JEE 2023 — Mathematics PYQ

Given $f(x)+f(\pi -x)={\pi }^2$

Use property of definite integral

$I={\int }_0^{\pi }f(\pi -x)\sin (\pi -x)dx$

Add (1) and (2)

$2I={\int }_0^{\pi }[f(x)+f(\pi -x)]\sin xdx$

$\therefore 2I={\int }_0^{\pi }{\pi }^2\sin xdx$

$\left[\because \int \sin xdx=-\cos x+C\right]$

$I={\pi }^2$