Question

Prove that: ${\int }_0^{\pi /2}\frac{dx}{1+\sqrt{\tan x}}=\frac{\pi }{4}$.

Answer 1

Consider

$I={\int }_0^{\pi /2}\frac{dx}{1+\sqrt{\tan x}}..............\left(1\right)$

$I={\int }_0^{\pi /2}\frac{dx}{1+\sqrt{\tan (\pi /2+0-x)}}$

$I={\int }_0^{\pi /2}\frac{dx}{1+\sqrt{\cot x}}$

$I={\int }_0^{\pi /2}\frac{dx}{1+\sqrt{\frac{1}{\tan x}}}$

$I={\int }_0^{\pi /2}\frac{\sqrt{\tan x}}{\sqrt{\tan x}+1}dx.........\left(2\right)$

Addding $\left(1\right)\&\left(2\right)$

$2I={\int }_0^{\pi /2}\frac{1+\sqrt{\tan x}}{1+\sqrt{\tan x}}dx$

$2I={\int }_0^{\pi /2}dx$

$2I=\pi /2-0$

$⟹I=\pi /4$

Hence proved.