${\int }_0^{2\mathrm{\pi }} (\sin \mathrm{x}+|\sin \mathrm{x}\left|\right)\mathrm{dx}$ is equal to

$\begin{array}{l}{\int }_0^{2\mathrm{\pi }} (\sin \mathrm{x}+|\sin \mathrm{x}\left|\right)\mathrm{dx}\\ ={\int }_0^{\mathrm{\pi }} (\sin \mathrm{x}+\sin \mathrm{x})\mathrm{dx}+{\int }_{\mathrm{\pi }}^{2\mathrm{\pi }} (\sin \mathrm{x}-\sin \mathrm{x})\mathrm{dx}\\ ={\int }_0^{\mathrm{\pi }} 2\sin \mathrm{xdx}+{\int }_{\mathrm{\pi }}^{2\mathrm{\pi }} 0\mathrm{dx}=2[-\cos \mathrm{x}{]}_0^{\mathrm{\pi }}+0\\ =-2(\cos \mathrm{\pi }-\cos 0)=-2(-1-1)=4\end{array}$