Definite Integration Question 54
Question: $ \int _{\pi /6}^{\pi /3}{\frac{dx}{1+\sqrt{\cot x}}} $ is
[DCE 2001]
Options:
A) $ \pi /3 $
B) $ \pi /6 $
C) $ \pi /12 $
D) $ \pi /2 $
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Answer:
Correct Answer: C
Solution:
$ I=\int _{\pi /6}^{\pi /3}{\frac{dx}{1+\sqrt{\cot x}}=}\int _{\pi /6}^{\pi /3}{\frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}}}dx $ …..(i)
$ I=\int _{\pi /6}^{\pi /3}{\frac{\sqrt{\cos x}}{\sqrt{\cos x}+\sqrt{\sin x}}}dx $ ……(ii)
Adding (i) and (ii), $ 2I=\int _{\pi /6}^{\pi /3}{dx} $ ; $ I=\frac{1}{2}( \frac{\pi }{3}-\frac{\pi }{6} )=\frac{\pi }{12} $ .
BETA
