JEE Main 2022 · 26 Jun · Shift 1, Mathematics question 28

King's rule. With $g(x)=\frac{3\pi x^2}2-x^3$, one finds $g(x)+g(\pi-x)=\frac{\pi^3}2$, while $\frac{\sin x}{1+\cos^2x}$ is unchanged by $x\to\pi-x$.

So $2I=\dfrac{\pi^3}2\displaystyle\int_0^\pi\frac{\sin x\,dx}{1+\cos^2x}=\frac{\pi^3}2\cdot\frac\pi2$, i.e. $I=\dfrac{\pi^4}8$.

Value. $\dfrac{48}{\pi^4}\cdot\dfrac{\pi^4}8=6$.

Answer: 6