Question

Evaluate each of the following integral:

\[\int_\frac{\pi}{6}^\frac{\pi}{3} \frac{\sqrt{\tan x}}{\sqrt{\tan x} + \sqrt{\cot x}}dx\]

Answer 1

\[\text{Let I} = \int_\frac{\pi}{6}^\frac{\pi}{3} \frac{\sqrt{\tan x}}{\sqrt{\tan x} + \sqrt{\cot x}}dx................\left( 1 \right)\]

Then,

\[I = \int_\frac{\pi}{6}^\frac{\pi}{3} \frac{\sqrt{\tan\left( \frac{\pi}{3} + \frac{\pi}{6} - x \right)}}{\sqrt{\tan\left( \frac{\pi}{3} + \frac{\pi}{6} - x \right)} + \sqrt{\cot\left( \frac{\pi}{3} + \frac{\pi}{6} - x \right)}}dx .....................\left[ \int_a^b f\left( x \right)dx = \int_a^b f\left( a + b - x \right)dx \right]\]
\[ = \int_\frac{\pi}{6}^\frac{\pi}{3} \frac{\sqrt{\tan\left( \frac{\pi}{2} - x \right)}}{\sqrt{\tan\left( \frac{\pi}{2} - x \right)} + \sqrt{\cot\left( \frac{\pi}{2} - x \right)}}dx\]
\[ = \int_\frac{\pi}{6}^\frac{\pi}{3} \frac{\sqrt{\cot x}}{\sqrt{\cot x} + \sqrt{\tan x}}dx ................\left( 2 \right)\]

Adding (1) and (2), we get

\[2I = \int_\frac{\pi}{6}^\frac{\pi}{3} \frac{\sqrt{\tan x} + \sqrt{\cot x}}{\sqrt{\tan x} + \sqrt{\cot x}}dx\]
\[ \Rightarrow 2I = \int_\frac{\pi}{6}^\frac{\pi}{3} dx\]
\[ \Rightarrow 2I = \left.x\right|_\frac{\pi}{6}^\frac{\pi}{3} \]
\[ \Rightarrow 2I = \frac{\pi}{3} - \frac{\pi}{6} = \frac{\pi}{6}\]
\[ \Rightarrow I = \frac{\pi}{12}\]