Find a probability integral For $0 <t<1$  express the integral
$$
\int_0^\pi \int_{\max\left\{-1;\cos x - \frac{t}{\sin x}\right\}}^{\cos x}  \frac{dy}{\sqrt{1-y^2}} \, dx
$$
as a function of $t$ (without any  integrals).
I am sorry, but actually I don't know  any attempt  to do it.
 A: This is no full answer, but too long for a comment, maybe somebody can use it.
Consider $f_t(x):=\cos x -\frac{t}{\sin x}$. 
From a discussion of $f_t(x)$ we see, that for any $0<t<1$ it has a unique maximum greater $-1$ in $[0,\pi]$ and from there towards both interval boundaries $x\to 0,\pi$ it goes to $f_t(x)\to-\infty$ monotonously. Hence $\forall t$ there are two points $0<x_-(t)<x_+(t)<\pi$ with $f_t(x_\pm (t))=-1$ where the integration can be decomposed like
$$\int_0 ^\pi dx \int_{\min(-1,f_t(x))} ^{\cos x} dy (\cdot)=
  \left(\int_0 ^{x_-(t)}dx+\int_{x_+(t)}^\pi dx\right)\int_{-1} ^{\cos x}dy(\cdot) +  
\int_{x_-(t)} ^{x_+(t)}dx\int_{f_t(x)} ^{\cos x}dy(\cdot)$$
I will derive $x_\pm (t)$, the only missing piece of your puzzle is then to find $\int dx \arcsin(f_t(x))$, which I was not able to do.
We want to solve 
$-1=f_t(x):=\cos x -\frac{t}{\sin x}$. 
Multiply by $\sin x$, substitute $c=\cos x$, use $\sin x=\sqrt{1-c^2}$ and finally square it. You end up with
$$(1-c^2)(1+c)^2=t^2$$
Now comes the ugly work of solving this fourth order equation giving four solutions. Two of them are purely real, which read:
$$c_\pm(t)=\pm\frac{A(t)}{2}+\frac{B(t)}{2\sqrt{3}}-\frac{1}{2}$$ where
$$A(t)=\sqrt{\frac{A_1(t)-z(t)-4t^2}{3z(t)}+2}$$
$$A_1(t)=\frac{2\sqrt{3}}{B(t)}$$
$$B(t)=\sqrt{\frac{3z(t)^2+3z(t)+4t^2}{z(t)}}$$
$$z(t)=\left(2t^2+\frac{2t^2\sqrt{27-16t^2}}{3^{3/2}}\right)^{1/3}$$
The values of $x_\pm(t)$ are then given by
$$x_\pm (t)=\arccos(c_\pm(t))$$
A: Sorry, I'm not allowed to comment, haven't found a PM here either.
This guy claims that he has found a beautiful answer to your problem after working on it for two months.
Scroll down here (you can also see the Google translated version).
If everything else fails you can try contacting him.
