let $\epsilon >0$, I tried to evaluate $\int\_{0+\epsilon}^{1-\epsilon}\left(\sqrt{1-x^2}^{\sqrt{1-x^2}^{\cdots}}\right) dx$ , using the fact $x= \cos t$ yield to have integrand using $\sin $ function seems is not easy to get such closed form by this variable change , For one iteration by means $\int\_{0}^{1}\left(\sqrt{1-x^2}^{\sqrt{1-x^2}}\right) dx$ we have the integrand [converge approximately](https://www.wolframalpha.com/input/?i=integral%28%28+sqrt%281-x%5E2%29%29%5E%28%28sqrt%281-x%5E2%29%29%29%29+%2C+x%3D0+to+1) to $\frac{\sqrt{3}}{{2}}$, For some odd iterations we have $l=0.89..$ and for even iterations we have $l=0.9..$ , Now if we fixe $\epsilon$ at at some small value for example $\epsilon=0.05$ such that $x$ [lie at a least between](https://www.wolframalpha.com/input/?i=e%5E%28-e%29%3Csqrt%281-x%5E2%29%3Ce%5E%281%2Fe%29) $(0+\epsilon,0.99782-\epsilon)$ to get convergence , My question here is : Is it possible to express the titled integral in elementary functions ?


**Note**:The Copy of this question is posted yesterday [here in MSE](https://math.stackexchange.com/q/3757128/156150), And I bielive that integrand has a closed form because the integrand is of the trigonometric form



Is it possible to express $\int_{0+\epsilon}^{1-\epsilon}\left(\sqrt{1-x^2}^{\sqrt{1-x^2}^{\cdots}}\right) dx$ in elementary functions?
I think this is
$$
-\int\_{0+\epsilon}^{1-\epsilon}
{\frac {2\;{\rm W} \left(-\frac12\,\ln \left( 1-{x
}^{2} \right) \right)}{\ln \left( 1-{x}^{2} \right) }}\,{\rm d}x
$$
not elementary.


