A) 2
B) \[-2\]
C) 4
D) \[-4\]
Correct Answer: C
Solution :
[c] \[I=\int\limits_{0}^{2\pi }{\sqrt{\frac{1-\cos 2x}{2}}.dx}=\int\limits_{0}^{2\pi }{\sqrt{\frac{1.{{\sin }^{2}}x}{2}}.dx}=\int\limits_{0}^{2\pi }{\sqrt{\left| \sin x \right|}.dx}\]\[=\int\limits_{0}^{2\pi }{\sqrt{\left| \sin x \right|}.dx}+\int\limits_{0}^{2\pi }{\sqrt{\left| \sin x \right|}.dx}=\int\limits_{0}^{2\pi }{\sin x.dx}-\int\limits_{0}^{2\pi }{\sin x.dx}\]\[=[-cosx]{{_{0}^{\pi }}^{-}}[-cosx]{{_{\pi }^{2\pi }}^{=}}-(-1-1)+(cos2\pi -cos\pi )\]\[=2+\left( 1-\left( -1 \right) \right)=2+2=4\] Hence, option [c] is correct.
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