Let $I=\displaystyle\int_{-\pi/2}^{\pi/2}\dfrac{\sin^2x}{1+2^x}dx$. Use the property $\int_{-a}^a f(x)dx$: replace $x\to-x$:

$I=\displaystyle\int_{-\pi/2}^{\pi/2}\dfrac{\sin^2x}{1+2^{-x}}dx=\displaystyle\int_{-\pi/2}^{\pi/2}\dfrac{2^x\sin^2x}{1+2^x}dx$

Adding: $2I=\displaystyle\int_{-\pi/2}^{\pi/2}\sin^2x\,dx=\displaystyle\int_{-\pi/2}^{\pi/2}\dfrac{1-\cos2x}{2}dx=\dfrac{\pi}{2}$

$I=\dfrac{\pi}{4}$