The value of $$\int_{-\pi/2}^{\pi/2} \frac{\cos^2 x}{1 + 3^x} dx$$ is:

We evaluate $$I = \int_{-\pi/2}^{\pi/2} \frac{\cos^2 x}{1 + 3^x} \, dx$$.

Using the substitution $$x \to -x$$ (King's property), we get $$I = \int_{-\pi/2}^{\pi/2} \frac{\cos^2(-x)}{1 + 3^{-x}} \, dx = \int_{-\pi/2}^{\pi/2} \frac{\cos^2 x \cdot 3^x}{1 + 3^x} \, dx$$.

Adding the two expressions for $$I$$: $$2I = \int_{-\pi/2}^{\pi/2} \cos^2 x \left(\frac{1}{1 + 3^x} + \frac{3^x}{1 + 3^x}\right) dx = \int_{-\pi/2}^{\pi/2} \cos^2 x \, dx$$.

Using the identity $$\cos^2 x = \frac{1 + \cos 2x}{2}$$: $$2I = \int_{-\pi/2}^{\pi/2} \frac{1 + \cos 2x}{2} \, dx = \frac{1}{2}\left[x + \frac{\sin 2x}{2}\right]_{-\pi/2}^{\pi/2} = \frac{1}{2}\left[\left(\frac{\pi}{2} + 0\right) - \left(-\frac{\pi}{2} + 0\right)\right] = \frac{\pi}{2}$$.

Therefore $$I = \frac{\pi}{4}$$.