arc length of $\dfrac{e^x - e^{-x}}{2}$ how can I please calculate an arc length of $\dfrac{e^x-e^{-x}}{2}$. I tried to substitute $\dfrac{e^x-e^{-x}}{2}=\sinh x$, which leads to $\int\sqrt{1+\cosh^2x}dx$, which unfortunately I can't solve.
Thank you very much.
C.
 A: There is no solution for arbitrary integration limits in elementary terms: we face an elliptic integral:
Wolfram alpha output
A: I find Wolfram Alpha's solution a bit ugly, in that it returns complex results for a real entity, so I'll roll my own solution here.
We start with
$$\int\sqrt{1+\cosh^2 x}\mathrm dx=\frac1{\sqrt{2}}\int\sqrt{3+\cosh\;2x}\;\mathrm dx$$
which can be turned into
$$\frac1{\sqrt{2}}\int\sqrt{3+\frac{1+\tanh^2 x}{1-\tanh^2 x}}\;\mathrm dx$$
(recognize Weierstrass? ;) )
Let
$$\tanh\;x=\mathrm{sn}\left(v|\frac12\right)$$
where $\mathrm{sn}(v|m)$ is a Jacobian elliptic function, resulting in
$$\frac1{\sqrt{2}}\int\sqrt{3+\frac{1+\mathrm{sn}^2\left(v|\frac12\right)}{1-\mathrm{sn}^2\left(v|\frac12\right)}}\mathrm{dc}\left(v|\frac12\right)\;\mathrm dv=\sqrt{2}\int\mathrm{dc}^2\left(v|\frac12\right)\;\mathrm dv$$
where $\mathrm{dc}(v|m)=\frac{\mathrm{dn}(v|m)}{\mathrm{cn}(v|m)}$ is a Jacobian elliptic function.
From this formula, we obtain
$$\sqrt{2}\left(v-E\left(\mathrm{am}\left(v|\frac12\right)|\frac12\right)+\mathrm{sn}\left(v|\frac12\right)\mathrm{dc}\left(v|\frac12\right)\right)$$
or, by undoing the transformation with $v=F\left(\arcsin\left(\tanh\;x\right)|\frac12\right)$,
$$\sqrt{2}\left(F\left(\arcsin\left(\tanh\;x\right)|\frac12\right)-E\left(\arcsin\left(\tanh\;x\right)|\frac12\right)+\tan\left(\arcsin\left(\tanh\;x\right)\right)\sqrt{1-\frac12\sin^2\left(\arcsin\left(\tanh\;x\right)\right)}\right)$$
which simplifies to
$$\sqrt{2}\left(F\left(\arcsin\left(\tanh\;x\right)|\frac12\right)-E\left(\arcsin\left(\tanh\;x\right)|\frac12\right)\right)+\tanh\;x\sqrt{1+\cosh^2 x}$$
to which an arbitrary constant can be added.

As an alternative, one can start with the Mathematica result
$$\frac{\sqrt{2}}{i}E\left(ix|\frac12\right)$$
and simplify (get rid of the complex stuff) accordingly, using the second relation in formula 19.7.7 in the DLMF. Note that $\arctan\sinh\;x=\arcsin\tanh\;x$.
