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Integral question

  1. Jun 15, 2009 #1
    [tex]
    \int \frac{r^2dr}{\sqrt{r^2+x^2}}=\int{rdt}\\
    [/tex]
    [tex]
    t=\sqrt{r^2+x^2}\\
    [/tex]
    [tex]
    dt=\frac{2rdr}{2\sqrt{r^2+x^2}}
    [/tex]

    i tried to solve it like that
     
  2. jcsd
  3. Jun 15, 2009 #2

    Cyosis

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    If you do it like that the r^2 in the numerator will pose a problem. Try the substitution [itex]r=x \sinh u[/itex].
     
    Last edited: Jun 15, 2009
  4. Jun 15, 2009 #3

    tiny-tim

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    No, it would be [tex]
    \int \frac{r^2dr}{\sqrt{r^2+x^2}}=\int{\sqrt{t^2 - x^2}dt}\\
    [/tex]
    erm :redface: … xsinhu :wink:
     
  5. Jun 15, 2009 #4

    Cyosis

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    Whoops, will fix it.
     
  6. Jun 15, 2009 #5
    what do i do after
    [tex]
    \int \frac{r^2dr}{\sqrt{r^2+x^2}}=\int{\sqrt{t^2 - x^2}dt}\\
    [/tex]
    whats the substitution
    ??

    and i dont know hyperbolic stuff
    its not on the course
     
  7. Jun 15, 2009 #6

    Cyosis

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    That integral has a sign ambiguity, because if [itex]t=\sqrt{r^2+x^2}[/itex] then [itex]r=\pm \sqrt{t^2-x^2}[/itex]. Which one do you take?

    Do you know of a good substitution if the integrand had been [itex]\sqrt{1-x^2}[/itex]? Try to find a trigonometric substitution for which 1-(...)^2=(...)^2.
     
  8. Jun 15, 2009 #7
    there is no trigonometric substitution for it
     
  9. Jun 15, 2009 #8

    Cyosis

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    There is and it is the most famous trig identity at that. Any ideas?
     
  10. Jun 15, 2009 #9
    tangence goes when there is no square root on the denominator
    1/(x^2+1) type

    so i dont have any clue
     
  11. Jun 15, 2009 #10

    Cyosis

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    Tangent works fine for the square root case as well. But you're jumping in between integrals again. It will be helpful if you stick to one integral.
     
  12. Jun 15, 2009 #11

    Mark44

    Staff: Mentor

    You speak with considerable authority here, but it is unwarranted. Draw a right triangle with the horizontal leg labelled x and the vertical leg labelled r and the hypotenuse labelled sqrt(r^2 + x^2). If the acute angle is labelled u, then sec u = sqrt(r^2 + x^2)/x and tan u = r/x, and sec u * du = dr/x.
     
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