Question

If $I=\underset{-\pi /2}{\overset{\pi /2}{\int }}\frac{{\sin }^{2}x}{1+4^x}dx,$ then the value of $\frac{24}{\pi }I$ is

Answer 1

$I=\underset{-\pi /2}{\overset{\pi /2}{\int }}\frac{{\sin }^{2}x}{1+4^x}dx\cdots \left(1\right)$
$\Rightarrow I=\underset{-\pi /2}{\overset{\pi /2}{\int }}\frac{{\sin }^2(-x)}{1+4^{-x}}dx$
$\Rightarrow I=\underset{-\pi /2}{\overset{\pi /2}{\int }}\frac{{\sin }^{2}x}{1+4^{-x}}dx$
$\Rightarrow I=\underset{-\pi /2}{\overset{\pi /2}{\int }}\frac{4^x{\sin }^{2}x}{1+4^x}dx\cdots \left(2\right)$
Adding $\left(1\right)$ and $\left(2\right),$ we get
$2I=\underset{-\pi /2}{\overset{\pi /2}{\int }}{\sin }^{2}xdx$
$\Rightarrow 2I=2\underset{0}{\overset{\pi /2}{\int }}{\sin }^{2}xdx$
$\Rightarrow I=\underset{0}{\overset{\pi /2}{\int }}{\sin }^{2}xdx=\frac{\pi }{4}$