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Integration involving ArcCos

  1. Jan 19, 2010 #1
    I was trying to derive the formula for the moment of inertia of a rectangular prism at its centre of mass. And ended up with this:


    Due to the limitation of my knowledge in Calculus, I was stuck here. Can anyone help me solve these two integral:

    [tex]\int arccos(\frac{a^{2}}{2r^{2}}-1)rdr[/tex]

    [tex]\int (\frac{\pi}{2}-arccos\frac{b/2}{r}-arccos\frac{a/2}{r})rdr[/tex]
  2. jcsd
  3. Jan 19, 2010 #2
  4. Jan 20, 2010 #3
    Still have no idea. First,by "reduce", do you mean to expand the first integral ? Second, I have not learnt partial integration.
  5. Jan 20, 2010 #4
    By reduce I mean simplify. Do it by trig identities (wikipedia has a great list of these):

    [tex]\arccos\alpha \pm \arccos\beta = \arccos(\alpha\beta \mp \sqrt{(1-\alpha^2)(1-\beta^2)})[/tex]

    [tex] \arcsin(x)+\arccos(x)=\pi/2[/tex]

    For taking the integral of these trig functions, here is the trick: first learn the derivatives, then use integration by parts

    [tex]\int u\, \frac{dv}{dx}\; dx=uv-\int v\, \frac{du}{dx} \; dx\![/tex]

    Notice that if you know the derivative of u, [tex]\frac{du}{dx}[/tex], you can evalute the integral of u wrt x, [tex]\int u \; dx[/tex], using the derivative instead.
    Last edited: Jan 20, 2010
  6. Jan 21, 2010 #5
    It seems impossible now. Since we know the derivative of arccos but not the integral of it, eg. for the first integral: [tex]
    \int arccos(\frac{a^{2}}{2r^{2}}-1)rdr


    then,from [tex]
    \int u\, \frac{dv}{dx}\; dx=uv-\int v\, \frac{du}{dx} \; dx\!
    and [tex]\frac{d}{dx}arccos(u)=\frac{-1}{\sqrt{1-u^{2}}}\frac{du}{dx}[/tex]

    we have
    [tex]\int arccos(\frac{a^{2}}{2r^{2}}-1)rdr=arccos(\frac{a^{2}}{2r^{2}}-1)\frac{r^{2}}{2}+\int \frac{1}{\sqrt{1-(\frac{a^{2}}{2r^{2}}-1)^{2}}}[\frac{d}{dr}(\frac{a^{2}}{2r^{2}}-1)]\frac{r^{2}}{2}dr[/tex]

    when I simplified it:
    [tex]\int arccos(\frac{a^{2}}{2r^{2}}-1)rdr=arccos(\frac{a^{2}}{2r^{2}}-1)\frac{r^{2}}{2}-[arcsin\frac{a}{2r}](ar)-a\int arcsin\frac{a}{2r}dr[/tex]

    Now I have to integrate ArcSin. Isn't this impossible? I am close to dying. I kind of started on the second integral, and it seem like it would get twice as long as this.
  7. Jan 21, 2010 #6
    Does anyone know how to derive the formula for the moment of inertia of a rectangular prism?
  8. Jan 21, 2010 #7

    Char. Limit

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    Gold Member

    Wait, where did that arcsin come in? It looks like you had everything algebraic before...
  9. Jan 21, 2010 #8
    Oopsss, I misused one formula. Now I corrected it. From this one:

    \int arccos(\frac{a^{2}}{2r^{2}}-1)rdr=arccos(\frac{a^{2}}{2r^{2}}-1)\frac{r^{2}}{2}+\int \frac{1}{\sqrt{1-(\frac{a^{2}}{2r^{2}}-1)^{2}}}[\frac{d}{dr}(\frac{a^{2}}{2r^{2}}-1)]\frac{r^{2}}{2}dr

    By using this formula:

    [tex]\int\frac{1}{\sqrt{u^{2}\pm a^{2}}}du=ln\left|u+\sqrt{u^{2}\pm a^{2}}\right|+C[/tex]

    and further integration by parts, I got this:

    [tex]\int arccos(\frac{a^{2}}{2r^{2}}-1)rdr=arccos(\frac{a^{2}}{2r^{2}}-1)\frac{r^{2}}{2}-\frac{ar}{2}ln\left|r+\sqrt{r^{2}-(a/2)^{2}}\right|+\frac{a}{2}\int ln\left|r+\sqrt{r^{2}-(a/2)^{2}}\right|dr[/tex]
  10. Jan 21, 2010 #9

    Char. Limit

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    Gold Member

  11. Jan 22, 2010 #10
    Damn!!! I made a major mistake!!!! The two [tex] rdr[/tex] should be [tex] r^{3}dr[/tex].
    I should make sure before I start next time. Thanks for both of your help though. I appreciate it.
  12. Jan 23, 2010 #11

    Char. Limit

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    Gold Member

    Don't give up now! Just work through it carefully. Try showing (and checking) each step. I will not allow you to give up!

    Joking, nothing I can say or do would require you to do anything. Still, you should have that attitude for yourself, sometimes.
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