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

We need to evaluate the integral: $$\int_{-\pi/2}^{\pi/2} \frac{\sin^2 x}{1+2^x} dx.$$

Notice that the limits are symmetric about 0. Define the function: $$f(x) = \frac{\sin^2 x}{1+2^x}.$$

Compute $$f(-x)$$: $$f(-x) = \frac{\sin^2 (-x)}{1+2^{-x}} = \frac{(-\sin x)^2}{1+2^{-x}} = \frac{\sin^2 x}{1+2^{-x}}.$$ Since $$2^{-x} = \frac{1}{2^x}$$, rewrite the denominator: $$1 + 2^{-x} = 1 + \frac{1}{2^x} = \frac{2^x + 1}{2^x}.$$ Substitute: $$f(-x) = \frac{\sin^2 x}{\frac{2^x + 1}{2^x}} = \sin^2 x \cdot \frac{2^x}{2^x + 1} = \frac{\sin^2 x \cdot 2^x}{1 + 2^x}.$$

Now, add $$f(x)$$ and $$f(-x)$$: $$f(x) + f(-x) = \frac{\sin^2 x}{1+2^x} + \frac{\sin^2 x \cdot 2^x}{1+2^x} = \frac{\sin^2 x (1 + 2^x)}{1+2^x} = \sin^2 x.$$

Split the original integral: $$\int_{-\pi/2}^{\pi/2} f(x) dx = \int_{-\pi/2}^{0} f(x) dx + \int_{0}^{\pi/2} f(x) dx.$$ For the first integral, substitute $$u = -x$$, so $$du = -dx$$. When $$x = -\pi/2$$, $$u = \pi/2$$; when $$x = 0$$, $$u = 0$$. Thus: $$\int_{-\pi/2}^{0} f(x) dx = \int_{\pi/2}^{0} f(-u) (-du) = \int_{0}^{\pi/2} f(-u) du = \int_{0}^{\pi/2} f(-x) dx.$$ Therefore: $$\int_{-\pi/2}^{\pi/2} f(x) dx = \int_{0}^{\pi/2} f(-x) dx + \int_{0}^{\pi/2} f(x) dx = \int_{0}^{\pi/2} [f(x) + f(-x)] dx = \int_{0}^{\pi/2} \sin^2 x dx.$$

Now evaluate: $$\int_{0}^{\pi/2} \sin^2 x dx.$$ Use the identity: $$\sin^2 x = \frac{1 - \cos 2x}{2}.$$ So: $$\int_{0}^{\pi/2} \sin^2 x dx = \int_{0}^{\pi/2} \frac{1 - \cos 2x}{2} dx = \frac{1}{2} \int_{0}^{\pi/2} (1 - \cos 2x) dx.$$ Split the integral: $$\frac{1}{2} \left[ \int_{0}^{\pi/2} 1 dx - \int_{0}^{\pi/2} \cos 2x dx \right].$$

The first integral is: $$\int_{0}^{\pi/2} 1 dx = x \Big|_{0}^{\pi/2} = \frac{\pi}{2} - 0 = \frac{\pi}{2}.$$

For the second integral, substitute $$u = 2x$$, so $$du = 2 dx$$ and $$dx = \frac{du}{2}$$. When $$x = 0$$, $$u = 0$$; when $$x = \pi/2$$, $$u = \pi$$. Thus: $$\int_{0}^{\pi/2} \cos 2x dx = \int_{0}^{\pi} \cos u \cdot \frac{du}{2} = \frac{1}{2} \int_{0}^{\pi} \cos u du = \frac{1}{2} \left[ \sin u \right]_{0}^{\pi} = \frac{1}{2} (\sin \pi - \sin 0) = \frac{1}{2} (0 - 0) = 0.$$

Combine: $$\frac{1}{2} \left[ \frac{\pi}{2} - 0 \right] = \frac{1}{2} \times \frac{\pi}{2} = \frac{\pi}{4}.$$

Hence, the value of the integral is $$\frac{\pi}{4}$$. Comparing with the options:

A. $$\pi$$

B. $$\frac{\pi}{2}$$

C. $$4\pi$$

D. $$\frac{\pi}{4}$$

So, the correct answer is Option D.