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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asked Jun 10, 2014 at 1:30
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6$\begingroup$ Have you tried $tan(x)=u^2$ ? $\endgroup$– user146010Commented Jun 10, 2014 at 1:34
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1$\begingroup$ Wait, nevermind. I got it. Thanks for the hint. $\endgroup$– A is for AmbitionCommented Jun 10, 2014 at 1:36
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5$\begingroup$ @user155812: You should have obtained: $\int \frac{2u^2}{u^4+1}\mathrm{d} u$, after which you use partial fractions, via $(u^4+1) = (u^2+u\sqrt 2 +1)(u^2-u\sqrt 2 +1)$ $\endgroup$– Graham KempCommented Jun 10, 2014 at 1:44
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3$\begingroup$ 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. $\endgroup$– heropupCommented Jun 10, 2014 at 1:53
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1$\begingroup$ The funny part is that all the answers lead to different closed formulas. $\endgroup$– polkovnikov.phCommented Jun 29, 2017 at 12:56
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7 Answers
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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?? $\endgroup$ Commented Dec 15, 2017 at 0:56
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$\begingroup$ @GabrielSandoval I think it's a typo, it should be just $ds$ instead of $s' ds$ $\endgroup$– ItachiCommented May 7, 2020 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 $(2)$:
$$I = \frac{1}{\sqrt{2}} \sin^{-1}(\sin x - \cos x) - \frac{1}{\sqrt{2}} \ln\left|\sin x + \cos x + \sqrt{\sin 2x} \vphantom{x^{x^x}} \right| + \mathbb{C}$$
answered Jun 10, 2014 at 2:22
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1$\begingroup$ Wait, do you mean $\int(I+J) = \int (\sqrt{\tan x}+\sqrt{\cot x})dx$? $\endgroup$– AddemCommented Aug 9, 2015 at 18:31
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2$\begingroup$ actually Here $\displaystyle I = \int\sqrt{\tan x}dx$ and $\displaystyle J = \int\sqrt{\cot x}dx$ $\endgroup$ Commented Aug 9, 2015 at 18:59
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2$\begingroup$ 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. $\endgroup$ Commented Mar 3, 2016 at 15:56
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3$\begingroup$ @MarkViola not really, not everyone (for example highschool students) is acquainted with hyperbolic functions $\endgroup$– ArcherCommented Jul 19, 2018 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 $$I=\frac{1}{\sqrt{2}}\tan^{-1}\left(\frac{\sqrt{\tan x}-\frac{1}{\sqrt{\tan x}}}{\sqrt{2}}\right)+\frac{1}{2\sqrt{2}}\ln\left(\frac{\sqrt{\tan x}+\frac{1}{\sqrt{\tan x}}-\sqrt{2}}{\sqrt{\tan x}+\frac{1}{\sqrt{\tan x}}+\sqrt{2}}\right)+C$$ $$=\frac{1}{\sqrt{2}}\tan^{-1}\left(\frac{\sqrt{\tan x}-\sqrt{\cot x}}{\sqrt{2}}\right)+\frac{1}{2\sqrt{2}}\ln\left(\frac{\sqrt{\tan x}+\sqrt{\cot x}-\sqrt{2}}{\sqrt{\tan x}+\sqrt{\cot x}+\sqrt{2}}\right)+C$$ $$=\frac{1}{\sqrt{2}}\tan^{-1}\left(\frac{\sin x-\cos x}{\sqrt{\sin 2x}}\right)+\frac{1}{2\sqrt{2}}\ln\left(\frac{\sin x+\cos x-\sqrt{\sin 2x}}{\sin x+\cos x+\sqrt{\sin 2x}}\right)+C$$
answered May 22, 2015 at 19:36
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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
answered Dec 8, 2016 at 8:35
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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 $a=i$ $$I=\frac 1{\sqrt a}\left(\tan ^{-1}\left(\frac{u}{\sqrt{a}}\right)-\tanh ^{-1}\left(\frac{u}{\sqrt{a}}\right)\right)=-\frac{1-i}{\sqrt{2}}\left(\tan ^{-1}\left((-1)^{3/4} u\right)-\tanh ^{-1}\left((-1)^{3/4} u\right)\right)$$
answered Aug 11, 2020 at 3:36