Evaluate: \int_{\pi/6}^{\pi/3} \frac{dx}{1+\sqrt{\cot x}}

Let I = \int_{\pi/3}^{\pi/6} \frac{dx}{1+\sqrt{\cot x}} = \int_{\pi/3}^{\pi/6} \frac{\sqrt{\sin x}}{\sqrt{\sin x} + \sqrt{\cos x}} dx
\int_{\pi/3}^{\pi/6} \frac{\sqrt{\sin(\pi/3-\pi/6-x)}}{\sqrt{\sin(\pi/3-\pi/6-x)} + \sqrt{\cos(\pi/3-\pi/6-x)}} dx
I = \int_{\pi/3}^{\pi/6} \frac{\sqrt{\cos x}}{\sqrt{\cos x} + \sqrt{\sin x}} dx
Adding both equations we get,
2I = \int_{\pi/3}^{\pi/6} dx = [x]_{\pi/3}^{\pi/6} = \frac{\pi}{6} - \frac{\pi}{3} = -\frac{\pi}{6}
Therefore, I = -\frac{\pi}{12}