JEE Main 2018 (Offline)Question 1
The value of $\int_{-\pi/2}^{\pi/2}\frac{\sin^2x}{1+2^x}dx$ is
Definite Integration
Solution
Concept. Add the integral to itself with $x\to-x$; the fractions with $2^{\pm x}$ add up to $1$. Formulas used. (1) $\dfrac{1}{1+2^x}+\dfrac{1}{1+2^{-x}}=\dfrac{1}{1+2^x}+\dfrac{2^x}{1+2^x}=1$(2) $\int_{-\pi/2}^{\pi/2}\sin^2x\,dx=\dfrac\pi2$Step 1 - Add. $\sin^2x$ is even, so with $x\to-x$ and (1), $$2I=\int_{-\pi/2}^{\pi/2}\sin^2x\,dx=\frac\pi2\quad\text{by (2)}.$$ Step 2 - Value. $I=\dfrac\pi4$. Answer: (A)