Question

\[\int\limits_{\pi/6}^{\pi/3} \frac{1}{1 + \sqrt{\tan x}} dx\]

Answer 1

\[Let\ I = \int_\frac{\pi}{6}^\frac{\pi}{3} \frac{1}{1 + \sqrt{\tan x}} d x .................(1)\]
\[ = \int_\frac{\pi}{6}^\frac{\pi}{3} \frac{1}{1 + \sqrt{\tan\left( \frac{\pi}{3} + \frac{\pi}{6} - x \right)}} d x\]
\[ = \int_\frac{\pi}{6}^\frac{\pi}{3} \frac{1}{1 + \sqrt{cotx}}dx ....................(2)\]
\[\text{Adding (1) and (2)}\]
\[2I = \int_\frac{\pi}{6}^\frac{\pi}{3} \left( \frac{1}{1 + \sqrt{\tan x}} + \frac{1}{1 + \sqrt{cotx}} \right) d x \]
\[ = \int_\frac{\pi}{6}^\frac{\pi}{3} \left( \frac{1 + \sqrt{cotx} + 1 + \sqrt{\tan x}}{1 + \sqrt{cotx} + \sqrt{\tan x} + \sqrt{\tan x cotx}} \right) dx\]
\[ = \int_\frac{\pi}{6}^\frac{\pi}{3} \frac{2 + \sqrt{cotx} + \sqrt{\tan x}}{2 + \sqrt{cotx} + \sqrt{\tan x}} dx\]
\[ = \int_\frac{\pi}{6}^\frac{\pi}{3} dx = \left[ x \right]_\frac{\pi}{6}^\frac{\pi}{3} \]
\[ = \frac{\pi}{3} - \frac{\pi}{6}\]
\[ \therefore 2I = \frac{\pi}{6}\]
\[Hence\ I = \frac{\pi}{12}\]