$\int\{\sqrt{\tan x}+\sqrt{\cot x}\} d x$ is equal to

$\sqrt{2} \sin ^{-1}(\sin x-\cos x)+C$

We have,

$I=\int\{\sqrt{\tan x}+\sqrt{\cot x}\} d x=\int \frac{\sin x+\cos x}{\sqrt{\sin x \cos x}} d x$

$\Rightarrow I=\sqrt{2} \int \frac{\sin x+\cos x}{\sqrt{2 \sin x \cos x}} d x$

$\Rightarrow I=\sqrt{2} \int \frac{1}{\sqrt{1-(\sin x-\cos x)^2}} d(-\cos x+\sin x)$

$\Rightarrow I=\sqrt{2} \sin ^{-1}(\sin x-\cos x)+C$