Evaluating the indefinite integral $ \int \sqrt{\tan x} ~ \mathrm{d}{x}. $ [closed]

I have been having extreme difficulties with this integral. I would appreciate any and all help.
$$
\int \sqrt{\tan x} ~ \mathrm{d}{x}.
$$

Answer

$$y=\int\sqrt{\tan x}\,\mathrm dx$$
$$g=\int\sqrt{\cot x}\,\mathrm dx$$

\begin{align} y+g&=\int\left(\sqrt{\tan x}+\sqrt{\cot x}\right)\,\mathrm dx
\\&=\sqrt2\int\frac{\sin x+\cos x}{\sqrt{\sin2x}}\mathrm dx \\&
=\sqrt2\int\frac{(\sin x-\cos x)’}{\sqrt{1-(\sin x-\cos x)^2}}\,\mathrm dx\\&
=\sqrt2\int\frac{1}{\sqrt{1-u^2}}\,\mathrm du \\&
=\sqrt2\sin^{-1}u \\&
=\sqrt2\sin^{-1}(\sin x-\cos x)\end{align}

\begin{align} y-g&=\int\left(\sqrt{\tan x}-\sqrt{\cot x}\right)\,\mathrm dx \\&
=\sqrt2\int\frac{\sin x-\cos x}{\sqrt{\sin2x}} \,\mathrm dx\\&
=-\sqrt2\int\frac{(\sin x+\cos x)’}{\sqrt{(\sin x+\cos x)^2-1}}\,\mathrm dx \\&
=-\sqrt2\int\frac{\mathrm ds}{\sqrt{s^2-1}} \\&
=-\sqrt2\cosh^{-1}(\sin x+\cos x) \end{align}

\begin{align}y&=\frac{(y-g)+(y+g)}2 \\&=
\frac{\sqrt2}2(\sin^{-1}(\sin x-\cos x)-\cosh^{-1}(\sin x+\cos x)) + C\end{align}

Attribution
Source : Link , Question Author : A is for Ambition , Answer Author : V.G

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