The value of $$\int_{-\pi/2}^{\pi/2} \frac{\sin^2 x}{1+2^x} dx$$ is:

Let us denote the required integral by $$I$$, so we have

$$I=\int_{-\pi/2}^{\pi/2}\frac{\sin^{2}x}{1+2^{x}}\;dx.$$

The limits are symmetric about the origin, therefore it is natural to look at the value of the integrand at $$-x$$. We perform the substitution $$x=-t$$ (so $$dx=-dt$$). Under this change:

When $$x=-\tfrac{\pi}{2}$$ we get $$t=\tfrac{\pi}{2}$$, and when $$x=\tfrac{\pi}{2}$$ we get $$t=-\tfrac{\pi}{2}$$. Swapping the limits produced by the minus sign restores the original order, giving

$$I=\int_{-\pi/2}^{\pi/2}\frac{\sin^{2}(-t)}{1+2^{-t}}\;dt.$$

Because $$\sin(-t)=-\sin t$$ and the square removes the sign, $$\sin^{2}(-t)=\sin^{2}t$$, so

$$I=\int_{-\pi/2}^{\pi/2}\frac{\sin^{2}t}{1+2^{-t}}\;dt.$$

Renaming the dummy variable $$t$$ back to $$x$$ for convenience, we have found that

$$I=\int_{-\pi/2}^{\pi/2}\frac{\sin^{2}x}{1+2^{-x}}\;dx.$$

Introduce a second integral

$$J=\int_{-\pi/2}^{\pi/2}\frac{\sin^{2}x}{1+2^{-x}}\;dx.$$

The previous equality shows directly that $$J=I$$. Now consider the sum $$I+J$$:

$$\begin{aligned} I+J&=\int_{-\pi/2}^{\pi/2}\sin^{2}x\!\left(\frac{1}{1+2^{x}}+\frac{1}{1+2^{-x}}\right)\!dx.\\[4pt] \end{aligned}$$

Inside the parentheses we simplify the second fraction. Since $$2^{-x}=\dfrac{1}{2^{x}}$$, we have

$$\frac{1}{1+2^{-x}}=\frac{1}{1+\dfrac{1}{2^{x}}}=\frac{2^{x}}{1+2^{x}}.$$

Therefore

$$\frac{1}{1+2^{x}}+\frac{1}{1+2^{-x}}=\frac{1}{1+2^{x}}+\frac{2^{x}}{1+2^{x}}=\frac{1+2^{x}}{1+2^{x}}=1.$$

Hence the integrand in $$I+J$$ reduces to simply $$\sin^{2}x$$, and we obtain

$$I+J=\int_{-\pi/2}^{\pi/2}\sin^{2}x\;dx.$$

But $$J=I$$, so

$$2I=\int_{-\pi/2}^{\pi/2}\sin^{2}x\;dx.$$

Now we evaluate the elementary integral on the right. First, recall the standard antiderivative

$$\int\sin^{2}x\;dx=\frac{x}{2}-\frac{\sin 2x}{4}+C.$$

Using this formula, compute the definite integral:

$$\begin{aligned} \int_{-\pi/2}^{\pi/2}\sin^{2}x\;dx &=\left[\frac{x}{2}-\frac{\sin 2x}{4}\right]_{-\pi/2}^{\pi/2}\\[4pt] &=\left(\frac{\pi/2}{2}-\frac{\sin\pi}{4}\right)-\left(\frac{-\pi/2}{2}-\frac{\sin(-\pi)}{4}\right)\\[4pt] &=\left(\frac{\pi}{4}-0\right)-\left(-\frac{\pi}{4}-0\right)\\[4pt] &=\frac{\pi}{4}+\frac{\pi}{4}\\[4pt] &=\frac{\pi}{2}. \end{aligned}$$

Substituting this back into $$2I=\dfrac{\pi}{2}$$ gives

$$2I=\frac{\pi}{2}\quad\Longrightarrow\quad I=\frac{\pi}{4}.$$

Thus the value of the original integral is $$\dfrac{\pi}{4}$$.

Hence, the correct answer is Option A.