$\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{dx}{1 + \sqrt{tanx}}$ is equal to

Let $I = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{dx}{1 + \sqrt{\tan x}}$ ... (1)

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Using King's Property: $\int_{a}^{b} f(x)dx = \int_{a}^{b} f(a+b-x)dx$

Here $a + b = \frac{\pi}{6} + \frac{\pi}{3} = \frac{\pi}{2}$

Replace $x$ with $\frac{\pi}{2} - x$:

$I = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{dx}{1 + \sqrt{\tan(\frac{\pi}{2} - x)}}$ ... (2)

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Using the identity $\tan(\frac{\pi}{2} - x) = \cot x = \frac{1}{\tan x}$:

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

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

Multiply numerator and denominator by $\sqrt{\tan x}$:

$I = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \dfrac{\sqrt{\tan x}}{\sqrt{\tan x} + 1} dx$ ... (3)

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Adding equations (1) and (3):

$2I = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \left(\dfrac{1}{1 + \sqrt{\tan x}} + \dfrac{\sqrt{\tan x}}{1 + \sqrt{\tan x}}\right) dx$

$2I = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \dfrac{1 + \sqrt{\tan x}}{1 + \sqrt{\tan x}} dx$

$2I = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} 1 \, dx$

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$2I = \left[x\right]_{\frac{\pi}{6}}^{\frac{\pi}{3}}$

$2I = \frac{\pi}{3} - \frac{\pi}{6}$

$2I = \frac{2\pi}{6} - \frac{\pi}{6} = \frac{\pi}{6}$

$I = \frac{\pi}{12}$

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$\boxed{\frac{\pi}{12}}$