7.2 definite integral — ગણિત ધોરણ 12 સાયન્સ — Question

$ \Rightarrow I = \int\limits_{ - \pi /2}^{\pi /2} {\frac{{{{\sin }^2}x}}{{1 - {2^{ - x}}}}dx} $ by replacing $x$ by

$\left(\frac{\pi}{2}-\frac{\pi}{2}-x\right)$

$ \Rightarrow I = \int\limits_{ - \pi /2}^{\pi /2} {\frac{{{2^x}.{{\sin }^2}x}}{{1 + {2^x}}}dx} $

Adding equations $(i)$ and $(ii),$ we get

$2I = \int\limits_{ - \pi /2}^{\pi /2} {{{\sin }^2}xdx} $

$ = \frac{1}{2}\int\limits_{ - \pi /2}^{\pi /2} {\left( {1 - \cos 2x} \right)dx} $

$\Rightarrow \quad \mathrm{I}=\frac{1}{4}\left[x+\frac{\sin 2 x}{2}\right]_{-\pi / 2}^{\pi / 2}$

$=\frac{1}{4}\left[\left(\frac{\pi}{2}+\frac{\sin \pi}{2}\right)-\left(-\frac{\pi}{2}+\frac{\sin (-\pi)}{2}\right)\right]$

$\Rightarrow \quad 1=\frac{1}{4}\left[\frac{\pi}{2}+\frac{\pi}{2}\right]=\frac{\pi}{4}$