$\text{I}=\int\limits_{\pi/6}^{\pi/3}\frac{\text{dx}}{1+\sqrt{\text{tan x}}}=\int\limits_{\pi/6}^{\pi/3}\frac{\sqrt{\text{cos x}}}{\sqrt{\text{cos x}}+\sqrt{\text{sin x}}}\text{dx}.....................\text{(i)}$
$\text{x}\rightarrow\Big(\pi/3+\pi/6-\text{x}\Big)$
$=\int\limits^{\pi/3}_{\pi/6}\frac{\sqrt{\cos(\pi/2-\text{x})}}{\sqrt{\cos(\pi/2-\text{x)}}+\sqrt{\sin(\pi/2-\text{x})}}\text{dx}=\int\limits^{\pi/3}_{\pi/6}\frac{\sqrt{\sin\text{x}}}{\sqrt{\sin\text{x}}+\sqrt{\cos\text{x}}}\text{dx}............\text{(ii)}$
Adding (i) and (ii) to get
2I $=\int\limits^{\pi/3}_{\pi/6}1.\text{ dx}=[\text{x}]^{\pi/3}_{\pi/6}=\pi/3-\pi/6=\pi/6$
$\Rightarrow\text{I}=\pi/12$.
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