Đặt $x=\pi -t$
$I=\int_0^{\pi} (\pi-t)\sin (\pi -t) \cos^2 (\pi-t)dt =\int_0^{\pi} \pi \sin t \cos^2 t -\int_0^{\pi} t\sin t \cos^2 t dt$
$=\pi\int_0^{\pi}\sin x \cos^2 x dx -\int_0^{\pi} x\sin x \cos^2 x dx$
$\Rightarrow 2I = \pi \int_0^{\pi} \sin x \cos^2 x dx =-\pi \int_0^{\pi} \cos^2 x d(\cos x) =-\dfrac{\pi}{3}\cos^3 x \bigg |_0^{\pi}=\dfrac{2\pi}{3} \Rightarrow I =\dfrac{\pi}{3}$