$\int_0^y e^{-\alpha\sqrt{x(1-x)}}~dx$
$=\int_0^y e^{-\alpha\sqrt{-(x^2-x)}}~dx$
$=\int_0^y e^{-\alpha\sqrt{-\left(x^2-x+\frac{1}{4}-\frac{1}{4}\right)}}~dx$
$=\int_0^y e^{-\alpha\sqrt{\frac{1}{4}-\left(x-\frac{1}{2}\right)^2}}~dx$
$=\int_{-\frac{1}{2}}^{y-\frac{1}{2}}e^{-\alpha\sqrt{\frac{1}{4}-x^2}}~dx$
$=\int_\pi^{\cos^{-1}(2y-1)}e^{-\alpha\sqrt{\frac{1}{4}-\left(\frac{\cos x}{2}\right)^2}}~d\left(\dfrac{\cos x}{2}\right)$
$=\dfrac{1}{2}\int_{\cos^{-1}(2y-1)}^\pi e^{-\frac{\alpha\sin x}{2}}\sin x~dx$
$=\int_{\cos^{-1}(2y-1)}^\pi\sum\limits_{n=0}^\infty\dfrac{\alpha^{2n}\sin^{2n+1}x}{2^{2n+1}(2n)!}dx-\int_{\cos^{-1}(2y-1)}^\pi\sum\limits_{n=0}^\infty\dfrac{\alpha^{2n+1}\sin^{2n+2}x}{4^{n+1}(2n+1)!}dx$
For $n$ is any non-negative integer,
$\int\sin^{2n+2}x~dx=\dfrac{(2n+2)!x}{4^{n+1}((n+1)!)^2}-\sum\limits_{k=0}^n\dfrac{(2n+2)!(k!)^2\sin^{2k+1}x\cos x}{4^{n-k+1}((n+1)!)^2(2k+1)!}+C$
This result can be done by successive integration by parts.
$\int\sin^{2n+1}x~dx$
$=-\int\sin^{2n}x~d(\cos x)$
$=-\int(1-\cos^2x)^n~d(\cos x)$
$=-\int\sum\limits_{k=0}^nC_k^n(-1)^k\cos^{2k}x~d(\cos x)$
$=\sum\limits_{k=0}^n\dfrac{(-1)^{k+1}n!\cos^{2k+1}x}{k!(n-k)!(2k+1)}+C$
$\therefore\int_{\cos^{-1}(2y-1)}^\pi\sum\limits_{n=0}^\infty\dfrac{\alpha^{2n}\sin^{2n+1}x}{2^{2n+1}(2n)!}dx-\int_{\cos^{-1}(2y-1)}^\pi\sum\limits_{n=0}^\infty\dfrac{\alpha^{2n+1}\sin^{2n+2}x}{4^{n+1}(2n+1)!}dx$
$=\left[\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^{k+1}n!\alpha^{2n}\cos^{2k+1}x}{2^{2n+1}(2n)!k!(n-k)!(2k+1)}\right]_{\cos^{-1}(2y-1)}^\pi-\left[\sum\limits_{n=0}^\infty\dfrac{\alpha^{2n+1}x}{2^{4n+3}n!(n+1)!}-\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(k!)^2\alpha^{2n+1}\sin^{2k+1}x\cos x}{2^{4n-2k+3}n!(n+1)!(2k+1)!}\right]_{\cos^{-1}(2y-1)}^\pi$
$=\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^kn!\alpha^{2n}((2y-1)^{2k+1}+1)}{2^{2n+1}(2n)!k!(n-k)!(2k+1)}+\sum\limits_{n=0}^\infty\dfrac{\alpha^{2n+1}(\cos^{-1}(2y-1)-\pi)}{2^{4n+3}n!(n+1)!}+\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(k!)^2\alpha^{2n+1}(2y-1)(1-(2y-1)^2)^{k+\frac{1}{2}}}{2^{4n-2k+3}n!(n+1)!(2k+1)!}$