# 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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Wait, do you mean $\int(I+J) = \int (\sqrt{\tan x}+\sqrt{\cot x})dx$? – Addem Aug 9 '15 at 18:31
actually Here $\displaystyle I = \int\sqrt{\tan x}dx$ and $\displaystyle J = \int\sqrt{\cot x}dx$ – juantheron Aug 9 '15 at 18:59
It might be more elegant to write the solution as a parallel construct using $\text{arsinh}(\sin(x)+\cos(x))$, rather than the log equivalent. – Dr. MV Mar 3 at 15:56

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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As already mentioned in some answers, let $t^2=\tan x \implies 2tdt=\sec^2x dx$ or $dx=\frac{2t}{t^4+1}$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 $$I=\frac{1}{\sqrt{2}}\tan^{-1}\left(\frac{\sqrt{\tan x}-\frac{1}{\sqrt{\tan x}}}{\sqrt{2}}\right)+\frac{1}{2\sqrt{2}}\ln\left(\frac{\sqrt{\tan x}+\frac{1}{\sqrt{\tan x}}-\sqrt{2}}{\sqrt{\tan x}+\frac{1}{\sqrt{\tan x}}+\sqrt{2}}\right)+C$$ $$=\frac{1}{\sqrt{2}}\tan^{-1}\left(\frac{\sqrt{\tan x}-\sqrt{\cot x}}{\sqrt{2}}\right)+\frac{1}{2\sqrt{2}}\ln\left(\frac{\sqrt{\tan x}+\sqrt{\cot x}-\sqrt{2}}{\sqrt{\tan x}+\sqrt{\cot x}+\sqrt{2}}\right)+C$$ $$=\frac{1}{\sqrt{2}}\tan^{-1}\left(\frac{\sin x-\cos x}{\sqrt{\sin 2x}}\right)+\frac{1}{2\sqrt{2}}\ln\left(\frac{\sin x+\cos x-\sqrt{\sin 2x}}{\sin x+\cos x+\sqrt{\sin 2x}}\right)+C$$

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$$y= \int \sqrt{tanx } \ dx \\ g= \int\sqrt{cotx}\ dx\\ y+g=\int \sqrt{tanx}+\sqrt{cotx} \ dx=\sqrt {2}\int\frac{sinx+cosx}{\sqrt{sin2x}}\\ y+g=\sqrt{2}\int \frac {(sinx-cosx)'}{\sqrt{1-(sinx-cosx)^2}}\\ y+g=\sqrt{2}\ \int\frac{u'}{\sqrt{1-u^2}}dx=\sqrt{2}\arcsin(u)\\ *y+g=\sqrt{2}arcsin(sinx-cosx)\\ y-g=\int\sqrt{tanx}-\sqrt{cotx } \ dx =\sqrt{2}\int\frac{sinx-cosx}{\sqrt{sin2x}}\\ y-g=-\sqrt{2}\int\frac{(sinx+cosx)'}{\sqrt{(sinx+cosx)^2-1}} dx =-\sqrt{2}\int\frac{s'}{\sqrt{s^2-1}}\ dx\\ y-g=-\sqrt{2}arcosh(sinx+cosx) * * \\ \frac{(y-g)+(y+g)}{2}=y\\ y=\frac{\sqrt{2}}{2}(arcsin(sinx-cosx)-arccosh(sinx+cos))$$
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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.

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