How to Integrate sqrt(tan x)

 

$$\begin{array}{l}\int \sqrt{tanx}\mathrm{d}x\\ \\ BySubstitution\\ z=\sqrt{tanx}\\ z^2=tanx\\ 2zdz=se{c}^{2}xdx\\ =(ta{n}^{2}x+1)dx=(z^4+1)dx\\ dx=\frac{2z}{z^4+1}dz\\ \int \frac{2z^2}{z^4+1}dz\\ \int \frac{(z^2+1)+\left(z^2-1\right)}{z^4+1}dz=\int \frac{z^2+1}{z^4+1}dz+\int \frac{z^2-1}{z^4+1}dz\\ \int \frac{1+\frac{1}{z^2}}{z^2+\frac{1}{z^2}}dz+\int \frac{1-\frac{1}{z^2}}{z^2+\frac{1}{z^2}}dz=\int \frac{1+\frac{1}{z^2}}{{\left(z-\frac{1}{z}\right)}^2+2}dz+\int \frac{1-\frac{1}{z^2}}{{\left(z+\frac{1}{z}\right)}^2-2}dz\\ bysubstitution\\ let\Rightarrow u=z-\frac{1}{z}du=\left(1+\frac{1}{z^2}\right)dz\\ andlet\Rightarrow v=z+\frac{1}{z}dv=\left(1-\frac{1}{z^2}\right)dz\\ \int \frac{du}{u^2+2}+\int \frac{dv}{v^2-2}=\frac{1}{\sqrt{2}}{\tan }^{-1}\frac{u}{\sqrt{2}}+\frac{1}{2\sqrt{2}}ln\left|\frac{v-\sqrt{2}}{v+\sqrt{2}}\right|+c\\ \frac{1}{\sqrt{2}}{\tan }^{-1}\frac{z-\frac{1}{z}}{\sqrt{2}}+\frac{1}{2\sqrt{2}}ln\left|\frac{z+\frac{1}{z}-\sqrt{2}}{z+\frac{1}{z}+\sqrt{2}}\right|+c\\ \frac{1}{\sqrt{2}}{\tan }^{-1}\frac{\sqrt{\tan x}-\frac{1}{\sqrt{\tan x}}}{\sqrt{2}}+\frac{1}{2\sqrt{2}}ln\left|\frac{\sqrt{\tan x}+\frac{1}{\sqrt{\tan x}}-\sqrt{2}}{\sqrt{\tan x}+\frac{1}{\sqrt{\tan x}}+\sqrt{2}}\right|+c\end{array}$$