Question
The value of $$\int\limits_{\frac{\pi }{2}}^{\frac{\pi }{2}} {\frac{{{x^2}\cos \,x}}{{1 + {e^x}}}dx} $$ is equal to-
A.
$$\frac{{{\pi ^2}}}{4} - 2$$
B.
$$\frac{{{\pi ^2}}}{4} + 2$$
C.
$${\pi ^2} - {e^{\frac{\pi }{2}}}$$
D.
$${\pi ^2} + {e^{\frac{\pi }{2}}}$$
Answer :
$$\frac{{{\pi ^2}}}{4} - 2$$
Solution :
$$I = \int_{ - \frac{\pi }{2}}^{\frac{\pi }{2}} {\frac{{{x^2}\cos \,x}}{{1 + {e^x}}}\,dx} .....(1)$$
Applying $$\int_a^b {f\left( x \right)dx = \int_a^b {f\left( {a + b - x} \right)dx,} } $$ we get
$$\eqalign{
& I = \int_{ - \frac{\pi }{2}}^{\frac{\pi }{2}} {\frac{{{e^x}{x^2}\cos \,x}}{{1 + {e^x}}}\,dx} .....(2) \cr
& {\text{Adding }}(1){\text{ and }}(2),{\text{ we get}} \cr
& 2I = \int_{ - \frac{\pi }{2}}^{\frac{\pi }{2}} {{x^2}\,\cos \,x\,dx} = 2\int\limits_0^{\frac{\pi }{2}} {{x^2}\cos \,x\,dx} \cr
& I = \left[ {{x^2}\,\sin \,x + 2x\,\cos \,x - 2\,\sin \,x} \right]_0^{\frac{\pi }{2}}\, = \frac{{{\pi ^2}}}{4} - 2 \cr} $$