JEE Main 2018 — Definite Integration Question with Solution

$I= \underset{-\frac{\pi }{2}}{\overset{\frac{\pi }{2}}{\int }}\frac{{\sin }^{2}x}{1+2^x} dx ...\left(\mathrm{i}\right)$

As $\underset{a}{\overset{b}{\int }}f\left(x\right)dx=\underset{a}{\overset{b}{\int }}f(a+b-x)dx$

$I=\underset{-\frac{\pi }{2}}{\overset{\frac{\pi }{2}}{\int }}\frac{{\sin }^2\left(-x\right)}{1+2^{-x}}dx$

$I= \underset{-\frac{\pi }{2}}{\overset{\frac{\pi }{2}}{\int }}\frac{2^x{\sin }^{2}x}{1+2^x}dx ...\left(\mathrm{ii}\right)$ 

By adding Equations $\left(\mathrm{i}\right)$ and $\left(\mathrm{ii}\right)$,

$2I= \underset{-\frac{\pi }{2}}{\overset{\frac{\pi }{2}}{\int }}\frac{\left(1+2^x\right){\sin }^{2}x}{\left(1+2^x\right)}dx$

$2I=2\underset{0}{\overset{\frac{\pi }{2}}{\int }}{\sin }^{2}xdx$

$I=\underset{0}{\overset{\frac{\pi }{2}}{\int }}{\sin }^{2}xdx ...\left(iii\right)$

$I=\underset{0}{\overset{\frac{\pi }{2}}{\int }}{\sin }^2\left(\frac{\pi }{2}-x\right)dx$

$I=\underset{0}{\overset{\frac{\pi }{2}}{\int }}{\cos }^{2}xdx ...\left(iv\right)$

On adding $\left(iii\right) \& \left(iv\right)$, we get

$2I= \underset{0}{\overset{\frac{\pi }{2}}{\int }}\left({\sin }^{2}x+{\cos }^{2}x\right)dx$

$2I= \underset{0}{\overset{\frac{\pi }{2}}{\int }}1 dx$

$2I= {\left[x\right]}_0^{\frac{\pi }{2}}$
$2I=\frac{\pi }{2}-0$
$I=\frac{\pi }{2}\times \frac{1}{2}=\frac{\pi }{4}$