Mathematics - Definite Integration Question with Solution | TestHub

$\mathrm{I}={\int }_{\pi /2}^{\pi /2}\frac{{\sin }^{2}x}{1+2^x}dx$ $$\Rightarrow \mathrm{I}={\int }_{-\pi /2}^{\pi /2}\frac{{\sin }^{2}x}{1+2^{-x}}dx\text{, by replacing }x\text{ by }$$ $$\begin{array}{rl}& \left(\frac{\pi }{2}-\frac{\pi }{2}-x\right)\\ \Rightarrow \mathrm{I}& ={\int }_{-\pi /2}^{\pi /2}\frac{2^x\cdot {\sin }^{2}x}{1+2^x}dx\end{array}$$ Adding equations (i) and (ii), we get $$\begin{array}{rl}2& I={\int }_{-\pi /2}^{\pi /2}{\sin }^{2}xdx=\frac{1}{2}{\int }_{-\pi /2}^{\pi /2}(1-\cos 2x)dx\\ \Rightarrow \mathrm{I}& =\frac{1}{4}{\left[x+\frac{\sin 2x}{2}\right]}_{-\pi /2}^{\pi /2}\\ & =\frac{1}{4}\left[\left(\frac{\pi }{2}+\frac{\sin \pi }{2}\right)-\left(-\frac{\pi }{2}+\frac{\sin (-\pi )}{2}\right)\right]\\ \Rightarrow \mathrm{I}& =\frac{1}{4}\left[\frac{\pi }{2}+\frac{\pi }{2}\right]=\frac{\pi }{4}\end{array}$$