Question

The area bounded by the curve $y=\frac{1}{1+\sqrt{\tan x}}$ and $x$-axis between the ordinates $x=\frac{\pi }{6}$ & $x=\frac{\pi }{3}$ is

• A. $\frac{\pi }{4}$
• B. $\frac{\pi }{2}$
• C. $\frac{\pi }{5}$
• D. $\frac{\pi }{12}$

Answer 1

The correct option is D $\frac{\pi }{12}$

$I={\int }_{\pi /6}^{\pi /3}\frac{dx}{1+\sqrt{\tan x}}={\int }_{\pi /6}^{\pi /3}\frac{\sqrt{\cos x}}{\sqrt{\cos x}+\sqrt{\sin x}}dx$ ...(1)
$={\int }_{\pi /6}^{\pi /3}\frac{\sqrt{\cos (\frac{\pi }{2}-x)}}{\sqrt{\cos (\frac{\pi }{2}-x)}+\sqrt{\sin (\frac{\pi }{2}-x)}}dx$
$\Rightarrow I={\int }_{\pi /6}^{\pi /3}\frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}}$ ...(2)
From (1) and (2), we get
$2I={\int }_{\pi /6}^{\pi /3}dx\Rightarrow I=\frac{\pi }{12}$