Official solution
Let
I = \[\int\limits_{ - {\pi \over 2}}^{{\pi \over 2}} {{{\sin }^4}} x\left( {1 + \log \left( {{{2 + \sin x} \over {2 - \sin x}}} \right)} \right)dx\] ......(1)
\[ \Rightarrow \] I = \[\int\limits_{ - {\pi \over 2}}^{{\pi \over 2}} {{{\sin }^4}}(- x)\left( {1 + \log \left( {{{2 + \sin (- x)} \over {2 - \sin(- x)}}} \right)} \right)dx\]
[ As \[\int\limits_a^b { f \left( x \right)dx} = \int\limits_a^b {f\left( {a + b - x} \right)} dx\] ]
= \[\int\limits_{ - {\pi \over 2}}^{{\pi \over 2}} {{{\sin }^4}} x\left( {1 + \log \left( {{{2 - \sin x} \over {2 + \sin x}}} \right)} \right)dx\]
= \[\int\limits_{ - {\pi \over 2}}^{{\pi \over 2}} {{{\sin }^4}} x\left( {1 - \log \left( {{{2 + \sin x} \over {2 - \sin x}}} \right)} \right)dx\] .......(2)
Adding equation (1) and (2) we get,
2I = 2\[\int\limits_{ - {\pi \over 2}}^{{\pi \over 2}} {{{\sin }^4}} xdx\]
2I = 4\[\int\limits_0^{{\pi \over 2}} {{{\sin }^4}} xdx\]
I = 2\[\int\limits_0^{{\pi \over 2}} {{{\sin }^4}} xdx\]
= \[{{2 \times {3 \over 2} \times {1 \over 2} \times \pi } \over {2 \times 2}}\] [ Using Gamma function ]
= \[{{3\pi } \over 8}\]