This is probably an elementary question but I still cannot convince myself or find a decent argument to believe.
Could anyone please explain to me why do we believe that the arcsine distribution (or chebyshev measure) $$d\mu_{[-1,1]}=\frac{dx}{\pi\sqrt{1-x^2}}$$ is the equilibrium measure on the interval $[-1,1]$ ? in the sense of potential theory, i.e., it minimizes the energy integral if we define the energy integral of a measure with support in a compact set $K\subseteq \mathbb{C}$, $$I(\mu)=-\iint \log|z-w|d\mu(z)d\mu(w).$$ That is, $$I(\mu_{[-1,1]})=\inf_{\substack{\mu\\ \mathrm{supp}(\mu)\subseteq K}}I(\mu)$$
One way to see this is to recall that if $K\subseteq{\mathbb{C}}$ is compact, then the equilibrium measure on $K$ is precisely harmonic measure from $\infty$ on $K$. (For details, see chapter III of "Harmonic Measure" by Garnett and Marshall.)
By symmetry, harmonic measure from infinity on the boundary of the unit disk $\overline{\mathbb{D}}\subseteq{\mathbb{C}}$ is just the uniform measure on the boundary, $\frac{d\theta}{2\pi}$. Harmonic measure is conformally invariant. Thus, if $\phi:\mathbb{C}\setminus{\mathbb{D}}\rightarrow{\mathbb{C}\setminus{K}}$ is conformal with $\phi(\infty)=\infty$ then the equilibrium measure on $K$ is just the pushforward under $\phi$ of the uniform measure $\frac{d\theta}{2\pi}$ on $\partial{\mathbb{D}}$.
Consider the map $z\mapsto{\frac{z+z^{-1}}{2}}$. This is a conformal map that sends $\mathbb{C}\setminus{\overline{\mathbb{D}}}$ to $\mathbb{C}\setminus{[-1,1]}$ and $\infty$ to $\infty$. Thus, the equilibrium measure on $[-1,1]$ is just the pushforward of the uniform measure on $\partial{\mathbb{D}}$ under this map. If you sit down and do the computation, you'll see that this is precisely $\frac{dx}{\pi\sqrt{1-x^{2}}}.$