The integral $\underset{0}{\overset{\pi }{\int }}\sqrt{1+4{\sin }^2\frac{x}{2}-4\sin \frac{x}{2}dx}$ equals

$$\begin{gathered} I=\underset{0}{\overset{\pi }{\int }}\sqrt{1+4{\sin }^2\frac{x}{2}-4\sin \frac{x}{2}dx} \\ =\underset{0}{\overset{\pi }{\int }}\sqrt{{\left(2\sin \frac{x}{2}-1\right)}^2}dx \\ I=\underset{0}{\overset{\pi }{\int }}\left|2\sin \frac{x}{2}-1\right|dx \\ make 2 \sin \frac{x}{2}-1=0 \\ \Rightarrow \sin \frac{x}{2}=1/2 \\ \Rightarrow \frac{x}{2}=\frac{\pi }{6}\Rightarrow x=\pi /3 \\ =\underset{0}{\overset{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$3$}\right.}{\int }}\left|2 \sin \frac{x}{2}-1\right|+\underset{\pi /3}{\overset{\pi }{\int }}\left|2 \sin \frac{x}{2}-1\right|dx \\ =\underset{0}{\overset{\pi /3}{\int }}-\left(2\sin \frac{x}{2}-1\right)dx+\underset{\pi /3}{\overset{\pi }{\int }}\left(2\sin \frac{x}{2}-1\right)dx \\ =\left(\sin ce 0<x<\frac{\pi }{3}\Rightarrow \left(2 \sin \frac{x}{2}-1\right)<0\right) \\ and \frac{\pi }{3}<x<\pi \Rightarrow \left(2 \sin \frac{x}{2}-1\right)>0 \\ ={\left(x+4\cos \frac{x}{2}\right)}_0^{\pi /3}+{\left(-4\cos \frac{x}{2}-x\right)}_{\pi /3}^{\pi } \\ =\left(\frac{\pi }{3}+\frac{4\sqrt{3}}{2}\right)-\left(0+4\right)+\left[\left(\frac{4\sqrt{3}}{2}+\frac{\pi }{3}\right)-\left(0+\pi \right)\right] \\ =4\sqrt{3}-4-\pi /3 \end{gathered}$$