Question

Evaluate:
${\int }_{-\pi /2}^{\pi /2}\frac{{\sin }^{2}x}{1+2^x}dx$

Answer 1

$I={\int }_{-\pi /2}^{\pi /2}\frac{{\sin }^{2}x}{1+2^x}dx$

$={\int }_{-\pi /2}^{\pi /2}\frac{{\sin }^2(0-x)}{1+2^{0-x}}dx$ using ${\int }_a^{b}f\left(x\right)dx={\int }_a^{b}f(a+b-x)dx$

$2I={\int }_{-\pi /2}^{\pi /2}\frac{{\sin }^{2}x}{1+2^x}+\frac{2^x{\sin }^{2}x}{2^x+1}dx$

$2I={\int }_{-\pi /2}^{\pi /2}\frac{{\sin }^{2}x(1+2^x)}{1+2^x}dx$

$2I={\int }_{-\pi }^{\pi }{\sin }^{2}xdx$

$2I=2{\int }_0^{\pi /2}{\sin }^{2}xdx$ using ${\int }_{-a}^{a}f\left(x\right)dx=2{\int }_0^{a}f\left(x\right)dx$ when $f\left(x\right)$ is even.

$={\int }_0^{\pi /2}(1-\cos 2x)dx$ using $\cos x=1-2{\sin }^{2}x$

$2I={[x-\frac{\sin 2x}{2}]}_0^{\pi /2}=(\frac{\pi }{2}-0)-(0,0)$

$\therefore 2I=\frac{\pi }{2}$

$\Rightarrow I=\frac{\pi }{4}$.