos x\,.\,(1)dx}$ [ $\because$ $\int_{ - a}^a {f(x)dx}$ $= 2\int_0^a {f(x)dx}$ , when $f( - x) = f(x)$ ]

$\Rightarrow 2I = 2\int_0^{\pi /2} {{x^2}\cos x\,dx}$

Using integration by parts, we get

$2I = 2[{x^2}(\sin x) - (2x)( - \cos x) + (2)( - \sin x)]_0^{\pi /2}$

$\Rightarrow 2I = 2\left[ {{{{\pi ^2}} \over 4} - 2} \right]$

$\therefore$ $I = {{{\pi ^2}} \over 4} - 2$