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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 at 1:34
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A Google search finds math.ucsd.edu/~wgarner/math20b/int_sqrt_tan.htm –  Robert Israel Jun 10 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. –  user155812 Jun 10 at 1:35
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@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 at 1:44
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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 at 1:53

2 Answers 2

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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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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