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In his Theoretical Physics, Joos condescends the following expression as "the familiar arc-cosine form":

$-\int \frac{1}{\sqrt{a+2 bx-hx^2}} \, dx=\frac{1}{\sqrt{h}}\arccos \left[\frac{b-hx}{\sqrt{a+b^2h}}\right]$

The only reason it is now "familiar" to me is that I have been staring at it for weeks, wondering how to derive it. I can prove it to be valid by substituting the terms into the form I do know how to derive.

$d\arccos (u)=-\frac{du}{\sqrt{1-u^2}}$

But that doesn't tell me how it was originally derived. I strongly suspect there is some geometric development which would illuminate the meaning of the variables and terms in the form Joos provides.

Any suggestions?

Steven Hatton
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    Start by completing the square in the integrand, then make a substitution to simplify. – DaveNine Mar 24 '16 at 23:32
  • Thanks. That gave me AN answer. I contacted my professor of introductory physics (which I never completed) some 33 years ago, and asked him about his geometric approach to a similar problem. He doesn't deny having presented it, but claims to not remember how he did it. He said he will get back with me on this. If I gain any insight, I will share it. – Steven Hatton Mar 25 '16 at 18:54

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$$ \begin{align} -\int\frac{\mathrm{d}x}{\sqrt{a+2bx-hx^2}} &=-\frac1{\sqrt{h}}\int\frac{\mathrm{d}x}{\sqrt{\frac{ah+b^2}{h^2}-\left(x-\frac bh\right)^2}}\tag{1}\\ &=-\frac1{\sqrt{h}}\int\frac{\mathrm{d}\frac{hx-b}{\sqrt{ah+b^2}}}{\sqrt{1-\left(\frac{hx-b}{\sqrt{ah+b^2}}\right)^2}}\tag{2}\\ &=\frac1{\sqrt{h}}\cos^{-1}\left(\frac{hx-b}{\sqrt{ah+b^2}}\right)+C\tag{3} \end{align} $$ Explanation:
$(1)$: complete the square and bring the $\frac1{\sqrt{h}}$ out front
$(2)$: move factors around to make it look like $\int\frac{\mathrm{d}u}{\sqrt{1-u^2}}$
$(3)$: standard integral

robjohn
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  • Yes, thank you. I had followed the previous advice and arrived at a much more long-winded version of the same. It's not the answer I was looking for, but is far more satisfying than not having an answer. – Steven Hatton Mar 25 '16 at 18:49
  • I am not sure what answer you were looking for. If you supply some more information, perhaps someone can help. – robjohn Mar 25 '16 at 22:03
  • I'm not as confident of this as I was when I first asked my question; but, I believe the variables in Joos's "familiar form" can be related to a geometric diagram which includes an ellipse. Or perhaps a circle. Something along these lines: http://math.caltech.edu/~nets/lecture19.pdf only with pictures. Perhaps with some kind of coordinate transformation. – Steven Hatton Mar 29 '16 at 17:56