Official solution
Let \[I = \int\limits_0^\pi {\sqrt {1 + 4{{\sin }^2}{x \over 2} - 4\sin {x \over 2}} } dx\]
\[ = \int\limits_0^\pi {\left| {2\sin {x \over 2} - 1} \right|} dx\]
\[ = \int\limits_0^{\pi /3} {\left( {1 - 2\sin {x \over 2}} \right)} dx + \int\limits_{\pi /3}^\pi {\left( {2\sin {x \over 2} - 1} \right)} dx\]
\[\left[ {} \right.\] As \[\sin {x \over 2} = {1 \over 2} \Rightarrow {x \over 2} = {\pi \over 6} \Rightarrow x = {\pi \over 3},\] \[{x \over 2}\]
\[\,\,\,\,\,\,\,\,\,\,\, = {{5\pi } \over 6} \Rightarrow x = {{5\pi } \over 3}\] \[\left. {} \right]\]
\[ = \left[ {x + 4\cos {x \over 2}} \right]_0^{\pi /3} + \left[ { - 4\cos {x \over 2} - x} \right]_{\pi /3}^\pi \]
\[ = {\pi \over 3} + 4{{\sqrt 2 } \over 2} - 4 + \left( {0 - \pi + 4{{\sqrt 3 } \over 2} + {\pi \over 3}} \right)\]
\[ = 4\sqrt 3 - 4 - {\pi \over 3}\]