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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.

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    It's not clear exactly what you're trying to compute, but it's likely there's a much easier method than this.2012-10-01

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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...).

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    Err... I think I'm off by a minus sign on the arccos derivative. Something to keep in mind if anyone can continue from here.2012-10-02