Answer:
Option C
Explanation:
Key Idea
Use property = $\int_{a}^{b} f( x) dx=\int_{a}^{b} f( a+b-x)dx$
Let $I=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{\sin ^{2}x}{1+2^{x}} dx$
$\Rightarrow$ I $= \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\frac{\sin^{2}(\frac{-\pi}{2}+\frac{\pi}{2}-x)}{1+2^{\frac{-\pi}{2}+\frac{\pi}{2}-x}} dx $
[ $\because $ $\int_{a}^{b} f( x) dx=\int_{a}^{b} f( a+b-x)dx$ ]
$\Rightarrow$ $I=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{\sin^{2}x}{1+2^{-x}}dx$
$\Rightarrow$ $I=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{2^{x}\sin^{2}x}{2^{x}+1^{}}dx$
$\Rightarrow$ $2I=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin^{2}x(\frac{2^{x}+1}{2^{x}+1})dx$
$\Rightarrow$ $2I=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin^{2}x dx$
$\Rightarrow$ $2I=2\int_{0}^{\frac{\pi}{2}}\sin^{2}x dx $ [ $\because\sin^{2}x$ is an even function ]
$\Rightarrow$ $I= \int_{0}^{\frac{\pi}{2}} \sin^{2}x dx$
$\Rightarrow$ $I= \int_{0}^{\frac{\pi}{2}} \cos^{2}x dx$
[ $\because \int_{0}^{a}f(x)dx=\int_{0}^{a}f(a-x)dx $]
$\Rightarrow$ 2I=$\int_{0}^{\frac{\pi}{2}} dx$
$\Rightarrow$ $2I= [ x] _{0}^{\frac{\pi}{2}}=I=\frac{\pi}{4}$
