JEE Main 2024 — 29 January, Shift 1 — Question 9

Simplify the integrand: $\frac{\csc x + \sin x}{\csc x \sec x + \tan x \sin^2 x} = \frac{(1+\sin^2 x)\cos x}{1+\sin^4 x}$. Substitute $t = \sin x$, $dt = \cos x\,dx$: integral becomes $\int \frac{1+t^2}{1+t^4} dt$. Divide numerator and denominator by $t^2$: $\int \frac{1/t^2 + 1}{(t - 1/t)^2 + 2} dt$. Let $u = t - 1/t$, then $du = (1 + 1/t^2) dt$.

Integral becomes $\int \frac{du}{u^2+2} = \frac{1}{\sqrt{2}} \tan^{-1}\left(\frac{u}{\sqrt{2}}\right) + C$. Back-substitute: $u = \sin x - \csc x$, so $y(x) = \frac{1}{\sqrt{2}} \tan^{-1}\left(\frac{\sin x - \csc x}{\sqrt{2}}\right) + C$. Use $\lim_{x\to (\pi/2)^-} y(x) = 0$: as $x\to \pi/2^-$, $\sin x - \csc x \to 0$, so $0 = \frac{1}{\sqrt{2}} \tan^{-1}(0) + C \Rightarrow C = 0$. Evaluate at $x = \pi/4$: $\sin(\pi/4) = 1/\sqrt{2}$, $\csc(\pi/4) = \sqrt{2}$, so $\sin x - \csc x = -1/\sqrt{2}$.

Then $\frac{\sin x - \csc x}{\sqrt{2}} = -1/2$. Hence $y(\pi/4) = \frac{1}{\sqrt{2}} \tan^{-1}\left(-\frac{1}{2}\right)$, which corresponds to option D.