Help with integral $\int\frac{1}{\sqrt{\tan x}}dx$

Question

I tried to solve by parts but it did not help.

Answer 1

Let $$u = \tan x \implies du = \sec^2 x dx \iff \,dx = \frac{du}{\sec^2 x} = \frac{du}{1 + \tan^2 x} = \frac{du}{1+ u^2}$$

Substituting gives us $$\int \dfrac{du}{(1 + u^2)u^{1/2}}$$

Answer 2

Here's a hint: set $\tan(x)=t^2.$

Answer 3

$u^2=\tan x\Longrightarrow 2udu=\sec^2xdx=(1+\tan^2x)dx=(1+u^4)dx\Longrightarrow dx=\frac{2u}{1+u^4}du$. So, $$\int\frac{1}{\sqrt\tan x}dx=2\int\frac{du}{1+u^4}=2\int\frac{du}{(1+\sqrt 2u+u^2)(1-\sqrt 2u+u^2)}$$ $$=2\int\frac{Au+B}{1+\sqrt 2u+u^2}du+2\int\frac{Cu+D}{1-\sqrt 2u+u^2}du:=2I+2J$$ where $$I=\int\frac{Au+B}{1+\sqrt 2u+u^2}du,\,J=\int\frac{Cu+D}{1-\sqrt 2u+u^2}du $$ with $A=-\frac{\sqrt 2}{4}$, $C=\frac{\sqrt 2}{4}$, $A=B=\frac{1}{2}$. $$I=-\frac{1}{4}\int\frac{\sqrt2u-2}{1+\sqrt 2u+u^2}du=-\frac{1}{4}\int\frac{(\sqrt2u+2-4)}{1+\sqrt 2u+u^2}du$$$$=-\frac{1}{4}\int\frac{(\sqrt2u+2)}{1+\sqrt 2u+u^2}du+\int\frac{du}{(u+\frac{\sqrt 2}{2})^2+\frac{1}{2}}$$ $$=-\frac{1}{4}\ln (1+\sqrt2u+u^2)+\sqrt2\arctan(u+\frac{\sqrt 2}{2}).$$ $J$ can be computed similarly.