Question
$\int_{ - \pi /2}^{\pi /2} {\sqrt {\frac{1}{2}(1 - \cos 2x)} } \,dx = $

Answer

b
(b) $\int_{ - \pi /2}^{\pi /2} {\sqrt {\frac{1}{2}(1 - \cos 2x)} } \,dx = 2\int_0^{\pi /2} {\,\,\,\,\,|\sin x|dx} $

$= 2[ - \cos x]_0^{\pi /2} = 2\left[ { - \cos \left( {\frac{\pi }{2}} \right) + \cos 0} \right] = 2$.

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