How to express the function $a(t)$ knowing a parametrization $a(eta)$ and $t(eta)$?

I have this function :
begin{equation}tag{1}
a(eta) = sqrt{sin{2 eta}},
end{equation}
and this time variable :
begin{equation}tag{2}
t(eta) = int_0^eta a(eta’) , deta’.
end{equation}
This integral is very difficult to express in an analytical way, because of the square-root.

I would like to know the function $a$ parametrized as a power series of $t$. How can I achieve this ?

When $eta$ is very small, I could get
begin{equation}tag{3}
a( , eta(t) , ) approx (3 , t)^{frac{1}{3}}.
end{equation}
When $eta$ isn’t so small, I’m expecting something like this (but I could be wrong) :
begin{equation}tag{4}
a(, eta(t) ,) = (3 , t)^{frac{1}{3}} f(t),
end{equation}
where $f(t)$ could be Taylor expanded (?). How to find this function using Mathematica ?

EDIT : To clarify a few things : I don’t know the function $f(t)$ defined above. This is what I’m looking for, as a Taylor series of $t$.

I can integrate the function (1) to get $t(eta)$ using Mathematica, as a power expansion :

FullSimplify[
    Series[
        Integrate[Sqrt[Abs[Sin[2 x]]], {x, 0, eta}, Assumptions -> 0 < eta < Pi/2],
    {eta, 0, 6}]
]

I then get this :
begin{equation}tag{5}
t(eta) approx sqrt{2} ; eta^{frac{3}{2}} big( frac{2}{3} – frac{2}{21} ; eta^2 + frac{1}{495} ; eta^4 – frac{1}{2835} ; eta^6 big).
end{equation}
Then I need to invert this, to get $a(t) equiv a( , eta(t) , )$, as a power expansion in $t$ (or maybe $t^{1/3}$ ?). It should be pretty basic. As I said above, I’m expecting something like
begin{equation}tag{6}
a( , eta(t) , ) = (3 , t)^{frac{1}{3}} f(t),
end{equation}
with $f(t)$ an unknown Taylor series.