JEE Main 2024 (Online) 29th January Morning ShiftQuestion 1
For $x\in\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$, if $y(x)=\int\frac{\operatorname{cosec}x+\sin x}{\operatorname{cosec}x\sec x+\tan x\sin^{2}x}dx$, and $\lim_{x\to\left(\frac{\pi}{2}\right)^{-}}y(x)=0$ then $y\left(\frac{\pi}{4}\right)$ is equal to
Indefinite Integration
Solution
Concept. Simplify to a function of $\sin x$ times $\cos x$; then the standard $\int\frac{1+s^2}{1+s^4}ds$ form appears. Formulas used. (1) $\displaystyle\int\frac{1+s^2}{1+s^4}ds=\int\frac{1+\frac1{s^2}}{\big(s-\frac1s\big)^2+2}ds=\frac1{\sqrt2}\tan^{-1}\frac{s^2-1}{\sqrt2\,s}+C$Step 1 - Simplify. Denominator: $\frac1{\sin x\cos x}+\frac{\sin^3x}{\cos x}=\frac{1+\sin^4x}{\sin x\cos x}$. Numerator: $\frac{1+\sin^2x}{\sin x}$. So the integrand is $$\frac{(1+\sin^2x)\cos x}{1+\sin^4x}.$$ Step 2 - Put $s=\sin x$. $ds=\cos x\,dx$. By (1), $$y=\frac1{\sqrt2}\tan^{-1}\frac{s^2-1}{\sqrt2\,s}+C.$$ Step 3 - Constant. As $x\to\frac\pi2^-$, $s\to1$ and $\frac{s^2-1}{\sqrt2 s}\to0$. So $C=0$. Step 4 - Value at $\frac\pi4$. $s^2=\frac12$, $\sqrt2s=1$, so $\frac{s^2-1}{\sqrt2s}=-\frac12$: $$y\Big(\frac\pi4\Big)=\frac1{\sqrt2}\tan^{-1}\Big(-\frac12\Big).$$ Answer: (B)