Prove Kendall's tau of a bivariate normal is given by
$$\rho_\tau (X_1,X_2)=\frac{2}{\pi}\arcsin\rho$$
I can derive the bivariate normal as
$$F(x_1,x_2)=\frac{1}{2\pi\sqrt{1-\rho^2}}e^{-\frac{1}{2}x_1\Sigma^{-1}x_2}$$
where, for simplicity, I am assuming $X_1$ and $X_2$ have mean $0$. I also am aware that the formula for Kendall's tau involves an integral over some domain involving $X_1$, $X_2$, and $\rho$, multiplied by $4$.
So it seems like all the pieces are there; the integral should result in the answer, if only the integral over the exponential term w.r.t. $X_1$ and $X_2$ were $1$. But this is not the case, so I am stuck.