Mathematics - Definite Integration Question with Solution | TestHub

Let $$\begin{array}{rl}I& ={\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}{\sin }^{4}x\left(1+\log \left(\frac{2+\sin x}{2-\sin x}\right)\right)dx\dots .(1)\\ \Rightarrow I& ={\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}{\sin }^4(-x))\left(1+\log \left(\frac{2+\sin (-x)}{2-\sin (-x)}\right)\right)\cdot dx\\ & =\left[\because {\int }_a^{b}f(x).dx={\int }_a^{b}f(a+b-x)\cdot dx\right]\end{array}$$ $$\begin{array}{rl}& ={\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}\left({\sin }^{4}x\right)\left(1+\log \left(\frac{2-\sin x}{2+\sin x}\right)\right)\cdot dx\\ & ={\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}{\sin }^{4}x\left(1-\log \left(\frac{2+\sin x}{2-\sin x}\right)\right)\cdot dx\dots .(2)\end{array}$$ After adding equation (1) and (2) we get, $$\begin{array}{rl}& 2I=2{\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}{\sin }^{4}x\cdot dx\\ & 2I=4{\int }_0^{\frac{\pi }{2}}{\sin }^{4}x\cdot dx\\ & I=2{\int }_0^{\frac{\pi }{2}}{\sin }^{4}x\cdot dx=\frac{\frac{3}{2}\times \frac{1}{2}\times \pi }{2\times 2}=\frac{3\pi }{8}\\ & [\text{ By Gamma function] }\end{array}$$