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

$\begin{array}{l}I=\underset{-\pi }{\overset{\pi }{\int }}\frac{{\cos }^{2}x}{1+a^x}dx\\ =\underset{0}{\overset{\pi }{\int }}[\frac{{\cos }^{2}x}{1+a^x}+\frac{{\cos }^2(-x)}{1+a^{-x}}]dx\\ [\therefore \underset{-a}{\overset{a}{\int }}f\left(x\right)dx=\underset{0}{\overset{a}{\int }}f\left(x\right)+f(-x)dx]\\ Iff\left(x\right)isneitherevennorodd\\ \underset{0}{\overset{\pi }{\int }}{\cos }^{2}x\left[\frac{1}{1+a^x}+\frac{1}{1+\frac{1}{a^x}}\right]dx\\ \underset{0}{\overset{\pi }{\int }}{\cos }^x\left(\frac{{\displaystyle 1+a^x}}{{\displaystyle 1+a^x}}\right)dx\\ =2\underset{0}{\overset{\pi /2}{\int }}{\cos }^{2}xdx\\ \left[\therefore \underset{0}{\overset{a}{\int }}f\left(x\right)dx=2\underset{0}{\overset{a/2}{\int }}f\left(x\right)dxiff(a-x)=f\left(x\right)\right]\\ =2\left[\frac{1}{2}.\frac{\pi }{2}\right]=\pi /2\end{array}$