# 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.
$$\int \sqrt{\tan x} ~ \mathrm{d}{x}.$$

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