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The following is identity no. 4.624 in Gradshteyn & Ryzhik's Table of Integrals, Series and Products:

[tex]

\begin{equation}

\int_0^{\pi}\int_0^{2\pi}f\left(\alpha\cos\theta+\beta\sin\theta\cos\phi+\gamma\sin\theta\sin\phi\right)\sin\theta d\theta d\phi=2\pi\int_0^{\pi}f\left(R\cos p\right)\sin p\, dp = 2\pi\int_{-1}^{1}f\left(Rt\right)dt\nonumber

\end{equation}

where

\begin{equation}

R = \sqrt{\alpha^2+\beta^2+\gamma^2}.\nonumber

\end{equation}

[/tex]

I would like to know how this can be proved, in particular, how to get from the first to the second integral. I know that a substitution of [tex]t=\cos p[/tex] will get you from the second to third integral.

Thanks.