JEE Main 2025 Apr 3 Shift 2, Mathematics Q9: Some Properties of Definite Integrals

Let $I={\int }_0^{\pi }\frac{8xdx}{4{\cos }^{2}x+{\sin }^{2}x}$. Apply the King property ${\int }_0^{\pi }f(x)dx={\int }_0^{\pi }f(\pi -x)dx$, noting that the denominator is invariant under $x\to \pi -x$:

$I={\int }_0^{\pi }\frac{8(\pi -x)dx}{4{\cos }^{2}x+{\sin }^{2}x}$.

Adding: $2I=8\pi {\int }_0^{\pi }\frac{dx}{4{\cos }^{2}x+{\sin }^{2}x}$. The integrand has period $\pi$ and is symmetric about $\pi /2$, so

$2I=8\pi \cdot 2{\int }_0^{\pi /2}\frac{dx}{4{\cos }^{2}x+{\sin }^{2}x}$.

Divide numerator and denominator by ${\cos }^{2}x$ and substitute $t=\tan x$:

${\int }_0^{\pi /2}\frac{{\sec }^{2}xdx}{4+{\tan }^{2}x}={\int }_0^{\infty }\frac{dt}{4+t^2}=\frac{1}{2}{\left[{\tan }^{-1}\frac{t}{2}\right]}_0^{\infty }=\frac{1}{2}\cdot \frac{\pi }{2}=\frac{\pi }{4}$.

So $2I=16\pi \cdot \frac{\pi }{4}=4{\pi }^2$, giving $I=2{\pi }^2$.