A) 0
B) 2
C) 1
D) \[-1\]
Correct Answer: B
Solution :
\[I=\int_{0}^{\pi }{\sqrt{\frac{1+\cos 2x}{2}}dx=\int_{0}^{\,\pi }{|\cos x|\,dx}}\] \[I=\int_{\,0}^{\,\pi /2}{\cos x\,dx}-\int_{\,\pi /2}^{\,\pi }{\cos x\,dx}=[\sin x]_{0}^{\pi /2}-[\sin x]_{\pi /2}^{\pi }\] \[I=\left[ \sin \frac{\pi }{2}-\sin 0 \right]-\left[ \sin \pi -\sin \frac{\pi }{2} \right]\]= 1 + 1 = 2.
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