Definite Integration Question 8
Question: $ \int_0^{\pi /4}{[\sqrt{\tan x}+\sqrt{\cot x}]dx} $ equals
[RPET 1997]
Options:
A) $ \sqrt{2}\pi $
B) $ \frac{\pi }{2} $
C) $ \frac{\pi }{\sqrt{2}} $
D) $ 2\pi $
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Answer:
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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