Question

I = $\underset{-\frac{\pi }{2}}{\overset{\frac{\pi }{2}}{\int }}\frac{{\sin }^{2}x}{1+2^x}$ dx    ..... (i)
using property  $\underset{a}{\overset{b}{\int }}$f(x) dx = $\underset{a}{\overset{b}{\int }}$f(a + b - x) dx
I = $\underset{-\frac{\pi }{2}}{\overset{\frac{\pi }{2}}{\int }}\frac{{\sin }^{2}x}{1+2^x}$ dx   ..... (ii)
adding (i) and (ii)
2I = $\underset{0}{\overset{\pi /2}{\int }}$sin²x dx
$\Rightarrow$ 2I = 2.$\underset{0}{\overset{\pi /2}{\int }}$sin²x dx
$\Rightarrow$ 2I = 2 $\times \frac{\mathrm{\pi }}{4}\Rightarrow I=\frac{\mathrm{\pi }}{4}$