To me, it looks like that $\arccos$ part is the source of all ugliness for this problem. I'm not sure using the substitution $x=R\sin\theta$ will simplify this $\arccos$ into something manageable. Therefore, I myself would try to eliminate the $\arccos$ through integration by parts.
$u=\arccos[z(R^2-x^2)^{-\frac12}],du=\frac{d[z(R^2-x^2)^{-\frac12}]}{\sqrt{1-z(R^2-x^2)^{-\frac12}}}=$
$\frac{-\frac12z(R^2-x^2)^{-\frac32}(-2x)dx}{\sqrt{1-z(R^2-x^2)^{-\frac12}}}$
Multiplying top and bottom by $(R^2-x^2)^\frac32$
$du=\frac{xzdx}{\sqrt{(R^2-x^2)^3-z(R^2-x^2)^\frac52}}$
$\int\arccos(\frac{z}{\sqrt{R^2-x^2}})dx=x\arccos(\frac{z}{\sqrt{R^2-x^2}})-\int\frac{zx^2dx}{\sqrt{(R^2-x^2)^3-z(R^2-x^2)^\frac52}}$
Ugly, I know, but at least we have some directions we can try. Let's try
$x=(R^2-u^2)^\frac12,dx=\frac{du}{2\sqrt{R^2-u^2}}$
$\int\frac{zx^2dx}{\sqrt{(R^2-x^2)^3-z(R^2-x^2)^\frac52}}=\int\frac{z(R^2-u^2)du}{2\sqrt{(R^2-u^2)(u^6-zu^5)}}=\frac z2\int\sqrt{\frac{R^2-u^2}{u^6-zu^5}}du$
Well, I gave it a valiant effort, but I appear to be stuck at this point. Maybe someone else can pick up where I left off (or more likely, someone will link to Wolfram making all of the above completely irrelevant...).