Question

Evaluate : ${\int }_{\pi /6}^{\pi /3}\frac{1}{1+\sqrt{\tan x}}dx$.

Answer 1

$\begin{array}{cc}& \text{Let}I={\int }_{x/6}^{x/3}\frac{dx}{1+\sqrt{\tan x}}\\ & I={\int }_{x/6}^{\pi /3}\frac{dx}{1+\sqrt{\tan (\frac{\pi }{6}+\frac{\pi }{3}-x)}}\\ & \left[\mathrm{By}u\sin g\text{property}{\int }_a^{b}f\right(x)dx={\int }_a^{b}f(a+b-x\left)dx\right]\dots \left(i\right)\\ & I={\int }_{\pi /6}^{\pi /3}\frac{dx}{1+\sqrt{\tan (\frac{\pi }{2}-x)}}\\ & I={\int }_{\pi /6}^{\pi /3}\frac{dx}{1+\frac{1}{\sqrt{{\cos }^2}}}={\int }_{\pi /6}^{\pi /3}\frac{dx}{1+\sqrt{\cot x}}\\ & I={\int }_{\pi /6}^{\pi /3}\frac{\sqrt{\tan x}}{1+\sqrt{\tan x}}dx\end{array}$

Adding (i) and (ii), we get

$\begin{array}{c}2I={\int }_{\pi /6}^{\pi /3}\frac{(1+\sqrt{\tan x})}{(1+\sqrt{\tan x})}dx\\ ={\int }_{\pi /6}^{\pi /3}dx=[x{]}_{x/6}^{\pi /3}=\frac{\pi }{3}-\frac{\pi }{6}=\frac{\pi }{6}\\ 2I=\frac{\pi }{6}\text{or}I=\frac{\pi }{12}\end{array}$