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

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Have you tried $tan(x)=u^2$ ? –  Edwin Jun 10 '14 at 1:34
A Google search finds math.ucsd.edu/~wgarner/math20b/int_sqrt_tan.htm –  Robert Israel Jun 10 '14 at 1:35
Yes. Using this substitution, I ended with the integral $\displaystyle \int\frac{2u}{u^4+1}\,du$, and I was unable to solve that. –  A is for Ambition Jun 10 '14 at 1:35
@user155812: You should have obtained: $\int \frac{2u^2}{u^4+1}\mathrm{d} u$, after which you use partial fractions, via $(u^4+1) = (u^2+u\sqrt 2 +1)(u^2-u\sqrt 2 +1)$ –  Graham Kemp Jun 10 '14 at 1:44
I rolled back the previous edit of the title because the use of "primitive" to mean "indefinite integral" is not universally understood in the mathematical literature. There was no reason to edit it given that the previous title was already unambiguously clear. –  heropup Jun 10 '14 at 1:53

Let $I = \sqrt{\tan x}\;\mathrm{d}x$ and $J = \sqrt{\cot x}\;\mathrm{d}x$.

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 $(2)$:

$$I = \frac{1}{\sqrt{2}} \sin^{-1}(\sin x - \cos x) - \frac{1}{\sqrt{2}} \ln\left|\sin x + \cos x + \sqrt{\sin 2x} \vphantom{x^{x^x}} \right| + \mathbb{C}$$

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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.

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