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

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

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

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