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Mathematics Question on Integrals of Some Particular Functions

If f(x)=2−xcos⁡x2+xcos⁡xf\left(x\right) = \frac{2- x\cos x}{2+x \cos x} and g(x)=log⁡ex.,(x>0)g\left(x\right) =\log_{e}x ., \left(x>0\right) then the value of integral ∫−π4π4g(f(x))dx\int\limits^{\frac{\pi}{4}}_{-\frac{\pi}{4}} g\left(f\left(x\right)\right)dx is :

A

log⁡e3\log_e 3

B

log⁡e2\log_e 2

C

log⁡ee\log_e e

D

log⁡e1\log_e 1

Answer

log⁡e1\log_e 1

Explanation

Solution

g(f(x))=ℓn(f(x))=ℓn(2−x.cos⁡x2+x.cos⁡x)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)
∴I=∫−π4π4ℓn(2−x.cos⁡x2+xcos⁡x)dx\therefore I = \int^{\frac{\pi}{4}}_{-\frac{\pi}{4}} \ell n\left(\frac{2-x.\cos x}{2+x\cos x}\right)dx
=∫0π4(ℓn(2−xcos⁡x2+x.cos⁡x)+ℓn(2+x.cos⁡x2−x.cos⁡x))dx= \int^{\frac{\pi}{4}}_{0} \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
=∫0π2(0)dx=0=log⁡e(1)= \int^{\frac{\pi}{2}}_{0} \left(0\right)dx =0 =\log_{e} \left(1\right)