Question

${\int }_{\pi /6}^{\pi /3}\frac{dx}{1+\sqrt{\tan x}}$ is equal to

• A. $\frac{\pi }{12}$
• B. $\frac{\pi }{2}$
• C. $\frac{\pi }{6}$
• D. $\frac{\pi }{4}$

Answer 1

Answer: (A)

$\frac{\pi }{12}$

Let $I={\int }_{\pi /6}^{\pi /3}\frac{dx}{1+\sqrt{\tan x}}$
$={\int }_{\pi /6}^{\pi /3}\left(\frac{\sqrt{\cos x}}{\sqrt{\sin x}+\sqrt{\cos x}}\right)dx$ ..... (i)
$={\int }_{\pi /6}^{\pi /3}\frac{\sqrt{\cos (\frac{\pi }{2}-x)}}{\sqrt{(\frac{\pi }{2}-x)}+\sqrt{\cos (\frac{\pi }{2}-x)}}dx$
$\Rightarrow I={\int }_{\pi /6}^{\pi /3}\frac{\sqrt{\sin x}}{\sqrt{\cos x}+\sqrt{\sin x}}dx$ ......(ii)
On adding Eqs. (i) and (ii), we get
$2I={\int }_{\pi /6}^{\pi /3}1dx=[x{]}_{\pi /6}^{\pi /3}$
$=\frac{\pi }{3}-\frac{\pi }{6}=\frac{\pi }{6}$
$\Rightarrow I=\frac{\pi }{12}$