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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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Try substituting $x = R \cos \theta$ (or $\sin$). – marty cohen Oct 1 '12 at 16:16
It's not clear exactly what you're trying to compute, but it's likely there's a much easier method than this. – Hurkyl Oct 1 '12 at 20:54

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.



Multiplying top and bottom by $(R^2-x^2)^\frac32$



Ugly, I know, but at least we have some directions we can try. Let's try


$$\int\frac{zx^2dx}{\sqrt{(R^2-x^2)^3-z(R^2-x^2)^\frac52}}=\int\frac{z(R^2-u^2)du}{2\sqrt{(R^2-u^2)(u^6-zu^5)}}=\frac z2\int\sqrt{\frac{R^2-u^2}{u^6-zu^5}}du$$

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. – Mike Oct 2 '12 at 0:18

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