Given $I={\int }_{-\pi /2}^{\pi /2}\frac{{\cos }^{2}x}{1+3^x}dx ...\left(i\right)$

Using ${\int }_a^{b}f\left(x\right)dx={\int }_a^{b}f\left(a+b-x\right)dx,$ we get

$I={\int }_{-\pi /2}^{\pi /2}\frac{{\cos }^{2}x}{1+3^{-x}}dx={\int }_{-\pi /2}^{\pi /2}\frac{3^x{\cos }^{2}x}{1+3^x}dx ...\left(ii\right)$

Adding $\left(i\right) \& \left(ii\right)$ we get, 

$2I={\int }_{-\pi /2}^{\pi /2}\frac{\left(1+3^x\right){\cos }^{2}x}{1+3^x}dx$

$2I={\int }_{-\pi /2}^{\pi /2}{\cos }^{2}xdx$

$2I=2{\int }_0^{\pi /2}{\cos }^{2}xdx$ 

$I={\int }_0^{\pi /2}{\cos }^{2}xdx$

$I=\frac{1}{2}{\int }_0^{\pi /2}\left(1+\cos 2x\right)dx=\frac{1}{2}{\left[x+\frac{\sin 2x}{2}\right]}_0^{\pi /2}$

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