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