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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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    $\begingroup$ Have you tried $tan(x)=u^2$ ? $\endgroup$
    – user146010
    Jun 10 '14 at 1:34
  • $\begingroup$ Yes. Using this substitution, I ended with the integral $\displaystyle \int\frac{2u}{u^4+1}\,du$, and I was unable to solve that. $\endgroup$ Jun 10 '14 at 1:35
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    $\begingroup$ Wait, nevermind. I got it. Thanks for the hint. $\endgroup$ Jun 10 '14 at 1:36
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    $\begingroup$ @user155812: You should have obtained: $\int \frac{2u^2}{u^4+1}\mathrm{d} u$, after which you use partial fractions, via $(u^4+1) = (u^2+u\sqrt 2 +1)(u^2-u\sqrt 2 +1)$ $\endgroup$ Jun 10 '14 at 1:44
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    $\begingroup$ 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. $\endgroup$
    – heropup
    Jun 10 '14 at 1:53
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$$y=\int\sqrt{\tan x}\,\mathrm dx$$ $$g=\int\sqrt{\cot x}\,\mathrm dx$$

\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}

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  • $\begingroup$ Why?? In the integral $y-g$ You have the substitution $s=\sin x + \cos x$ and the expression $\dfrac{s'}{\sqrt{s^2-1}}dx$ becomes $\dfrac{ds}{\sqrt{s^2-1}}$. Can you explain that step?? $\endgroup$ Dec 15 '17 at 0:56
  • $\begingroup$ @GabrielSandoval I think it's a typo, it should be just $ds$ instead of $s' ds$ $\endgroup$
    – Sebitas
    May 7 '20 at 5:10
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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 $(2)$:

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

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    $\begingroup$ Wait, do you mean $\int(I+J) = \int (\sqrt{\tan x}+\sqrt{\cot x})dx$? $\endgroup$
    – Addem
    Aug 9 '15 at 18:31
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    $\begingroup$ actually Here $\displaystyle I = \int\sqrt{\tan x}dx$ and $\displaystyle J = \int\sqrt{\cot x}dx$ $\endgroup$
    – juantheron
    Aug 9 '15 at 18:59
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    $\begingroup$ 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. $\endgroup$
    – Mark Viola
    Mar 3 '16 at 15:56
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    $\begingroup$ @MarkViola not really, not everyone (for example highschool students) is acquainted with hyperbolic functions $\endgroup$
    – Archer
    Jul 19 '18 at 6:59
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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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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 $$I=\frac{1}{\sqrt{2}}\tan^{-1}\left(\frac{\sqrt{\tan x}-\frac{1}{\sqrt{\tan x}}}{\sqrt{2}}\right)+\frac{1}{2\sqrt{2}}\ln\left(\frac{\sqrt{\tan x}+\frac{1}{\sqrt{\tan x}}-\sqrt{2}}{\sqrt{\tan x}+\frac{1}{\sqrt{\tan x}}+\sqrt{2}}\right)+C$$ $$=\frac{1}{\sqrt{2}}\tan^{-1}\left(\frac{\sqrt{\tan x}-\sqrt{\cot x}}{\sqrt{2}}\right)+\frac{1}{2\sqrt{2}}\ln\left(\frac{\sqrt{\tan x}+\sqrt{\cot x}-\sqrt{2}}{\sqrt{\tan x}+\sqrt{\cot x}+\sqrt{2}}\right)+C$$ $$=\frac{1}{\sqrt{2}}\tan^{-1}\left(\frac{\sin x-\cos x}{\sqrt{\sin 2x}}\right)+\frac{1}{2\sqrt{2}}\ln\left(\frac{\sin x+\cos x-\sqrt{\sin 2x}}{\sin x+\cos x+\sqrt{\sin 2x}}\right)+C$$

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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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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

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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 $a=i$ $$I=\frac 1{\sqrt a}\left(\tan ^{-1}\left(\frac{u}{\sqrt{a}}\right)-\tanh ^{-1}\left(\frac{u}{\sqrt{a}}\right)\right)=-\frac{1-i}{\sqrt{2}}\left(\tan ^{-1}\left((-1)^{3/4} u\right)-\tanh ^{-1}\left((-1)^{3/4} u\right)\right)$$

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