Question
$\int_{\,0}^{\,\pi } {\sqrt {\frac{{1 + \cos 2x}}{2}} \,dx} $ is equal to

Answer

b
(b) $I = \int_0^\pi {\sqrt {\frac{{1 + \cos 2x}}{2}} dx = \int_0^{\,\pi } {|\cos x|\,dx} } $

$I = \int_{\,0}^{\,\pi /2} {\cos x\,dx} - \int_{\,\pi /2}^{\,\pi } {\cos x\,dx} $

$= [\sin x]_0^{\pi /2} - [\sin x]_{\pi /2}^\pi $

$I = \left[ {\sin \frac{\pi }{2} - \sin 0} \right] - \left[ {\sin \pi - \sin \frac{\pi }{2}} \right] $

$=1+ 1 = 2.$

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