Question

Evaluate : ${\int }_{\frac{\pi }{6}}^{\frac{\pi }{3}}\frac{dx}{1+\sqrt{\tan x}}$

• A. $\frac{\pi }{4}$
• B. $\frac{\pi }{8}$
• C. $\frac{\pi }{12}$
• D. $\frac{\pi }{6}$

Answer 1

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

We know ${\int }_a^{b}f\left(x\right)dx={\int }_a^{b}f(a+b-x)dx$
Let $I={\int }_{\pi /6}^{\pi /3}\frac{dx}{1+\sqrt{\tan x}}={\int }_{\pi /6}^{\pi /3}\frac{dx}{1+\sqrt{\frac{\sin x}{\cos x}}}$
$={\int }_{\pi /6}^{\pi /3}\frac{\sqrt{\cos x}dx}{\sqrt{\cos x}+\sqrt{\sin x}}$ ....(1)
$I={\int }_{\pi /6}^{\pi /3}\frac{dx}{1+\sqrt{\tan (\frac{\pi }{3}+\frac{\pi }{6}-x)}}$
$={\int }_{\pi /6}^{\pi /3}\frac{dx}{1+\sqrt{\tan (\frac{\pi }{2}-x)}}$
$={\int }_{\pi /6}^{\pi /3}\frac{dx}{1+\sqrt{\cot x}}$
$={\int }_{\pi /6}^{\pi /3}\frac{\sqrt{\sin x}dx}{\sqrt{\sin x}+\sqrt{\cos x}}$ .....(2)
Adding (1) and (2), we get

$\Rightarrow 2I={\int }_{\pi /6}^{\pi /3}\frac{\sqrt{\sin x}+\sqrt{\cos x}}{\sqrt{\sin x}+\sqrt{\cos x}}dx$
$={\int }_{\pi /6}^{\pi /3}dx={\left(x\right)}_{\pi /6}^{\pi /3}=\frac{\pi }{3}-\frac{\pi }{6}=\frac{\pi }{6}\Rightarrow I=\frac{\pi }{12}$