Let
$$
\begin{aligned}
I &=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin ^4 x\left(1+\log \left(\frac{2+\sin x}{2-\sin x}\right)\right) d x \ldots .(1) \\
\Rightarrow I &\left.=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin ^4(-x)\right)\left(1+\log \left(\frac{2+\sin (-x)}{2-\sin (-x)}\right)\right) \cdot d x \\
&=\left[\because \int_a^b f(x) . d x=\int_a^b f(a+b-x) \cdot d x\right]
\end{aligned}
$$
$$
\begin{aligned}
&=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\left(\sin ^4 x\right)\left(1+\log \left(\frac{2-\sin x}{2+\sin x}\right)\right) \cdot d x \\
&=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin ^4 x\left(1-\log \left(\frac{2+\sin x}{2-\sin x}\right)\right) \cdot d x \ldots .(2)
\end{aligned}
$$
After adding equation (1) and (2) we get,
$$
\begin{aligned}
&2 I=2 \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin ^4 x \cdot d x \\
&2 I=4 \int_0^{\frac{\pi}{2}} \sin ^4 x \cdot d x \\
&I=2 \int_0^{\frac{\pi}{2}} \sin ^4 x \cdot d x=\frac{\frac{3}{2} \times \frac{1}{2} \times \pi}{2 \times 2}=\frac{3 \pi}{8} \\
&{[\text { By Gamma function] }}
\end{aligned}
$$