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