A) \[\frac{4\pi }{3}\]
B) \[\frac{2\pi }{3}\]
C) \[\pi \]
D) \[\frac{\pi }{3}\]
Correct Answer: D
Solution :
Let \[I=\int_{0}^{\pi /3}{\frac{\cos x+\sin x}{\sqrt{1+2\sin x\cos x}}}dx\] \[=\int_{0}^{\pi /3}{\frac{\cos x+\sin x}{\sqrt{{{\cos }^{2}}x+{{\sin }^{2}}x+2\sin x\cos x}}dx}\] \[=\int_{0}^{\pi /3}{\frac{\cos x+\sin x}{\sqrt{{{(\cos x+\sin x)}^{2}}}}}dx\] \[=\int_{0}^{\pi /3}{\frac{\cos x+\sin x}{\cos x+\sin x}dx=\int_{0}^{\pi /3}{1\,\,dx}=\frac{\pi }{3}}\] Alternate Method: \[I=\int_{0}^{\pi /3}{\frac{\cos x+\sin x}{\sqrt{1+2\sin x\cos x}}dx}\] \[=\int_{0}^{\pi /3}{\frac{\cos x+\sin x}{\sqrt{2-{{(\sin x-\cos x)}^{2}}}}dx}\] Let \[\sin x-\cos x=t\] \[\Rightarrow \] \[(\cos x+\sin x)dx=dt\] \[\therefore \] \[I=\left[ {{\sin }^{-1}}\frac{(\sin x-\cos x)}{\sqrt{2}} \right]_{0}^{\pi /3}\] \[=\left[ {{\sin }^{-1}}\sin \left( x-\frac{\pi }{4} \right) \right]_{0}^{\pi /3}\] \[=\left( x-\frac{\pi }{4} \right)_{0}^{\pi /3}=\frac{\pi }{3}-\frac{\pi }{4}-\left( 0-\frac{\pi }{4} \right)\] \[=\frac{\pi }{3}\]
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