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?

Question

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 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 $(0+\epsilon,0.99782-\epsilon)$ 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, And I bielive that integrand has a closed form because the integrand is of the trigonometric form

Answer 1

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.