If $f\left(x\right)=\frac{2-x\mathrm{c}\mathrm{o}\mathrm{s}x}{2+x\mathrm{c}\mathrm{o}\mathrm{s}x}$ and $g\left(x\right)={\log }_{e}x,$ then the value of the integral $\underset{-\frac{\pi }{4}}{\overset{\frac{\pi }{4}}{\int }}g\left(f\left(x\right)\right)dx$ is

Given,

$f\left(x\right)=\frac{2-x\mathrm{c}\mathrm{o}\mathrm{s}x}{2+x\mathrm{c}\mathrm{o}\mathrm{s}x},$ $g\left(x\right)={\log }_{e}x$

$\Rightarrow g\left(f\left(x\right)\right)=\mathrm{l}\mathrm{o}{\mathrm{g}}_{\mathrm{e}}\left(\frac{2-x\mathrm{c}\mathrm{o}\mathrm{s}x}{2+x\mathrm{c}\mathrm{o}\mathrm{s}x}\right)$

$\Rightarrow g\left(f\left(-x\right)\right)=\mathrm{l}\mathrm{o}{\mathrm{g}}_e\left(\frac{2-(-x)\mathrm{c}\mathrm{o}\mathrm{s}(-x)}{2+(-x)\mathrm{c}\mathrm{o}\mathrm{s}(-x)}\right)$

$\Rightarrow g\left(f\left(-x\right)\right)={\log }_e\left(\frac{2+x\cos x}{2-x\cos x}\right)$

$\Rightarrow g\left(f\left(-x\right)\right)=-\log \left(\frac{2-x\cos x}{2+x\cos x}\right)$

$\Rightarrow g\left(f\left(-x\right)\right)=–g\left(f\left(x\right)\right)$

Hence, $g\left(f\left(x\right)\right)$ is an odd function.

By using the property of definite integration, ${\int }_{-a}^{a}f\left(x\right)dx=\left\{\begin{array}{ll}2{\int }_0^a& f\left(x\right)dx, f\left(-x\right)=f\left(x\right)\\ 0,& f\left(-x\right)=-f\left(x\right)\end{array}\right.$, we can write
$\underset{-\frac{\pi }{4}}{\overset{\frac{\pi }{4}}{\int }}g\left(f\right(x\left)\right)dx=0={\log }_{e}1$