Evaluating the indefinite integral $ \int \sqrt{\tan x} ~ \mathrm{d}{x}. $ I have been having extreme difficulties with this integral. I would appreciate any and all help.
$$
\int \sqrt{\tan x} ~ \mathrm{d}{x}.
$$
 A: $$y=\int\sqrt{\tan x}\,\mathrm dx$$
$$g=\int\sqrt{\cot x}\,\mathrm dx$$
\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}
A: Let $I = \int\sqrt{\tan x}\;\mathrm{d}x$ and $J = \int\sqrt{\cot x}\;\mathrm{d}x$.
Now $$\begin{align}I + J 
&= \int\left(\sqrt{\tan x} + \sqrt{\cot x}\right) \;\mathrm{d}x \\
&= \sqrt{2} \int\frac{\sin x + \cos x}{\sqrt{\sin 2x}} \;\mathrm{d}x \\[5pt]
&= \sqrt{2} \int\frac{(\sin x - \cos x)'}{\sqrt{1-(\sin x - \cos x)^2}} \;\mathrm{d}x \\[5pt]
&= \sqrt{2} \sin^{-1}(\sin x - \cos x) + \mathbb{C_1} \tag{1} \\
\end{align}$$
and $$\begin{align}I - J
&= \int\left(\sqrt{\tan x} - \sqrt{\cot x}\right) \;\mathrm{d}x \\
&= \sqrt{2} \int\frac{(\sin x - \cos x)}{\sqrt{\sin 2x}} \;\mathrm{d}x \\
&= -\sqrt{2} \int\frac{(\sin x + \cos x)'}{\sqrt{(\sin x + \cos x)^2 - 1}} \;\mathrm{d}x \\
&= -\sqrt{2} \ln\left|(\sin x + \cos x) + \sqrt{(\sin x + \cos x)^2 - 1}\right| + \mathbb{C_2} \tag{2} \\
\end{align}$$
Now, adding $(1)$ and $(2)$:
$$I = \frac{1}{\sqrt{2}} \sin^{-1}(\sin x - \cos x) - \frac{1}{\sqrt{2}} \ln\left|\sin x + \cos x + \sqrt{\sin 2x} \vphantom{x^{x^x}} \right| + \mathbb{C}$$
A: Let $u = \sqrt{\tan x}$, then $u^2 = \tan x$. Thus $2u\;\mathrm{d}u = \sec^2 x\;\mathrm{d}x = (u^4 + 1)\mathrm{d}x$. Thus $\mathrm{d}x = \dfrac{2u\;\mathrm{d}u}{u^4 + 1}$. So:
$$\int\sqrt{\tan x}\;\mathrm{d}x = \int\frac{2u^2}{u^4+1}\;\mathrm{d}u$$
You can take it from here.
A: As already mentioned in some answers, let $t^2=\tan x \implies 2tdt=\sec^2x dx\implies dx=\frac{2tdt}{t^4+1}$. Now, We can easily reach to the final answer as follows $$I=\int \frac{2t^2 dt}{t^4+1}=\int \frac{2 dt}{t^2+\frac{1}{t^2}}=\int \frac{\left(1+\frac{1}{t^2}\right)+\left(1-\frac{1}{t^2}\right) dt}{t^2+\frac{1}{t^2}}$$ $$=\int \frac{\left(1+\frac{1}{t^2}\right) dt}{t^2+\frac{1}{t^2}}+\int \frac{\left(1-\frac{1}{t^2}\right) dt}{t^2+\frac{1}{t^2}}=\int \frac{\left(1+\frac{1}{t^2}\right) dt}{\left(t-\frac{1}{t}\right)^2+2}+\int \frac{\left(1-\frac{1}{t^2}\right) dt}{\left(t+\frac{1}{t}\right)^2-2}$$ $$=\int \frac{\left(1+\frac{1}{t^2}\right) dt}{\left(t-\frac{1}{t}\right)^2+(\sqrt{2})^2}+\int \frac{\left(1-\frac{1}{t^2}\right) dt}{\left(t+\frac{1}{t}\right)^2-(\sqrt{2})^2}$$ $$=\frac{1}{\sqrt{2}}\tan^{-1}\left(\frac{t-\frac{1}{t}}{\sqrt{2}}\right)+\frac{1}{2\sqrt{2}}\ln \left(\frac{\left(t+\frac{1}{t}\right)-\sqrt{2}}{\left(t+\frac{1}{t}\right)+\sqrt{2}}\right)+C$$ Now, substituting the value of $t$, we get $$I=\frac{1}{\sqrt{2}}\tan^{-1}\left(\frac{\sqrt{\tan x}-\frac{1}{\sqrt{\tan x}}}{\sqrt{2}}\right)+\frac{1}{2\sqrt{2}}\ln\left(\frac{\sqrt{\tan x}+\frac{1}{\sqrt{\tan x}}-\sqrt{2}}{\sqrt{\tan x}+\frac{1}{\sqrt{\tan x}}+\sqrt{2}}\right)+C$$ $$=\frac{1}{\sqrt{2}}\tan^{-1}\left(\frac{\sqrt{\tan x}-\sqrt{\cot x}}{\sqrt{2}}\right)+\frac{1}{2\sqrt{2}}\ln\left(\frac{\sqrt{\tan x}+\sqrt{\cot x}-\sqrt{2}}{\sqrt{\tan x}+\sqrt{\cot x}+\sqrt{2}}\right)+C$$ $$=\frac{1}{\sqrt{2}}\tan^{-1}\left(\frac{\sin x-\cos x}{\sqrt{\sin 2x}}\right)+\frac{1}{2\sqrt{2}}\ln\left(\frac{\sin x+\cos x-\sqrt{\sin 2x}}{\sin x+\cos x+\sqrt{\sin 2x}}\right)+C$$
A: 
Hint:

Let $\sqrt{\tan x}=u$, $\quad \frac{1}{2\sqrt{\tan x}}\sec^2 x dx=du$, $\frac{1+u^4}{2u}\ dx=du$
$$\int \sqrt{\tan x}\ dx=\int u \frac{2u}{1+u^4}\ du=\int \frac{2u^2}{1+u^4}\ du$$
Now, make partial fractions $$\frac{2u^2}{1+u^4}=\frac{u^2}{(u^2+u\sqrt 2+1)(u^2-u\sqrt 2+1)}$$ Carry on to get the answer
A: Just another way to do it.
$$I=\int\sqrt{\tan x}\,dx = \int\frac{2u^2}{u^4+1}\,du=\int \left(\frac{1}{u^2+a}+\frac{1}{u^2-a} \right)\,du$$ where $a=i$
$$I=\frac 1{\sqrt a}\left(\tan ^{-1}\left(\frac{u}{\sqrt{a}}\right)-\tanh ^{-1}\left(\frac{u}{\sqrt{a}}\right)\right)=-\frac{1-i}{\sqrt{2}}\left(\tan ^{-1}\left((-1)^{3/4} u\right)-\tanh ^{-1}\left((-1)^{3/4} u\right)\right)$$
A: A slight improvement: instead of $u^2=\tan\theta$, let $u^2=2\tan\theta$.  This gives
$$I=\frac1{\sqrt2}\int \frac{4u^2}{u^4+4}\,du
   =\frac1{\sqrt2}\int \frac{u}{u^2-2u+2}-\frac{u}{u^2+2u+2}\,du\ .$$
Observe that except for the constant out the front, no surds are involved.  Now substitute $v=u-1$ for the first bit and $v=u+1$ for the second bit.  You will need to be careful with the algebra, but it's not all that bad.
