Official solution
\[g\left( {f\left( x \right)} \right)\] = \[\ln \left( {f\left( x \right)} \right)\] = \[\ln \left( {{{2 - x\cos x} \over {2 + x\cos x}}} \right)\]
\[ \therefore \] I = \[\int\limits_{ - {\pi \over 4}}^{{\pi \over 4}} {\ln \left( {{{2 - x\cos x} \over {2 + x\cos x}}} \right)dx} \]
( Using property \[\int\limits_{ - a}^a {f\left( x \right)} dx = \int\limits_0^a {\left( {f\left( x \right) + f\left( { - x} \right)} \right)} dx\] )
I = \[\int\limits_0^{{\pi \over 4}} {\left( {\ln \left( {{{2 - x\cos x} \over {2 + x\cos x}}} \right) + \ln \left( {{{2 + x\cos x} \over {2 - x\cos x}}} \right)} \right)dx} \]
= \[\int\limits_0^{{\pi \over 4}} {\left( {\ln \left( 1 \right)} \right)dx} \]
= 0 = loge1