JEE Main 2026 · 5 Apr · Shift 1, Mathematics question 20

Substitute $t=\tan x$. $\operatorname{cosec}^2x=1+\dfrac1{t^2}$ and $\dfrac{dx}{\cos^4x}=(1+t^2)\,dt$, so
$$\int\left(3-\frac1{t^2}\right)(1+t^2)\,dt=\int\left(2+3t^2-\frac1{t^2}\right)dt=2t+t^3+\frac1t.$$

Limits $t=\tfrac1{\sqrt3}$ to $\sqrt3$: $\dfrac{16}{\sqrt3}-\dfrac{16}{3\sqrt3}=\dfrac{32}{3\sqrt3}$.

(C)