TITLE: An elementary way of simplifying a trigonometric triple integral?
QUESTION [7 upvotes]: By stressing my manipulative powers and a bit of help from Mathematica I was able to show that the triple integral
$$\frac1{8\pi}\int_0^{2\pi}\int_0^\pi\int_0^\pi\,(2\cos\theta\cos\varphi-\sin\theta\sin\varphi\cos\psi)^{2k}\sin\theta\sin\varphi \,\mathrm{d}\theta\mathrm{d}\varphi\mathrm{d}\psi $$
where k is a positive integer is equivalent to
$${\left(\frac{2^k}{2k+1}\right)^2}\sum_{j=0}^k\frac{\binom{2j}{j}}{\binom{2k}{k}}$$
by making use of the binomial expansion and then using a hypergeometric identity that degenerates to a ratio of gamma functions.
For some reason I think there was no need for Mathematica to invoke the Gauss hypergeometric function ${}_2 F_1\left(a,b;c;z\right)$ in performing the triple integration, so I'm wondering if there might be a way to show that this identity holds, using nothing more complicated than a gamma function.

REPLY [3 votes]: Here is a hint to an other approach: Search for the key words "convolution formula" "Legendre polynomial" "addition formula" and you will be en-lighted... 
--- A little explanation and some spherical harmonics ---
@J. M. You are totally right in particular when I was wrong in my suggestion! :/
What I thought about was convolution on the double coset space $X=K\backslash G/K$ of $G= SO(3)$ where $K=SO(2)$. The space $X$ may be identified with $[-1,1]$ and the convolution (as induced from the translation operator on $G$) is then given by
$$f*g(x)= \frac{1}{4\pi}\int_{-1}^1\int_{0}^{2\pi}f(y)g(xy+\sqrt{1-x^2}\sqrt{1-y^2}\cos\theta) d\theta dy.$$
An interesting fact is that $f*g=g*f$ even though $G$ is not Abelian.
Choosing $f(x)=1$ and $g(x)=x^{2k}$ we get
$$\frac{1}{8\pi}\int_{-1}^1\int_{-1}^1\int_{0}^{2\pi}(xy+\sqrt{1-x^2}\sqrt{1-y^2}\cos\theta)^{2k} d\theta dydx=$$
$$\int_{-1}^1f*g(x)dx/2 = \int_{-1}^1 g*f(x)dx/2=\int_{-1}^1x^{2k}dx/2= 1/(2k+1).$$
In our case the integrand is (after changing variables) 
$$(2xy + \sqrt{1-x²}\sqrt{1-y²}\cos\theta)^{2k}=$$
$$=\sum_{j=0}^{2k}{2k \choose j}(xy)^j(xy+\sqrt{1-x²}\sqrt(1-y²)\cos\theta)^{2k-j}$$
Using this we get an other convolution - but I can not see how this can be used directly. :(
Legendre polynomials might be good since they are the characters on the space $X$.