A) 0
B) 2
C) \[\frac{1}{2}\]
D) None of these
Correct Answer: B
Solution :
\[\int_{-\pi /2}^{\pi /2}{\sqrt{\frac{1}{2}(1-\cos 2x)}}\,dx=2\int_{0}^{\pi /2}{\,\,\,\,\,|\sin x|dx}\] = \[2[-\cos x]_{0}^{\pi /2}=2\left[ -\cos \left( \frac{\pi }{2} \right)+\cos 0 \right]=2\].
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