# Integral Transform with Hyperbolic Functions

I am at it with understanding the nitty-gritty of the integral transform suggested in a previous question of mine: Length of a Parabolic Curve

To solve this integral, you can use the substitution $\sqrt{1+4x^2}=\cosh u$, so $x=\frac12\sinh u$ and $\mathrm dx=\frac12\cosh u\,\mathrm du$, to get

$$L=\frac12\int_0^{\operatorname{arcosh}\sqrt5}\cosh^2 u\,\mathrm du=\frac14\left[x+\sinh x\cosh x\right]_0^{\operatorname{arcosh}\sqrt5}=\frac14\left(\arccos\sqrt5+2\sqrt5\right)\approx1.479$$

Since $\cosh^2u-\sinh^2u=1$ by definition, $\cosh u = \sqrt{1+\sinh^2u}$. If we posit conveniently $x=\frac12\sinh u$, $\cosh u = \sqrt{1+4x^2}$. $x$ is calculus-ed to get the derivative $\mathrm dx = \frac12 \cosh u \mathrm du$.

The integrand is now: $$\cosh u (\frac12\cosh u \mathrm du )$$ This is almost like the first expression for $L$. But I don't understand where $\operatorname{arcosh}{\sqrt5}$ comes from. The integral range must be dependent on the domain $0\leq x \leq 1$, then again, I don't know what the range is. Is it $u$ or $\mathrm du$ or somethinge else?

Another thing I need help with is the transition from the first to the second expression. Wolfram|Alpha says the integral of $\cosh^2(x)$ is: $$\frac12(x + \sinh(x)\cosh(x)) +C$$ But can we substitute back for $U=x$ in the second expression?

I understand the transition from the second to the third expression and to the numerical value.

Edit: I've just understood that if, $x=\frac12\sinh u$, then $u = \operatorname{sinh^{-1}}(2x)$.
For the domain $0\leq x \leq 1$,
$0 \leq u \leq 1.4436\dots$.
$1.4436\dots$ happens to be the value of $\operatorname{arcosh}\sqrt5$. Is there a proof by identity for this?
 Well, since $\mathrm{arsinh}(x)=\mathrm{arcosh}(\sqrt{1+x^2})$ (you'll have to remember a basic hyperbolic function identity to justify this), you do have $\mathrm{arsinh}(2)=\mathrm{arcosh}\sqrt 5$. – J. M. Nov 7 '12 at 3:15