In Griffith's Introduction to Electrodynamics 4th edition example 5.11 the solution for the vector potential of a uniformly charged spinning spherical shell is given.
Now let's assume that the surface charge density is given by $\sigma(\theta)=\sigma_0sin(\theta)$ and the sphere is rotating with a constant angular velocity $\omega$ about the $\hat{z}$ axis. So, unlike the figure above we have to put the axis of rotation on $\hat{z}$. The velocity vector will be:
$\vec{v}=\vec{\omega}\times\vec{r'}=\begin{vmatrix} \hat{x} & \hat{y} & \hat{z}\\ 0 & 0 & \omega\\ Rsin\theta'cos\phi' & Rsin\theta'sin\phi' & Rcos\theta' \end{vmatrix}$
After expanding we get
$\vec{v}=-R\omega sin\theta' sin\phi'\hat{x}+R\omega sin\theta' cos\phi'\hat{y}$
The surface current becomes
$\vec{K}=-R\omega\sigma (sin\theta')^2 sin\phi'\hat{x}+R\omega\sigma (sin\theta')^2 cos\phi'\hat{y}$
We can write it as $\vec{K}=-R\omega\sigma sin\theta' \hat{\phi}$
The vector potential can be written as
$\hat{A}=\frac{-\mu_0R^3\omega\sigma}{4\pi}\hat{\phi}\int_0^{\pi}\int_0^{2\pi}\frac{sin^2\theta'}{\sqrt{R^2+r^2-2Rrcos\alpha}}d\phi'd\theta'$
in which $\alpha$ is taken to be the angle between $\vec{r}$ and $\vec{r'}$ and we can express $cos\alpha$ using the polar and azimuthal angles in the problem as
$cos\alpha=cos\theta cos\theta' + sin\theta sin\theta' cos(\phi-\phi')$
Although I think the formulation of the solution is correct the integral for the vector potential is very difficult for me to solve. I also have tried to integrate using MATHEMATICA but gave up after half an hour since it didn't give me an answer. Is there a better way to solve this problem?