A Tricky Integral!!

Solution as described by Ishita Bhatia .

$\begin{aligned} I & = \int_0^{\frac{\pi}{2}} \frac{\sqrt{\cot \color{#3D99F6}{x}}}{\sqrt{\cot \color{#3D99F6}{x}}+ \sqrt{\tan \color{#3D99F6}{x}}} \, dx \quad \quad \small \color{#3D99F6}{\text{Let } x = \frac{\pi}{2} - u \implies dx = - du } \\ & = \int_0^{\frac{\pi}{2}} \frac{\sqrt{\color{#3D99F6}{\tan u}}}{\sqrt{\color{#3D99F6}{\tan u}}+ \sqrt{\color{#3D99F6}{\cot u}}} \, d\color{#3D99F6}{u} \quad \quad \small \color{#3D99F6}{\text{Replace } u \text{ with } x} \\ \implies 2I & = \int_0^{\frac{\pi}{2}} \frac{\sqrt{\cot x}}{\sqrt{\cot x}+ \sqrt{\tan x}} \, dx + \int_0^{\frac{\pi}{2}} \frac{\sqrt{\tan x}}{\sqrt{\cot x}+ \sqrt{\tan x}} \, dx \\ & = \int_0^{\frac{\pi}{2}} \frac{\sqrt{\cot x}+\sqrt{\tan x}}{\sqrt{\cot x}+ \sqrt{\tan x}} \, dx \\ & = \int_0^{\frac{\pi}{2}} \, dx = \frac{\pi}{2} \\ \implies I & = \boxed{\dfrac{\pi}{4}} \end{aligned}$

We first consider the given equation as equation one.

now we put x= 90-x and put it as equation 2

then we simply add both the equation so the numerator and the denominator

cancel out. we then integrate.