Evaluate $\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{dx}{1 + \sqrt{\tan x}}$.

Explanation

1. 01

   With $x \to \frac{\pi}{6} + \frac{\pi}{3} - x$, $\tan x$ becomes $\cot x$, and $\frac{1}{1 + \sqrt{\cot x}} = \frac{\sqrt{\tan x}}{\sqrt{\tan x} + 1}$.

2. 02

   Adding the two forms: $2I = \int_{\pi/6}^{\pi/3} 1\,dx = \frac{\pi}{6}$, so $I = \frac{\pi}{12}$.