The integral ${\int }_0^{\mathrm{\pi }} \sqrt{1+4{\sin }^2\frac{\mathrm{x}}{2}-4\sin \frac{\mathrm{x}}{2}}\mathrm{dx}$ equals

$\begin{array}{l}{\int }_0^{\mathrm{\pi }} \sqrt{1+4{\sin }^2\frac{\mathrm{x}}{2}-4\sin \frac{\mathrm{x}}{2}}\mathrm{dx}\\ ={\int }_0^{\mathrm{\pi }} \left|2\sin \frac{\mathrm{x}}{2}-1\right|\mathrm{dx}\left[\begin{array}{c}\because \sin \frac{\mathrm{x}}{2}=\frac{1}{2}\\ \Rightarrow \frac{\mathrm{x}}{2}=\frac{\mathrm{\pi }}{6}\to \mathrm{x}=\frac{\mathrm{\pi }}{3}\\ \frac{\mathrm{x}}{2}=\frac{5\mathrm{\pi }}{6}\to \mathrm{x}=\frac{5\mathrm{\pi }}{3}\end{array}\right]\\ ={\int }_0^{\mathrm{\pi }/3} \left(1-2\sin \frac{\mathrm{x}}{2}\right)\mathrm{dx}+{\int }_{\mathrm{\pi }/3}^{\mathrm{\pi }} \left(2\sin \frac{\mathrm{x}}{2}-1\right)\mathrm{dx}\\ ={\left[\mathrm{x}+4\cos \frac{\mathrm{x}}{2}\right]}_0^{\mathrm{\pi }/3}+{\left[-4\cos \frac{\mathrm{x}}{2}-\mathrm{x}\right]}_{\mathrm{\pi }/3}^{\mathrm{\pi }}\\ =\frac{\mathrm{\pi }}{3}+4\frac{\sqrt{3}}{2}-4+\left(0-\mathrm{\pi }+4\frac{\sqrt{3}}{2}+\frac{\mathrm{\pi }}{3}\right)\\ =4\sqrt{3}-4-\frac{\mathrm{\pi }}{3}\end{array}$