The value of the integral ∫_- π 2^ π 2 sin ^4 x (1+log ( 2+sin x 2-sin x ) ) d x is

Let $$\begin{aligned} I &=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin ^4 x\left(1+\log \left(\frac{2+\sin x}{2-\sin x}\right)\right) d x \ldots .(1) \\ \Rightarrow I &\left.=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin ^4(-x)\right)\left(1+\log \left(\frac{2+\sin (-x)}{2-\sin (-x)}\right)\right) \cdot d x \\ &=\left[\because \int_a^b f(x) . d x=\int_a^b f(a+b-x) \cdot d x\right] \end{aligned}$$I⇒I​=∫−2π​2π​​sin4x(1+log(2−sinx2+sinx​))dx….(1)=∫−2π​2π​​sin4(−x))(1+log(2−sin(−x)2+sin(−x)​))⋅dx=[∵∫ab​f(x).dx=∫ab​f(a+b−x)⋅dx]​ $$\begin{aligned} &=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\left(\sin ^4 x\right)\left(1+\log \left(\frac{2-\sin x}{2+\sin x}\right)\right) \cdot d x \\ &=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin ^4 x\left(1-\log \left(\frac{2+\sin x}{2-\sin x}\right)\right) \cdot d x \ldots .(2) \end{aligned}$$​=∫−2π​2π​​(sin4x)(1+log(2+sinx2−sinx​))⋅dx=∫−2π​2π​​sin4x(1−log(2−sinx2+sinx​))⋅dx….(2)​ After adding equation (1) and (2) we get, $$\begin{aligned} &2 I=2 \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin ^4 x \cdot d x \\ &2 I=4 \int_0^{\frac{\pi}{2}} \sin ^4 x \cdot d x \\ &I=2 \int_0^{\frac{\pi}{2}} \sin ^4 x \cdot d x=\frac{\frac{3}{2} \times \frac{1}{2} \times \pi}{2 \times 2}=\frac{3 \pi}{8} \\ &{[\text { By Gamma function] }} \end{aligned}$$​2I=2∫−2π​2π​​sin4x⋅dx2I=4∫02π​​sin4x⋅dxI=2∫02π​​sin4x⋅dx=2×223​×21​×π​=83π​[ By Gamma function] ​