If f(x)=\frac{2-x\cos x}{2+x\cos x} and g(x)=\log_{e}x,(x>0) then value of the integral \int_{-\pi/4}^{\pi/4}g(f(x))dx is :
 

  • Option 1)

    \log_{e}3

  • Option 2)

    \log_{e}1

  • Option 3)

    \log_{e}e

     

  • Option 4)

    \log_{e}2

 

Answers (1)

f(x)=\frac{2-x\cos x}{2+x\cos x}g(x)=\log_{e}x\; \; [x>0)

then \int_{-\pi/4}^{\pi/4}g(f(x))d\\g(f(x))=\log_{e}\left ( \frac{2-x\cos x}{2+x\cos x} \right )

g(f(-x))=-g(f(x))

g(f(x)) is an odd function

\Rightarrow \int_{-\pi/4}^{\pi/4}g(f(x))dx=0=\log_{e}1


Option 1)

\log_{e}3

Option 2)

\log_{e}1

Option 3)

\log_{e}e

 

Option 4)

\log_{e}2

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