Question

If $f\left(x\right)=\frac{2-x\cos x}{2+x\cos x}$ and $g\left(x\right)={\log }_{e}x.,\left(x>0\right)$ then the value of integral $\underset{-\frac{\pi }{4}}{\overset{\frac{\pi }{4}}{\int }}g\left(f\left(x\right)\right)dx$ is :

• A. ${\log }_{e}3$
• B. ${\log }_{e}2$
• C. ${\log }_{e}e$
• D. ${\log }_{e}1$

Answer 1

Answer: (D)

$g\left(f\left(x\right)\right)=\ell n\left(f\left(x\right)\right)=\ell n\left(\frac{2-x.\cos x}{2+x.\cos x}\right)$
$\therefore I={\int }_{-\frac{\pi }{4}}^{\frac{\pi }{4}}\ell n\left(\frac{2-x.\cos x}{2+x\cos x}\right)dx$
$={\int }_0^{\frac{\pi }{4}}\left(\ell n\left(\frac{2-x\cos x}{2+x.\cos x}\right)+\ell n\left(\frac{2+x.\cos x}{2-x.\cos x}\right)\right)dx$
$={\int }_0^{\frac{\pi }{2}}\left(0\right)dx=0={\log }_e\left(1\right)$