1708\hspace{-16}$Let $\bf{\mathbb{I} = \int \left(\sqrt{\tan x}+\sqrt{\cot x}\right)dx = \sqrt{2}\int\frac{(\sin x+\cos x)}{\sqrt{\sin 2x}}dx}$\\\\\\ Now Let $\bf{(sin x-\cos x)=t\Rightarrow 1-\sin 2x = t^2}$\\\\\\ So $\bf{\sin 2x = 1-t^2}$ and $\bf{(\cos x+\sin x)dx = dt}$\\\\\\ So $\bf{\mathbb{I} = \sqrt{2}\int\frac{1}{\sqrt{1-t^2}}dt = \sqrt{2}\cdot \sin^{-1}(t)+\mathbb{C}}$\\\\\\ So $\bf{\mathbb{I} = \int \left(\sqrt{\tan x}+\sqrt{\cot x}\right)dx =\sqrt{2}\cdot \sin^{-1}(\sin x-\cos x)+\mathbb{C}}$
