The value of $\int\limits_{ - \dfrac{\pi }{2}}^{\dfrac{\pi }{2}} {\dfrac{{{{\sin }^2}x}}{{1 + {2^x}}}dx} $ is
A) $4\pi $
B) $\dfrac{\pi }{4}$
C) $\dfrac{\pi }{8}$
D) $\dfrac{\pi }{2}$

Hint: If $I$ is given as $I = \int\limits_a^b {f\left( x \right)dx} $, then $x$ can be replaced by $a + b - x$, then the value of $I$ remains unchanged. So, $I = \int\limits_a^b {f\left( {a + b - x} \right)dx} $.
Then add both, we will get our value.

Complete step-by-step answer:
Let $I = \int\limits_{ - \dfrac{\pi }{2}}^{\dfrac{\pi }{2}} {\dfrac{{{{\sin }^2}x}}{{1 + {2^x}}}dx} $.................(1)
As we know that if $I = \int\limits_a^b {f\left( x \right)dx} $, then by using king property , $I = \int\limits_a^b {f\left( {a + b - x} \right)dx} $.
So, using king property in equation (1),
$\Rightarrow$$I = \int\limits_{ - \dfrac{\pi }{2}}^{\dfrac{\pi }{2}} {\dfrac{{{{\sin }^2}\left( {\dfrac{\pi }{2} - \dfrac{\pi }{2} - x} \right)}}{{1 + {2^{\left( {\dfrac{\pi }{2} - \dfrac{\pi }{2} - x} \right)}}}}} dx$
$\Rightarrow$$I = \int\limits_{ - \dfrac{\pi }{2}}^{\dfrac{\pi }{2}} {\dfrac{{{{\sin }^2}\left( { - x} \right)}}{{1 + {2^{ - x}}}}} dx$
As we know $\sin \left( { - x} \right) = - \sin x{\text{ , so, }}{\sin ^2}\left( { - x} \right) = {\sin ^2}x$.
$
\Rightarrow I = \int\limits_{ - \dfrac{\pi }{2}}^{\dfrac{\pi }{2}} {\dfrac{{{{\sin }^2}\left( x \right)}}{{1 + \dfrac{1}{{{2^x}}}}}} dx \\
  {\text{Now taking LCM of }}{2^x}, \\
\Rightarrow I = \int\limits_{ - \dfrac{\pi }{2}}^{\dfrac{\pi }{2}} {\dfrac{{{2^x}{{\sin }^2}\left( x \right)}}{{{2^x} + 1}}} dx…………………….{\text{ (2)}} \\
    \\
 $
Now, if we add equation (1) and (2),
$
\Rightarrow I + I = \int\limits_{ - \dfrac{\pi }{2}}^{\dfrac{\pi }{2}} {\left( {\dfrac{{{2^x}{{\sin }^2}\left( x \right)}}{{{2^x} + 1}} + \dfrac{{{{\sin }^2}x}}{{1 + {2^x}}}} \right)} dx \\
\Rightarrow 2I = \int\limits_{ - \dfrac{\pi }{2}}^{\dfrac{\pi }{2}} {{{\sin }^2}x\dfrac{{{2^x} + 1}}{{{2^x} + 1}}}
\Rightarrow 2I = \int\limits_{ - \dfrac{\pi }{2}}^{\dfrac{\pi }{2}} {{{\sin }^2}x} dx \\
 $
As we know that $\cos 2\theta = 1 - 2{\sin ^2}\theta $, so, ${\sin ^2}\theta = \dfrac{{1 - \cos 2\theta }}{2}$. Using this value we get,
$
\Rightarrow 2I = \int\limits_{ - \dfrac{\pi }{2}}^{\dfrac{\pi }{2}} {\dfrac{{1 - \cos 2x}}{2}dx} \\
\Rightarrow 2I = \dfrac{1}{2}\int\limits_{ - \dfrac{\pi }{2}}^{\dfrac{\pi }{2}} {\left( {1 - \cos 2x} \right)dx} \\
\Rightarrow 2I = \dfrac{1}{2}{\left[ {x - \dfrac{{\sin 2x}}{2}} \right]_{ - \dfrac{\pi }{2}}}^{\dfrac{\pi }{2}} \\
\Rightarrow I = \dfrac{1}{4}\left[ {\left( {\dfrac{\pi }{2} - \left( { - \dfrac{\pi }{2}} \right)} \right) - \left( {\dfrac{{\sin \pi }}{2} - \dfrac{{\sin \left( { - \pi } \right)}}{2}} \right)} \right] \\
\Rightarrow I = \dfrac{1}{4}\left[ {\pi - \sin \pi } \right] = \dfrac{\pi }{4} \\
 $
(As we know $\sin n\pi = 0$ ,where $n$ is an integer)
So, value of $\int\limits_{ - \dfrac{\pi }{2}}^{\dfrac{\pi }{2}} {\dfrac{{{{\sin }^2}x}}{{1 + {2^x}}}dx} $ is $\dfrac{\pi }{4}$

So, option B is the correct answer.

Note: As we know, if $f\left( x \right) = f\left( { - x} \right)$, then it is an even function and when $f\left( x \right)$ is an even function, then $\int\limits_{ - a}^a {f\left( x \right)dx = 2\int\limits_0^a {f\left( x \right)dx} } $. And if $f\left( x \right)$ is odd, means $f\left( x \right) = - f\left( { - x} \right)$, then $\int\limits_{ - a}^a {f\left( x \right)dx = 0} $.