A Decomposition into irreps of compact Lie group

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When decomposing a representation ##\rho## of a finite group ##G## into irreducible representations, we can find the number of times the representation contains a particular irrep ##\rho_0## through the character inner product
$$
\langle \chi, \chi_0\rangle = \frac{1}{|G|} \sum_{g\in G} \chi(g) \chi_0(g)^*$$
where ##\chi## and ##\chi_0## are the characters of ##\rho## and ##\rho_0##, respectively. Since all group elements in the same conjugacy class have the same characters, this may be rewritten as
$$
\langle \chi, \chi_0\rangle = \frac{1}{|G|} \sum_{C \in [G]} n(C) \chi(C) \chi_0(C)^*,
$$
where ##[G]## is the set of conjugacy classes of ##G## and ##n(C)## is the number of elements in ##C##.

This should generalise to compact Lie groups with the character inner product given by
$$
\langle \chi, \chi_0\rangle = \frac{1}{V_G} \int_{g\in G} \chi(g) \chi_0(g)^* d\mu_G
$$
where ##V_G = \int_{g\in G}d\mu_G## and ##d\mu_G## is the Haar measure on ##G##. The corresponding expression on the conjugacy class level should be
$$
\langle \chi, \chi_0\rangle = \frac{1}{V_G} \int_{C\in [G]} \chi(C) \chi_0(C)^* d\mu_{[G]}
$$
with ##d\mu_{[G]}## being a measure on the conjugacy classes determined by integrating the Haar measure over each conjugacy class. Right?

My particular problem is doing this for ##SO(3)##. More explicitly, the conjugacy classes of ##SO(3)## should be parametrised by the rotation angle ##\theta## (with the conjugate elements being rotations around different axis by the same ##\theta##). I am especially looking for the function ##f(\theta)## such that
$$
\langle \chi, \chi_0\rangle = \frac{1}{V_G} \int_{0}^\pi \chi(\theta) \chi_0(\theta)^* f(\theta) d\theta.$$

Any insights?
 
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fresh_42 said:
I have found the formula
$$
\int_{\operatorname{SU}(2)}f(\theta)\,d\theta = \dfrac{2}{\pi}\int_0^\pi (f\circ e)(t)\sin^2(t)\,dt
$$
including a proof, but in German. Maybe it helps anyway.

Source: https://wwwold.mathematik.tu-dortmund.de/~lschwach/SS10/Bachelor-Seminar/Averim.pdf
I think I found my issue. I was trying ##\sin^2(\theta)##, but when mapping from SU(2) to SO(3), the ##t## here is actually ##\theta/2##?

Using ##\sin^2(\theta)## I obtained that the fundamental representation should contain the trivial one once, which would be absurd. Using ##\sin^2(\theta/2)## gives zero as expected.

It also gives the correct result (2) for the representation ##\operatorname{Sym}^2(\operatorname{Sym}^2 V)## (with ##V## being the fundamental representation), which is what I was really looking for.
 
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