A) \[\pi /12\]
B) \[\pi /2\]
C) \[\pi /6\]
D) \[\pi /4\]
E) \[2\pi /3\]
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, \[2I=\int_{\pi /6}^{\pi /3}{\ dx}\] Þ \[I=\frac{1}{2}\left( \frac{\pi }{3}-\frac{\pi }{6} \right)=\frac{\pi }{12}\].
You need to login to perform this action.
You will be redirected in
3 sec
