= Solution
The triangular support can be written $0\le x\le\pi/2$, $\pi/2-x\le y\le\pi/2$. The density is nonnegative there, and the <normalization of a probability distribution> requires
$$
1=A\int_0^{\pi/2}\sin x(1-\cos x)\,dx=\frac A2.
$$
Hence $\boxed{A=2}$. Multiplying the <joint probability density> by the required function cancels $\sin x$ on the interior. Its <expectation> is
$$
\boxed{\mathbb E\left[\frac1{\sin X}\right]=2\int_0^{\pi/2}(1-\cos x)\,dx=\pi-2.}
$$
The singularity at $x=0$ lies at a support vertex of zero probability; the displayed <integral> establishes integrability. Assigning any value to the random function at that single vertex does not change the <expectation>.
Back to article page
