Inverting a function given by an integral

Question

I'm not sure if it's even possible to do this in general, but I'd like to find a function $f^{-1}$ which is the inverse of $$f(t) = \int^t_0\frac{ds}{\sqrt{4s(1-s^2)}}$$

Answer 1

For $s\in (0,1)$, the integrand is well-defined and always positive, so $f$ is strictly increasing and hence injective. Moreover, the limit $L= \lim_{t\to 1}f(t)$ does exist, so we may set $f(0)=0$ and $f(1)=L$ to obtain that $f:[0,1]\longrightarrow [0,L]$ is a continuous bijection that is also differentiable on $(0,1)$, with derivative given by the integrand.

This guarantees the existence of $f^{-1}:[0,L]\longrightarrow[0,1]$, but in general it needs not have a 'nice' expression, if that's what you're looking for.

Using Abdallah's idea, one can describe $g=f^{-1}$ by the initial value problem:

\begin{equation} \left\{\begin{array}{} g(0)=0\\ g'(t)=2\sqrt{g(t)\left(1-{g(t)}^2\right)} \end{array}\right. \end{equation}

Answer 2

Hint

We have

$$f'(t)=\frac{ 1 }{ \sqrt{ 4t(1-t^2) }}$$

$$(f^{-1})'(t)=\frac{1}{ f'(f^{-1}(t))}$$

$$=\sqrt{4f^{-1}(t)(1-(f^{-1}(t))^2)}$$