Definite Integration Question 125
Question: $ \int _{\pi /6}^{\pi /3}{\frac{dx}{1+\sqrt{\tan x}}=} $
[Kerala (Engg.) 2005]
Options:
A) $ \pi /12 $
B) $ \pi /2 $
C) $ \pi /6 $
D) $ \pi /4 $
E) $ 2\pi /3 $
Show Answer
Answer:
Correct Answer: A
Solution:
$ I=\int _{\pi /6}^{\pi /3}{\frac{dx}{1+\sqrt{\tan x}}} $
$ =\int _{\pi /6}^{\pi /3}{\frac{\sqrt{\cos x}}{\sqrt{\cos x}+\sqrt{\sin x}}\ dx} $ …..(i) $ I=\int _{\pi /6}^{\pi /3}{\frac{\sqrt{\sin x}}{\sqrt{\cos x}+\sqrt{\sin x}}\ } $ ……(ii) (Since $ \int_a^{b}{f(x)dx}=\int_a^{b}{f(a+b-x)dx} $ )
Adding (i) and (ii), we get,
$ 2I=\int _{\pi /6}^{\pi /3}{\ dx} $
therefore $ I=\frac{1}{2}( \frac{\pi }{3}-\frac{\pi }{6} )=\frac{\pi }{12} $ .
BETA
