Question

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

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

Answer 1

The correct option is C ${\log }_{e}1$

Given,
$f\left(x\right)=\frac{2-x\cos x}{2+x\cos x}$ & $g\left(x\right)={\log }_{e}x$
$I=\underset{-\pi /4}{\overset{\pi /4}{\int }}g\left(f\right(x\left)\right)dx$
$I=\underset{-\pi /4}{\overset{\pi /4}{\int }}{\log }_e\left(\frac{2-x\cos x}{2+x\cos x}\right)dx...\left(1\right)$

Applying $\underset{b}{\overset{a}{\int }}f\left(x\right)dx=\underset{b}{\overset{a}{\int }}f(a+b-x)dx$
$I=\underset{-\pi /4}{\overset{\pi /4}{\int }}{\log }_e\left(\frac{2+x\cos x}{2-x\cos x}\right)dx...\left(2\right)$

Adding equation (1) & (2)
$2I=\underset{-\pi /4}{\overset{\pi /4}{\int }}{\log }_e\left(1\right)dx$
$2I=0$
$\therefore I=0$