Evaluate $\displaystyle \int \limits_{\frac{\pi}{6}}^{\frac{\pi}{3}} \dfrac{1}{1 + \sqrt{\cot x}} \; dx$
$\begin{aligned}
\text{Let } I & = \displaystyle \int \limits_{\frac{\pi}{6}}^{\frac{\pi}{3}} \dfrac{1}{1 + \sqrt{\cot x}} \; dx \;\;\; \cdots \; (1) \\\\
& = \displaystyle \int \limits_{\frac{\pi}{6}}^{\frac{\pi}{3}} \dfrac{1}{1 + \sqrt{\cot \left(\dfrac{\pi}{6} + \dfrac{\pi}{3} - x\right)}} \; dx \\\\
& \left[\text{Note: } \int \limits_{a}^{b} f\left(x\right) \; dx = \int \limits_{a}^{b} f \left(a + b - x\right) \; dx\right] \\\\
& = \int \limits_{\frac{\pi}{6}}^{\frac{\pi}{3}} \dfrac{1}{1 + \sqrt{\cot \left(\dfrac{\pi}{2} - x\right)}} \; dx \\\\
& = \int \limits_{\frac{\pi}{6}}^{\frac{\pi}{3}} \dfrac{1}{1 + \sqrt{\tan x}} \; dx \\\\
& = \int \limits_{\frac{\pi}{6}}^{\frac{\pi}{3}} \dfrac{1}{1 + \dfrac{1}{\sqrt{\cot x}}} \; dx \\\\
& = \int \limits_{\frac{\pi}{6}}^{\frac{\pi}{3}} \dfrac{\sqrt{\cot x}}{\sqrt{\cot x} + 1} \; dx \;\;\; \cdots \; (2)
\end{aligned}$
Adding equations $(1)$ and $(2)$ gives
$2 I = \displaystyle \int \limits_{\frac{\pi}{6}}^{\frac{\pi}{3}} \dfrac{1 + \sqrt{\cot x}}{1 + \sqrt{\cot x}} \; dx$
i.e. $2 I = \int \limits_{\frac{\pi}{6}}^{\frac{\pi}{3}} dx = \left[x\right]_{\pi / 6}^{\pi / 3} = \dfrac{\pi}{3} - \dfrac{\pi}{6} = \dfrac{\pi}{6}$
$\implies$ $I = \dfrac{\pi}{12}$