The value of \(\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{\sin ^2 x}{1+2^x} \mathrm{~d} x\) is

Let \(\mathrm{I}=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{\sin ^2 x}{1+2^x} \mathrm{~d} x\)...(i)
\(\therefore \quad \mathrm{I}=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{\sin ^2 x}{1+2^{-x}} \mathrm{~d} x\)...(ii)
\(\cdots\left[\because \int_{\mathrm{a}}^{\mathrm{b}} \mathrm{f}(x) \mathrm{d} x=\int_{\mathrm{a}}^{\mathrm{b}} \mathrm{f}(\mathrm{a}+\mathrm{b}-x) \mathrm{d} x\right]\)
Adding (i) and (ii), we get
\(\begin{aligned}
& 2 \mathrm{I}=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin ^2 x \mathrm{~d} x \\
& \Rightarrow 2 \mathrm{I}=2 \int^{\frac{\pi}{2}} \sin ^2 x \mathrm{~d} x
\end{aligned}\)
\(\left[\begin{array}{l}\because \int_{-a}^a f(x) \mathrm{d} x=2 \int_0^a \mathrm{f}(x) \mathrm{d} x \\ \quad \text { if } \mathrm{f}(x) \text { is an even function }\end{array}\right]\)
\(\begin{aligned} \Rightarrow \mathrm{I} & =\int_0^{\frac{\pi}{2}}\left(\frac{1-\cos 2 x}{2}\right) \mathrm{d} x \\ & =\frac{1}{2}\left[x-\frac{\sin 2 x}{2}\right]_0^{\frac{\pi}{2}} \\ & =\frac{1}{2}\left(\frac{\pi}{2}-0\right)=\frac{\pi}{4}\end{aligned}\)