JEE Main 2013 — Mathematics Calculus

$\displaystyle \mathrm{I}=\int_{\pi / 2}^{\pi / 2} \frac{\sin ^2 x}{1+2^x} d x$I=∫π/2π/2​1+2xsin2x​dx $$\Rightarrow \mathrm{I}=\int_{-\pi / 2}^{\pi / 2} \frac{\sin ^2 x}{1+2^{-x}} d x \text {, by replacing } x \text { by }$$⇒I=∫−π/2π/2​1+2−xsin2x​dx, by replacing x by  $$\begin{aligned} & \left(\frac{\pi}{2}-\frac{\pi}{2}-x\right) \\ \Rightarrow \quad \mathrm{I} & =\int_{-\pi / 2}^{\pi / 2} \frac{2^x \cdot \sin ^2 x}{1+2^x} d x \end{aligned}$$⇒I​(2π​−2π​−x)=∫−π/2π/2​1+2x2x⋅sin2x​dx​ Adding equations (i) and (ii), we get $$\begin{array}{rl} 2 & I=\int_{-\pi / 2}^{\pi / 2} \sin ^2 x d x=\frac{1}{2} \int_{-\pi / 2}^{\pi / 2}(1-\cos 2 x) d x \\ \Rightarrow \quad \mathrm{I} & =\frac{1}{4}\left[x+\frac{\sin 2 x}{2}\right]_{-\pi / 2}^{\pi / 2} \\ & =\frac{1}{4}\left[\left(\frac{\pi}{2}+\frac{\sin \pi}{2}\right)-\left(-\frac{\pi}{2}+\frac{\sin (-\pi)}{2}\right)\right] \\ \Rightarrow \quad \mathrm{I} & =\frac{1}{4}\left[\frac{\pi}{2}+\frac{\pi}{2}\right]=\frac{\pi}{4} \end{array}$$2⇒I⇒I​I=∫−π/2π/2​sin2xdx=21​∫−π/2π/2​(1−cos2x)dx=41​[x+2sin2x​]−π/2π/2​=41​[(2π​+2sinπ​)−(−2π​+2sin(−π)​)]=41​[2π​+2π​]=4π​​