JEE Advanced 2019 Paper 1Question 1
If $$I=\frac{2}{\pi}\int_{-\pi/4}^{\pi/4}\frac{dx}{(1+e^{\sin x})(2-\cos 2x)}$$ then $27I^2$ equals ____
Definite Integration
Answer$4$
Solution
Concept. Add the integral to itself with $x\to-x$; the factors with $e^{\pm\sin x}$ add up to $1$. Formulas used. (1) $\dfrac{1}{1+e^{s}}+\dfrac{1}{1+e^{-s}}=1$(2) $2-\cos2x=1+2\sin^2x$, and dividing by $\cos^2x$: $\dfrac{\sec^2x}{1+3\tan^2x}$Step 1 - Use (1). With $x\to-x$ the factor $2-\cos2x$ does not change, so $$2I=\frac2\pi\int_{-\pi/4}^{\pi/4}\frac{dx}{2-\cos2x}\ \Longrightarrow\ I=\frac2\pi\int_0^{\pi/4}\frac{dx}{2-\cos2x}.$$ Step 2 - Use (2). With $t=\tan x$: $$\int_0^1\frac{dt}{1+3t^2}=\frac{1}{\sqrt3}\tan^{-1}\sqrt3=\frac{\pi}{3\sqrt3}.$$ Step 3 - Value. $I=\dfrac2\pi\cdot\dfrac{\pi}{3\sqrt3}=\dfrac{2}{3\sqrt3}$, so $27I^2=27\cdot\dfrac{4}{27}=4$. Answer: 4