JEE Main 2021 (Online) 26th August Evening ShiftQuestion 1
The value of $\int_{-\pi/2}^{\pi/2}\left(\frac{1+\sin^{2}x}{1+\pi^{\sin x}}\right)dx$ is :
Definite Integration
Solution
Concept. $\sin x$ is odd and $1+\sin^2x$ is even; adding the two forms removes the exponential. Formulas used. (1) $\dfrac1{1+\pi^{s}}+\dfrac{\pi^{s}}{1+\pi^{s}}=1$Step 1 - $x\to-x$. $$I=\int_{-\pi/2}^{\pi/2}\frac{(1+\sin^2x)\,\pi^{\sin x}}{1+\pi^{\sin x}}dx.$$ Step 2 - Add. By (1), $$2I=\int_{-\pi/2}^{\pi/2}(1+\sin^2x)\,dx=\pi+\frac\pi2=\frac{3\pi}2.$$ Step 3 - Value. $I=\frac{3\pi}4$. Answer: (B)