MathematicsMedium210×since 2002Q3350The integral ∫0π1+4sin2x2−4sinx2 dx\int\limits_0^\pi {\sqrt {1 + 4{{\sin }^2}{x \over 2} - 4\sin {x \over 2}{\mkern 1mu} } } dx0∫π1+4sin22x−4sin2xdx equals:A43−44\sqrt 3 - 443−4B43−4−π34\sqrt 3 - 4 - {\pi \over 3}43−4−3πCπ−4\pi - 4π−4D2π3−4−43{{2\pi } \over 3} - 4 - 4\sqrt 332π−4−43Check answerSkip