If $f(x)=\frac{2-x \cos x}{2+x \cos x}$ and $g(x)=\log _{e} x,(x>0)$ then the
value of the integral $\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} g(f(x)) d x$ is :
Correct Option: , 4
$g(f(x))=\log \left(\frac{2-x \cos x}{2+x \cos x}\right), x>0$
Let $I=\int_{-\pi / 4}^{\pi / 4} \log \left(\frac{2-x \cos x}{2+x \cos x}\right) d x$ .......(1)
Use the property $\int_{a}^{b} f(x) d x=\int_{a}^{b} f(a+b-x) d x$
Then, equation (1) becomes,
$I=\int_{-\pi / 4}^{\pi / 4} \log \left(\frac{2+x \cos x}{2-x \cos x}\right) d x$....(2)
Adding (1) and (2)
$2 I=\int_{-\pi / 4}^{\pi / 4} \log \left(\frac{2-x \cos x}{2+x \cos x} \cdot \frac{2+x \cos x}{2-x \cos x}\right) d x$
$2 I=\int_{-\pi / 2}^{\pi / 2} \log (1) d x=0$
$\Rightarrow I=0=\log 1$
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