Area of Lit Part of a Half Sphere

A light beam from the point $P$ that just touches the hemisphere's rim at $(\cos\theta,\sin\theta,0)$ touches the bottom of the bowl at the point $\left(\frac{2(1-\cos\theta)}{3-2\cos\theta}\,,\,\frac{2\sin\theta}{3 - 2\cos\theta}\,,\,\frac{1 - 2\cos\theta}{3 - 2\cos\theta}\right)$ provided that $\cos\theta > \tfrac12$ (if $\cos\theta< \tfrac12$ , the light beam would have met the upper half of the sphere first, had it been there). These points are precisely the points of intersection of the hemisphere with the plane $2x - z = 1$ . Thus the region $R_1$ of the hemisphere that is lit on the inside is defined by the inequality $2x - z \le 1$ .

The region of the hemisphere that is lit from the outside is the region of the hemisphere that is inside the cone with vertex the point $P$ that is tangent to the sphere. Thus the region $R_2$ that is lit from outside is defined by the inequality $2x + z \ge 1$ . Thus the region of the hemisphere that is unlit is defined by the inequality $|z| \ge |2x-1|$ . Thus the area of the unshaded region is $\begin{aligned} A & = \; 2\iint_{x^2 + z^2 \le 1, z \le 0, |2x-1| \le |z|} \frac{dx\,dz}{\sqrt{1-x^2-z^2}} \; = \; 2\int_0^{\frac45}\left(\int_{-\sqrt{1-x^2}}^{-|2x-1|}\frac{dz}{\sqrt{1-x^2-z^2}}\right)\,dx \\ & = \; 2\int_0^{\frac45} \left(\tfrac12\pi - \sin^{-1}\left(\frac{|2x-1|}{\sqrt{1-x^2}}\right)\right)\,dx \; = \; 2\int_0^{\frac45}\sin^{-1}\left(\sqrt{\frac{x(4-5x)}{1-x^2}}\right)\,dx \end{aligned}$

I have not worked out how to do this one yet, but Mathematica tells me that $A \; = \; \tfrac{8}{\sqrt{5}} \tan^{-1}\left(\sqrt{\tfrac35}\right) - 2\pi\big(\tfrac{1}{\sqrt{5}} - \tfrac13\big)$ and this makes the lit proportion equal to $1 - \tfrac{A}{2\pi} = 0.738606$ , so the answer is $\boxed{73.8606}$ .