STD 12 - 7.2 definite integral — JEE कक्षा 11 साइन्स — Question

Using, $\int_{a}^{b} f(x) d x=\int_{a}^{b} f(a+b-x) d x,$ we get :

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

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

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

$ \Rightarrow 2I.\int\limits_0^{{\rm{x}}/2} {{{\sin }^2}xdx} $

$2 \mathrm{I}=2 \times \frac{\pi}{4} \Rightarrow \mathrm{I}=\frac{\pi}{4}$