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

$\begin{array}{l}\mathrm{Let} \mathrm{I}={\int }_{-\mathrm{\pi }}^{\mathrm{\pi }} \frac{{\cos }^2\mathrm{x}}{1+{\mathrm{a}}^{\mathrm{x}}}\mathrm{dx} \to \left(1\right) \\ By u\sin g \boxed{{\int }_a^{b}f\left(x\right) dx={\int }_a^{b}f(a+b-x) dx} \\ \\ I={\int }_{-\mathrm{\pi }}^{\mathrm{\pi }} \frac{{\cos }^2(-\mathrm{x})}{1+{\mathrm{a}}^{-\mathrm{x}}}\mathrm{dx} \\ I={\int }_{-\pi }^{\pi }\frac{{\cos }^{2}x}{1+a^{-x}}dx \to \left(2\right) \\ \\ By adding \left(1\right) and \left(2\right) ;\\ 2\mathrm{I}={\int }_{-\mathrm{\pi }}^{\mathrm{\pi }} \frac{\left({\mathrm{a}}^{\mathrm{x}}+1\right)}{\left({\mathrm{a}}^{\mathrm{x}}+1\right)}{\cos }^2\mathrm{xdx}\\ ={\int }_{-\mathrm{\pi }}^{\mathrm{\pi }} \frac{\cos 2\mathrm{x}+1}{2}\mathrm{dx}\\ \\ 2I=\frac{1}{2}\left[{\left(\frac{\sin 2x}{2}\right)}_{-\pi }^{\pi }+{\left(x\right)}_{-\pi }^{\pi }\right] \\ \\ 2I=\frac{1}{2}\left(2\pi \right)\end{array}$

$\mathrm{I}=\frac{\mathrm{\pi }}{2}$