Integral calculus What is the integral 
$$\int \arccos(z/\sqrt{R^2-x^2})dx$$
This is one of 4 equations for integrating the area of a major sector of a circle within a sphere between limits to find its volume. Two of the functions are easily integrated (first and last), but the above and $\int\arccos(x^2(z/\sqrt{R^2-x^2}))dx$ are difficult to do. I also need the integration of this equation too.
The full equation to be integrated is: 
$$
Area = \pi(R^2-x^2)-(R^2-x^2)\arccos(z/\sqrt{R^2-x^2}+z*\sqrt{R^2-x^2-z^2}.
$$
The function $z$ is the distance to the major segment chord from the center.
 A: 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...).
