JEE Main 2022 · 26 Jul · Shift 2, Mathematics question 19

Substitute $t=\sin x+\cos x$, $dt=(\cos x-\sin x)dx$, $\sin2x=t^2-1$.

Denominator $=\frac2{\sqrt3}\left(t^2-k^2\right)$ with $k^2=\frac{2-\sqrt3}2$, so $k=\frac{\sqrt3-1}2$.

Integrate. $\frac{\sqrt3-1}2\int\frac{dt}{t^2-k^2}=\frac12\ln\left|\frac{t-k}{t+k}\right|$.

Check option A. Its derivative is $\frac12\left[\frac1{\sin(x+\pi/6)}-\frac1{\sin(x+\pi/3)}\right]$, which equals the integrand ($1-\frac1{\sqrt3}$ at $x=0$, and 0 at $x=\frac\pi4$, like $\cos x-\sin x$).

(A)