Mathematics - Definite Integration Question with Solution | TestHub

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