= Solution
Normalize <Haar measure> on $\mathrm{SU}(2)$ to total mass one. For a <continuous> class function $F$, the <Weyl integration formula for SU2> is
$$
\boxed{\int_{\mathrm{SU}(2)}F(g)\,dg
=\frac2\pi\int_0^\pi
F(\operatorname{diag}(e^{i\theta},e^{-i\theta}))\sin^2\theta\,d\theta}.
$$
Equivalently it is $\pi^{-1}\int_0^{2\pi}F(t_\theta)\sin^2\theta\,d\theta$. For arbitrary <continuous> $F$, replace $F(t_\theta)$ on the right by its conjugacy average $\int_{\mathrm{SU}(2)}F(ht_\theta h^{-1})\,dh$.
Here is a direct proof. Identify $\mathrm{SU}(2)$ with unit quaternions, the sphere $S^3$. Multiplication by a unit quaternion is a transformation given by an <orthogonal matrix> of $\mathbb R^4$, so normalized round volume is bi-invariant and is <Haar measure>. Write a quaternion as $\cos\theta+\mathbf n\sin\theta$, $0\le\theta\le\pi$, $\mathbf n\in S^2$. The round volume element is $\sin^2\theta\,d\theta\,dA(\mathbf n)$, and the total volume is $2\pi^2$. Conjugation rotates the unit imaginary <vector> $\mathbf n$, so a class function depends only on $\theta$. Integrating over the sphere of area $4\pi$ gives the factor $4\pi/(2\pi^2)=2/\pi$. The endpoint degeneracies have measure zero. Conjugacy averaging and Haar invariance prove the more general version.
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