Here's a line integral that turned out I couldn't find its closed form:


where curve C: { $x=t\sqrt{t}$, $y=arctan(\sqrt{t})$ and $t\in[0,1]$ };



=> $I=\int_{0}^{1}[{arcsin(arctan(\sqrt{t})){\frac{3}{2}\sqrt{t}}}+{\frac{t^4}{2(1+t^2)}]dt}$,

Let us note:




There shouldn't be any problems solving $I_2$. But there are some problems (as I see it) regarding integral $I_1$.

My approach to it was setting $t=tan^2{(u)};$

$t=0\rightarrow$ $u=0$; $t=1 \rightarrow u=\frac{\pi}{4}$


So, $I_1=\int_{0}^{\frac{\pi}{4}}{\arcsin(arctan{(|\tan{u}|)}}{\cdot}3{\ |\tan{u}|}{\cdot}\tan(u)\tan{'}(u)du$

I assumed that $tan{(u)}>0$ for $u {\in}[0, {\frac{\pi}{4}}]$ which implies $|\tan{(u)}|=\tan{(u)}$,

Therefore, we would have:



Let $J=\int_{0}^{\frac{\pi}{4}}{\frac{\tan^3{(u)}}{\sqrt{1-u^2}}}du$

So what would $J$ round up to? Can we end up with a closed form solution? Or is it bound to be numerically integrated or approximated with Taylor series? Or maybe we could get it in form of Struve/Bessel functions? If one of these possibilites is available, how can it be done? If not, then what?

To be honest with you, I think (to be interpreted as 'I'm not so sure') it shouldn't be something too complicated given the fact that after all, it is a line integral.

If I did something wrong I will be most eager to know what and learn from it. Thank you for your time invested in it.

Best regards.

  • $\begingroup$ I tried three CAS for $I_1$ : no success. $\endgroup$ Jul 13, 2019 at 3:14
  • $\begingroup$ @ClaudeLeibovici Thank you for your time and implication anyways! Have a great day, sir. $\endgroup$
    – Vlad Botis
    Jul 14, 2019 at 1:32

1 Answer 1


This is not an answer.

Considering $$I=\int \sqrt{t} \sin ^{-1}\left(\tan ^{-1}\left(\sqrt{t}\right)\right)\,dt$$ none of the CAS I tried has been able to provide any expression (even using special functions).

Considering the numerical integration $$J=\int_0^1 \sqrt{t} \sin ^{-1}\left(\tan ^{-1}\left(\sqrt{t}\right)\right)\,dt=0.4632123107582015077125384493582154662529123249$$ none of the inverse symbolic calculators has been able to identify the result.

In any manner, in my humble opinion, for numerical integration, it would be better to leave it as written since, over the range, the integrand is almost linear (this is not at all the case after the change of variable $t=\tan^2{(u)}$).

If you make $t=u^2$, the integrand fo $I$ becomes $2 u^2 \sin ^{-1}\left(\tan ^{-1}(u)\right)$ which, expanded as a Taylor series, gives $$2 u^2 \sin ^{-1}\left(\tan ^{-1}(u)\right)=2 \sum_{n=1}^\infty \frac{a_n}{b_n}u ^{2n+1}$$ where the $a_n$ and $b_n$ respectively correspond to sequences $A096717$ and $A096718$ at $OEIS$. You should notice that no asymptotics are provided.

  • $\begingroup$ This is an interesting approach, and rather insightful I would say. This could actually mean the ultimatum to this 'issue'. I appreciate it very much and I think this is the most helpful in actually finding the solution. Again, thank you for your time and implication. Have a really great day, sir! $\endgroup$
    – Vlad Botis
    Jul 16, 2019 at 3:18

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