JEE Main 2024 (Online) 6th April Morning ShiftQuestion 1
$\int_{0}^{\pi/4}\frac{\cos^{2}x\sin^{2}x}{\left(\cos^{3}x+\sin^{3}x\right)^{2}}dx$ is equal to
Definite Integration
Solution
Concept. Divide top and bottom by $\cos^6x$; everything is in $\tan x$ and $\sec^2x\,dx=dt$. Formulas used. (1) $\int\dfrac{t^2\,dt}{(1+t^3)^2}=-\dfrac{1}{3(1+t^3)}$Step 1 - Rewrite. $$\frac{\cos^2x\sin^2x}{(\cos^3x+\sin^3x)^2}=\frac{\tan^2x\sec^2x}{(1+\tan^3x)^2}.$$ Step 2 - Put $t=\tan x$. $t:0\to1$. By (1), $$I=\Big[-\frac{1}{3(1+t^3)}\Big]_0^1=-\frac16+\frac13=\frac16.$$ Answer: (B)