In the interval $\frac{\pi}{3}$ to $\frac{\pi}{2},[\mathrm{x}]=1$ $\begin{aligned} \therefore \mathrm{I} &=\int_{\pi / 3}^{\pi / 2} \mathrm{x} \sin (\pi-\mathrm{x}) \mathrm{dx}=\int_{\pi / 3}^{\pi / 2} \mathrm{x} \sin \mathrm{x} \mathrm{d} \mathrm{x} \\ &=[-\mathrm{x} \cos \mathrm{x}+\sin \mathrm{x}]_{\pi / 3}^{\pi / 2}=1-\frac{\sqrt{3}}{2}+\frac{\pi}{6} \end{aligned}$