For a > 0, the value of $\underset{-\pi }{\overset{\pi }{\int }}\frac{{\cos }^{2}x}{1+a^x}dx$ is 

$Let I=\underset{-\pi }{\overset{\pi }{\int }}\frac{{\cos }^{2}x}{1+a^x}dx,a>0 ..........\left(1\right)$

$\therefore I=\underset{-\pi }{\overset{\pi }{\int }}\frac{{\cos }^{2}x}{1+a^{-x}}dx=\underset{-\pi }{\overset{\pi }{\int }}\frac{a^x{\cos }^{2}x}{1+a^x}dx ............\left(2\right)$

Adding (1) and (2) we get

$2I=\underset{-\pi }{\overset{\pi }{\int }}{\cos }^{2}x\Rightarrow {}^1\cancel{2}I={}^2 \cancel{4}\underset{0}{\overset{\pi /2}{\int }}{\cos }^{2}xdx$

$I=\cancel{2}\left[\frac{\pi }{2}.\frac{1}{\cancel{2}}\right]=\frac{\pi }{2}$