A) \[\sqrt{2}\pi \]
B) \[\frac{\pi }{2}\]
C) \[\frac{\pi }{\sqrt{2}}\]
D) \[2\pi \]
Correct Answer: C
Solution :
\[I=\int_{0}^{\pi /4}{[\sqrt{\tan x}+\sqrt{\cot x]}}dx=\int_{0}^{\pi /4}{\frac{\sin x+\cos x}{\sqrt{\sin x\cos x}}dx}\] \[=\sqrt{2}\int_{0}^{\pi /4}{\frac{\sin x+\cos x}{\sqrt{1-{{(\sin x-\cos x)}^{2}}}}dx}\] Put \[\sin x-\cos x=t\]; \[(\cos x+\sin x)dx=dt\] \[\therefore \,\,\,I=\sqrt{2}\int_{-1}^{0}{\frac{dt}{\sqrt{1-{{t}^{2}}}}}\] \[I=\sqrt{2}[{{\sin }^{-1}}t]_{-1}^{0}=\sqrt{2}[0-(-\pi /2)]=\frac{\pi }{\sqrt{2}}\].
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