Integration of $\sqrt {\tan x}$ I have tried many ways to integrate $\sqrt {\tan x}$ including integration by parts but didn't get to any final result.
I also assumed,
$$ \tan x = t^2 $$
$$ \int  \sqrt {\tan x} \,dx $$
$$⇒\int \frac{2t^2}{1+t^4}dt$$
but it's getting a bit complicated further, kindly help. Also, are there any simpler ways to integrate this.
Answer,
$$ \frac{1}{\sqrt 2} \tan^{-1}\left[\frac {\sqrt {\tan x}-\sqrt {\cot x}}{\sqrt{2}}\right] +\frac{1}{2\sqrt 2}\ln\left[\frac {\sqrt {\tan x}+\sqrt {\cot x}-\sqrt {2}}{\sqrt {\tan x}+\sqrt {\cot x}+\sqrt {2}}\right] +C $$
 A: This is an integral blackpenredpen did on YouTube some time ago. To somewhat summarize his solution (i.e. I claim no originality in this solution), we begin with where you left off. I will rewrite the integral in terms of $u$ to follow the video solution too, for ease of following along.
$$\mathcal I := \int \frac{2u^2}{u^4 + 1} du$$
First, multiply by $1/u^2$ on top and bottom. Then
$$\mathcal I = \int \frac{2}{u^2 + u^{-2}}du$$
We can sort of complete the square on the bottom: it bears resemblance to part of a perfect trinomial, just sans a third term. We note the following pair of identities:
$$\begin{align}
\left( u + \frac 1 u \right)^2 &= u^2 + 2 + \frac{1}{u^2}\\
\left( u - \frac 1 u \right)^2 &= u^2 - 2 + \frac{1}{u^2}
\end{align}$$
The second becomes more useful in a bit; for now, we use the first (though the decision is ultimately arbitrary). Thus, subtracting $2$ from both sides from the first, we get a nice substitution:
$$\mathcal I =\int \frac{2}{(u+u^{-1})^2 - 2}du$$
This suggests a substitution: $t = u + u^{-1}$. Notice that, if we were to make this substitution, then $dt = (1 - u^{-2})du$. We would like that latter expression to pop up in our integrand somewhere. We notice, then,
$$2 = 1 - \frac{1}{u^2} + 1 + \frac{1}{u^2}$$
We then split up our integral into two at this point as well:
$$\mathcal I = \int \frac{1 - u^{-2}}{(u+u^{-1})^2 - 2}du + \int \frac{1 + u^{-2}}{(u+u^{-1})^2 - 2}du$$
Return to our pair of identities from earlier, and notice that
$$\left( u + \frac 1 u \right)^2 -2 = u^2 + \frac{1}{u^2} = \left( u - \frac 1 u \right)^2 + 2$$
This equality is used in our second integral, bringing us to this point:
$$\mathcal I = \int \frac{1 - u^{-2}}{(u+u^{-1})^2 - 2}du + \int \frac{1 + u^{-2}}{(u-u^{-1})^2 + 2}du$$
So, we go forward with out $t$-substitution we suggested earlier for the first integral, and follow it up with a comparable $w$-substitution for the second integral:
$$\begin{align}
t = u+ \frac 1 u &\implies dt = \left( 1 - \frac{1}{u^2} \right)du \\
w = u - \frac 1 u &\implies dw = \left(1 + \frac{1}{u^2} \right)du
\end{align}$$
Each substitution negates the numerator, bringing us to
$$\mathcal I = \int \frac{1}{t^2 - 2}dt + \int \frac{1}{w^2 + 2} du$$
These are fairly standard integrals to calculate, and what remains from here is just to back-substitute our way through each integral and do whatever simplification we desire. I'll leave those calculations up to you.
A: You can factor the denominator as the product of two quadratics (finding the [necessarily complex] roots can help there), then use partial fractions. It gets ugly, but Wofram Alpha gives the same ugly answer as you get by doing this, so I assume there isn't a simpler way.
BTW, are you sure about the numerator? I have vague memories of it being $\frac{1}{1+t^4}$, but it was quite a while ago that I did it.
