JEE Main 2023 (Online) 1st February Evening ShiftQuestion 1
The value of the integral $\int_{-\frac\pi4}^{\frac\pi4}\frac{x+\frac\pi4}{2-\cos2x}dx$ is :
Definite Integration
Solution
Concept. $\frac{x}{2-\cos2x}$ is odd and drops out; the constant part is done with $t=\tan x$. Formulas used. (1) $\int_{-a}^a(\text{odd})=0$, $\int_{-a}^a(\text{even})=2\int_0^a$(2) $t=\tan x$: $2-\cos2x=\dfrac{1+3t^2}{1+t^2}$, $dx=\dfrac{dt}{1+t^2}$Step 1 - Use (1). $$I=\frac\pi4\cdot2\int_0^{\pi/4}\frac{dx}{2-\cos2x}=\frac\pi2\int_0^{\pi/4}\frac{dx}{2-\cos2x}.$$ Step 2 - Use (2). $t:0\to1$: $$\int_0^1\frac{dt}{1+3t^2}=\frac1{\sqrt3}\tan^{-1}\sqrt3=\frac{\pi}{3\sqrt3}.$$ Step 3 - Value. $I=\frac\pi2\cdot\frac\pi{3\sqrt3}=\frac{\pi^2}{6\sqrt3}$. Answer: (D)