JEE Main 2025 Apr 7 Shift 1, Mathematics Q10: Some Properties of Definite Integrals

Let $I={\int }_0^{\pi }\frac{(x+3)\sin x}{1+3{\cos }^{2}x}dx$.

Apply the king property $x\to \pi -x$ (note $\sin (\pi -x)=\sin x$ and ${\cos }^2(\pi -x)={\cos }^{2}x$):

$I={\int }_0^{\pi }\frac{(\pi -x+3)\sin x}{1+3{\cos }^{2}x}dx$

Add the two expressions for $I$:

$2I={\int }_0^{\pi }\frac{(\pi +6)\sin x}{1+3{\cos }^{2}x}dx=(\pi +6){\int }_0^{\pi }\frac{\sin x}{1+3{\cos }^{2}x}dx$

Evaluate the standard integral. Substitute $t=\sqrt{3}\cos x\Rightarrow dt=-\sqrt{3}\sin xdx$; when $x=0$, $t=\sqrt{3}$; when $x=\pi$, $t=-\sqrt{3}$.

${\int }_0^{\pi }\frac{\sin x}{1+3{\cos }^{2}x}dx=\frac{1}{\sqrt{3}}{\int }_{-\sqrt{3}}^{\sqrt{3}}\frac{dt}{1+t^2}=\frac{1}{\sqrt{3}}\cdot 2\arctan (\sqrt{3})=\frac{1}{\sqrt{3}}\cdot \frac{2\pi }{3}=\frac{2\pi }{3\sqrt{3}}$

Assemble the answer:

$2I=(\pi +6)\cdot \frac{2\pi }{3\sqrt{3}}\Rightarrow I=\frac{\pi (\pi +6)}{3\sqrt{3}}$

$\therefore$ Option (3).