Integral from JEE Main 2018

Using the identity $\displaystyle \int_{-a}^a \frac {f(x)}{1+c^{g(x)}}\ dx = \int_0^a f(x)\ dx$ , where $f(x)$ is even, $g(x)$ is odd and $a$ and $c$ are constants (see proof ), we have:

$\begin{aligned} \int_{-\frac \pi 2}^\frac \pi 2 \frac {\sin^2 x}{1+2^x}\ dx & = \int_0^\frac \pi 2 \sin^2 x \ dx \\ & = \int_0^\frac \pi 2 \frac {1-\cos (2x)}2 \ dx \\ & = \frac x2 - \frac {\sin (2x)}4 \ \bigg|_0^\frac \pi 2 \\ & = \boxed{\frac \pi 4} \end{aligned}$

$\begin{aligned}\text{Let, I}&=\int_{-\frac\pi2}^{\frac\pi2}\frac{\sin^2x}{1+2^x}\ dx \\ \text{then, I}&=\int_{-\frac\pi2}^{\frac\pi2}\frac{\sin^2x}{1+2^{-x}}\ dx &\because \int_{-a}^af(x)\ dx= \int_{-a}^af(-x)\ dx\\ \therefore 2I&=\int_{-\frac\pi2}^{\frac\pi2}\sin^2x\left(\frac{1}{1+2^x}+\frac{1}{1+2^{-x}}\right)\ dx\\ \text{so, } I&=\frac12\int_{-\frac\pi2}^{\frac\pi2}\sin^2x\left(\frac{1+2^x}{1+2^x}\right)\ dx\\ &= \frac12\int_{-\frac\pi2}^{\frac\pi2}\sin^2x\ dx\\ &= \frac14\int_{-\frac\pi2}^{\frac\pi2}\left(1-\cos{2x}\right)\ dx\\ &= \frac14\left(x-\frac{\sin{2x}}{2}\right)\rvert_{-\frac\pi2}^{\frac\pi2}\\ &= \boxed{\frac\pi4}\end{aligned}$