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

$\frac{\pi}{12}$

The correct answer is Option (3) → $\frac{\pi}{12}$

Let $I = \int\limits_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{dx}{1 + \sqrt{\tan x}} = \int\limits_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sqrt{\cos x} \, dx}{\sqrt{\cos x} + \sqrt{\sin x}} \quad \dots (1)$

Then, by $P_3$

$I = \int\limits_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sqrt{\cos (\frac{\pi}{3} + \frac{\pi}{6} - x)} \, dx}{\sqrt{\cos (\frac{\pi}{3} + \frac{\pi}{6} - x)} + \sqrt{\sin (\frac{\pi}{3} + \frac{\pi}{6} - x)}}$

$= \int\limits_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sqrt{\sin x}}{\sqrt{\sin x} + \sqrt{\cos x}} \, dx \quad \dots (2)$

Adding (1) and (2), we get

$2I = \int\limits_{\frac{\pi}{6}}^{\frac{\pi}{3}} dx = [x]_{\frac{\pi}{6}}^{\frac{\pi}{3}} = \frac{\pi}{3} - \frac{\pi}{6} = \frac{\pi}{6} \text{. Hence } I = \frac{\pi}{12}$