Does $\int_{-\infty }^{\ln \pi} \sqrt{\sin e^t} \, \rm{d}t$ have closed form? Background
I am studying the area of the implicit function curve
$$y^2=\sin \left(e^x\right).$$
I found that I need to calculate this integral:
$$I=\int_{-\infty }^{\ln \pi} \sqrt{\sin e^t} \, \rm{d}t.$$
Try 1
let $t = \sin e^x$, then
$$\int \frac{\sqrt{t}}{\sqrt{1-t^2} }\frac{1}{\arcsin t}\,\rm{d}t$$
I can't figure out how to calculate this.
Try 2
let $t = e^x$, then
$$\int_{0}^{\pi} \frac{\sqrt{\sin (t)}}{t}\,\rm{d}t$$
This form looks simple, but I still don't know how to calculate.
Question
Does $I$ have closed form?
 A: Another not-an-answer regarding the computational matter. As commented, $$I=\int_0^\pi\frac{\sqrt{\sin x}}{x}\,dx=\int_0^\pi\frac{\sqrt{\sin x}}{\pi-x}\,dx
=\frac\pi2\int_0^\pi\frac{\sqrt{\sin x}}{x(\pi-x)}\,dx$$ which becomes, after the substitution $x=(1+t)\pi/2$, $$I=\int_{-1}^1\frac{f(t)\,dt}{\sqrt{1-t^2}},\qquad f(t)=\sqrt{\frac1{1-t^2}\cos\frac{\pi t}2}.$$ This choice of $f(t)$ "eats out" the singularities at $t=\pm1$, so that $$f(t)=\sum_{n=0}^\infty f_n\,t^{2n}\implies I=\pi\sum_{n=0}^\infty\frac{(2n-1)!!}{(2n)!!}f_n,$$ and the radius of convergence of the power series equals $3$, so that $f_n\sim 9^{-n}$ as $n\to\infty$.
Thus $I_N:=\pi\sum_{n=0}^N(\ldots)$ are approximations of the given integral:
\begin{align*}
I_0&=\pi,\\
I_1&=\frac{5\pi}{4}-\frac{\pi^3}{32},\\
I_2&=\frac{89\pi}{64}-\frac{11\pi^3}{256}-\frac{\pi^5}{4096},\\
I_3&=\frac{381\pi}{256}-\frac{103\pi^3}{2048}-\frac{17\pi^5}{49152}-\frac{19\pi^7}{1179648},
\end{align*}
and $I_{35}=2.96168593796665631486155938224615651\ldots$ (with these digits correct).
Still, this rate of convergence is much slower than what double-exponential methods give.
