Avg Area of triangle 
Consider $3$ points in the Complex plane namely, $i,e^{i\alpha},e^{i\beta}$. Find the average area of the triangle formed as $\alpha,\beta$ vary over $[0,2\pi]$. They are uniformely distributed

Using determinant method, I calculated the are to be $$\triangle=\frac12|\sin(\alpha-\beta)+cos(\beta)-cos(\alpha)|$$
And to find average Area, I integrated it with $\alpha\in[0,\beta]$ then with $\beta\in[0,2\pi]$
i.e. Average Area = $$\left|\frac1{2\times2\pi}\int_0^{2\pi}\left(\frac1\beta\int_0^{\beta}\left(\sin(\alpha-\beta)+cos(\beta)-cos(\alpha)\right) d\alpha \right) d\beta\right|=0.306$$
But the answer is $0.477$. Is this a conceptual or calculational mistake? I used Mathematica
Basically, I'm looking for a way to remove mod from $$\frac1{8\pi^2}\int_0^{2\pi}\int_0^{2\pi}\left|\sin(\alpha-\beta)+cos(\beta)-cos(\alpha)\right| d\alpha\,\,\, d\beta$$

You have to look at the integrand piecewise: the only relationship between $\int \lvert f \rvert$ and $\lvert \int f \rvert$ is that the former is at least as big as the latter (by the triangle inequality). I also don't know where you've got that $1/\beta$ from: the density function should just be $1/(4\pi^2)$.
In fact, by rotational invariance due to the distributions being uniform, it suffices to consider $1,e^{i\alpha},e^{i\beta}$ as the vertices: this simply makes the area
$$ A(\alpha,\beta) = \frac{1}{2} \lvert \sin{(\alpha-\beta)} - \sin{\beta} + \sin{\alpha} \rvert. $$
The latter two sines can be written as $\cos{\frac{1}{2}(\alpha+\beta)}\sin{\frac{1}{2}(\alpha-\beta)}$, and combining this with the double-angle formula and the corresponding formula for the cosine eventually gives
$$ A(\alpha,\beta) = 2\lvert \sin{\tfrac{1}{2}\alpha}\sin{\tfrac{1}{2}\beta}\sin{\tfrac{1}{2}(\beta-\alpha)} \rvert. $$
This tells you that the contents of the absolute value are positive for $0<\beta-\alpha < 2\pi $, and again the symmetry gives
$$ E[A] = \frac{1}{4\pi^2} 2\int_0^{2\pi} \int_0^{\beta} A(\alpha,\beta) \, d\alpha \, d\beta = \frac{1}{4\pi^2} \int_0^{2\pi} \int_0^{\beta} \left( \sin{(\alpha-\beta)} - \sin{\beta} + \sin{\alpha} \right)d\alpha \, d\beta, $$
which gives $3/(2\pi) \approx 0.477 $.
