Question

The value of \[\int\limits_{ - \dfrac{\pi }{2}}^{\dfrac{\pi }{2}} {\dfrac{{{x^2}\cos x}}{{1 + {e^x}}}} dx\] is equal to
(A) $\dfrac{{{\pi ^2}}}{4} - 2$
(B) $\dfrac{{{\pi ^2}}}{4} + 2$
(C) ${\pi ^2} - {e^{\dfrac{\pi }{2}}}$
(D) ${\pi ^2} + {e^{\dfrac{\pi }{2}}}$

Answer 1

Hint: To solve this integral we use some property of integral
1. $\int\limits_{ - a}^a {f(x)dx = \int\limits_{ - a}^a {f( - x)dx} } $
2. $\int\limits_{ - a}^a {f(x)dx = 2\int\limits_0^a {f(x)dx} } $ this property holds only when the given function is even.
3. Integration by part:
$\int {(u.v)dx} = u\int {vdx - \int {\left( {\dfrac{{du}}{{dx}} \times \int {vdx} } \right)} } dx$

Complete step-by-step answer:
Given function $I$= \[\int\limits_{ - \dfrac{\pi }{2}}^{\dfrac{\pi }{2}} {\dfrac{{{x^2}\cos x}}{{1 + {e^x}}}} dx\] - - - - - - equation (1)
Now by using property $\int\limits_{ - a}^a {f(x)dx = \int\limits_{ - a}^a {f( - x)dx} } $
By using this we can write
$I$= \[\int\limits_{ - \dfrac{\pi }{2}}^{\dfrac{\pi }{2}} {\dfrac{{{{( - x)}^2}\cos ( - x)}}{{1 + {e^{ - x}}}}} dx\]
Now arrange this equation
We get $I$=\[\int\limits_{ - \dfrac{\pi }{2}}^{\dfrac{\pi }{2}} {\dfrac{{{x^2}{e^x}\cos x}}{{1 + {e^x}}}} dx\] - - - - - - - - -equation (2)
$\because \cos ( - x) = \cos x$
Now add equation 1 and equation 2
$2I$= \[\int\limits_{ - \dfrac{\pi }{2}}^{\dfrac{\pi }{2}} {\dfrac{{{x^2}\cos x}}{{1 + {e^x}}}} dx\] + \[\int\limits_{ - \dfrac{\pi }{2}}^{\dfrac{\pi }{2}} {\dfrac{{{x^2}{e^x}\cos x}}{{1 + {e^x}}}} dx\]
Now we arrange this
$2I$= \[\int\limits_{ - \dfrac{\pi }{2}}^{\dfrac{\pi }{2}} {\dfrac{{{x^2}\cos x(1 + {e^x})}}{{1 + {e^x}}}} dx\]
$2I$= \[\int\limits_{ - \dfrac{\pi }{2}}^{\dfrac{\pi }{2}} {{x^2}\cos x} dx\]
Now by using property second but first we check the function $f(x) = {x^2}\cos x$ is even or not
So when a function is even it follow following property
$f( - x) = f(x)$
So $f( - x)$ = ${( - x)^2}\cos ( - x)$
So $f( - x) = {x^2}\cos x$
So we can say that the given function is an even function
So $2I$= \[\int\limits_{ - \dfrac{\pi }{2}}^{\dfrac{\pi }{2}} {{x^2}\cos x} dx\] can be written as
$2I$= \[2\int\limits_0^{\dfrac{\pi }{2}} {{x^2}\cos x} dx\] by using property 2 given in hint.
So from here
$I$= \[\int\limits_0^{\dfrac{\pi }{2}} {{x^2}\cos x} dx\]
Now by using integration by part theorem
$I = {x^2}\int\limits_0^{\dfrac{\pi }{2}} {\cos x} - \int\limits_0^{\dfrac{\pi }{2}} {\dfrac{d}{{dx}}} {x^2} \times \int {\cos xdx} $
$I = \left[ {{x^2}\sin x} \right]_0^{\dfrac{\pi }{2}} - \int\limits_0^{\dfrac{\pi }{2}} {2x.\sin x} $$dx$
Now we solve $\int\limits_0^{\dfrac{\pi }{2}} {2x.\sin x} dx$
Now we use integration by part theorem
$\int\limits_0^{\dfrac{\pi }{2}} {2x.\sin x} dx$ = $2x\int {\sin x - \int {\dfrac{d}{{dx}}} } (2x)\int {\sin xdx} $
So from this we get
\[2x( - \cos x) - \int {2( - \cos x)dx} \]
$ - 2x\cos x + 2\sin x$
Now from this we can write
$I = [{x^2}\sin x + 2x\cos x - 2\sin x]_0^{\dfrac{\pi }{2}}$
Now putting limit
We get
$I = \left[ {\dfrac{{{\pi ^2}}}{4}\sin \dfrac{\pi }{2} + 2 \times \dfrac{\pi }{2}\cos \dfrac{\pi }{2} - 2\sin \dfrac{\pi }{2}} \right] - \left[ {0 \times \sin 0 + 2 \times 0 \times \cos 0 - 2\sin 0} \right]$
now after solving we get
 $I = $ $\dfrac{{{\pi ^2}}}{4} + 0 - 2$
$\because \sin \dfrac{\pi }{2} = \cos 0 = 1$ and $\sin 0 = 0$
So from here we can say that
$I = \dfrac{{{\pi ^2}}}{4} - 2$
So option A is correct.

Note: $\int\limits_{ - a}^a {f(x) = 0} $ when the $f(x)$ is an odd function. This property is always true for odd functions.
A function $f(x)$ is said to be an odd function if $f( - x) = - f(x)$ .