I have two questions about the value of this integral

$$ \int_0^\pi\sqrt{1+t^2}\,\mathrm dt $$

First question: I would like to know if my computations below are correct.

Second question: I would like to know if there is a more simple closed form for it value than what I found.

Let $t=\sinh x$, then

$$ \int_0^\pi\sqrt{1+t^2}\,\mathrm dt=\int_0^{\operatorname{arsinh}\pi}\cosh^2 x\,\mathrm dx=\int_0^{\operatorname{arsinh}\pi}\mathrm dx+\int_0^{\operatorname{arsinh}\pi}\sinh^2x\,\mathrm dx\\=\operatorname{arsinh}\pi+\sinh x\cosh x\big|_0^{\operatorname{arsinh}\pi}-\int_0^{\operatorname{arsinh}\pi}\cosh^2 x\,\mathrm dx $$

Thus I found that

$$ \int_0^\pi\sqrt{1+t^2}\,\mathrm dt=\frac12(\operatorname{arsinh}\pi+\pi\cosh\left(\operatorname{arsinh}\pi\right)) $$

and because $\cosh x=\sqrt{\sinh^2 x+1}$ and $\operatorname{arsinh}x=\ln\left(x+\sqrt{1+x^2}\right)$ (for $x\in\Bbb R$ at least) then finally I get the expression

$$ \int_0^\pi\sqrt{1+t^2}\,\mathrm dt=\frac12(\ln(\pi+\sqrt{1+\pi^2})+\pi\sqrt{1+\pi^2}) $$

  • $\begingroup$ For your second question, no. $\endgroup$ – Weijun Zhou Jan 23 '18 at 18:50
  • $\begingroup$ $1.$ Your computations are correct. $2.$ Why don't you use the direct formula for $\int {\sqrt{1+x^2}.dx}= \frac{x \sqrt{1+x^2}+ln|x+\sqrt{1+x^2}|}{2}+c$? $\endgroup$ – I am Back Jan 23 '18 at 18:53
  • $\begingroup$ @VidyanshuMishra I dont knew this "direct formula", where you see it? $\endgroup$ – Masacroso Jan 23 '18 at 18:54
  • $\begingroup$ Well I didn't see it anywhere, I derived it using integration by part. (Also I am not the only one or first one to derive this XD) $\endgroup$ – I am Back Jan 23 '18 at 18:58
  • 1
    $\begingroup$ math.stackexchange.com/a/590407/363566 see this answer. $\endgroup$ – I am Back Jan 23 '18 at 19:01

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