The integral $\int\limits_0^\pi {\sqrt {1 + 4{{\sin }^2}{x \over 2} - 4\sin {x \over 2}{\mkern 1mu} } } dx$ equals:

Question

$4\sqrt 3 - 4$

$4\sqrt 3 - 4 - {\pi \over 3}$

$\pi - 4$

${{2\pi } \over 3} - 4 - 4\sqrt 3 $

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