let $a<b\in \overline {\mathbb R}$ such that $\lim_{x\to a}f(x)=0$, Let $f:(a,b) \to \mathbb R$ be continuous and positive on $(a,b)$ $$L\{f(x)\}^{-1}=\int^x_a \frac{dx}{f(x)}$$ Where $f(x)^{-1}$ is the inverse of $f(x)$.
$\int^x_a \frac{dx}{f(x)}$ is defined on $[a,b]$ and $L\{f(x)\}$ is defined on $\Big[0,\int^b_a \frac{dx}{f(x)}\Big]$.
This construction, applied on the equation of a circle, i.e. to $f(x)=\sqrt{1-x^2}$ gives the sine function, as shown below $$ L\Big\{\sqrt{1-x^2}\Big\}=\left(\int^x_{-1} \frac{dx}{\sqrt{1-x^2}}\right)^{-1}=\sin(x), $$ For $f(x)=\sqrt{(1-k^2x^2)(1-x^2)}$, it gives the elliptic function $$ L\Big\{\sqrt{1-k^2x^2}\Big\}=\left(\int^x_{\frac{1}{k}}\frac{dx}{\sqrt{(1-k^2x^2)(1-x^2)}}\right)^{-1}$$ Applied to the equation of the hyperbola it gives hyperbolic functions $$ L\Big\{\sqrt{1+x^2}\Big\}=\left(\int^x_{0}\frac{dx}{\sqrt{1+x^2}}\right)^{-1} $$ Question: does this construction has a geometric representation?
I am asking this since, due to the known property $$ \frac{dr}{dx}= \frac{1}{\dfrac{dx}{dr}}\iff f'(x)= \frac{1}{\big(f^{-1}(x)\big)'}, $$ if $f(x)$ is integrable, calling $F(x)$ an antiderivative of $f(x)$, this implies $$ L\big\{f(x)\big\} = F(p)\big|^{p=f^{-1}(x)}_{p=f^{-1}(0)} $$