The value of the integral ∫ - π 2 π 2 sin 4 ⁡ x 1 + ln ⁡ 2 + sin ⁡ x 2 -…

$I=\underset{-\frac{\pi }{2}}{\overset{\frac{\pi }{2}}{\int }}{\sin }^{4}x\left(1+\ln \left(\frac{2+\sin x}{2-\sin x}\right)\right)dx......\left(i\right)$

$I=2\underset{0}{\overset{\frac{\mathrm{\pi }}{2}}{\int }}{\sin }^{4}xdx ............\left(i\right),\left(\because {\sin }^{4}x \text{is an even function}\right)$

$\left(\because \underset{-\frac{\pi }{2}}{\overset{\frac{\pi }{2}}{\int }}\left({\sin }^{4}x\right)\ln \left(\frac{2+\sin x}{2-\sin x}\right)dx=0, \because \left({\sin }^{4}x\right)\ln \left(\frac{2+\sin x}{2-\sin x}\right) \text{is odd function}\right)$

Let $\underset{0}{\overset{\frac{\mathrm{\pi }}{2}}{\int }}{\sin }^{4}x dx=m$

$\Rightarrow m=\underset{0}{\overset{\frac{\mathrm{\pi }}{2}}{\int }}{\sin }^4\left(\frac{\mathrm{\pi }}{2}-x\right)dx=\underset{0}{\overset{\frac{\mathrm{\pi }}{2}}{\int }}{\cos }^4\left(x\right)dx$

Adding both, we get

$\Rightarrow 2m=\underset{0}{\overset{\frac{\mathrm{\pi }}{2}}{\int }}{\sin }^4\left(x\right)dx+\underset{0}{\overset{\frac{\mathrm{\pi }}{2}}{\int }}{\cos }^4\left(x\right)dx$

$\Rightarrow 2m=\underset{0}{\overset{\frac{\mathrm{\pi }}{2}}{\int }}\left\{{\sin }^4\left(x\right)+{\cos }^4\left(x\right)\right\}dx$

$\Rightarrow 2m={\left({\sin }^2\left(x\right)+{\cos }^2\left(x\right)\right)}^2-2{\sin }^2\left(x\right){\cos }^2\left(x\right)$

$\Rightarrow 2m=1-\frac{{\sin }^2\left(2x\right)}{2}$

$\Rightarrow 2m=\underset{0}{\overset{\frac{\mathrm{\pi }}{2}}{\int }}\left\{1-\frac{1}{2}\left(\frac{1-\cos 4x}{2}\right)\right\}dx=\underset{0}{\overset{\frac{\mathrm{\pi }}{2}}{\int }}\left\{\frac{3}{4}+\frac{\cos 4x}{4}\right\}dx$

$\Rightarrow 2m={\left[\frac{3x}{4}+\frac{\sin 4x}{16}\right]}_0^{\frac{\mathrm{\pi }}{2}}=\frac{3\mathrm{\pi }}{8}$

$\Rightarrow m=\frac{3\mathrm{\pi }}{16}$

Substituting in equation $\left(i\right)$, we get

$I=2m=\frac{3\pi }{8}$