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

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

Let $I=\int \sqrt{\tan x}+\sqrt{\cot x} dx$

$=\sqrt{2} \int \frac{\sin x+\cos x}{\sqrt{2 \sin x . \cos x}} dx$

$=\sqrt{2} \int \frac{(\sin x+\cos x)}{\sqrt{\sin 2 x}} dx$

Let t = sin x – cos x

$\Rightarrow t^2=1-\sin 2 x$

$\Rightarrow dt=(\cos x+\sin x) dx$

∴ $I=\sqrt{2} \int \frac{d t}{\sqrt{1-t^2}}=\sqrt{2} \sin ^{-1}(t)+c$

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

Hence (3) is the correct answer.