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Integral help from S.R. and G.R. physicsforums

  1. Mar 1, 2008 #1


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    Would someone evaluate this integral:
    See Physics Forums -Special and General Relativity above.
    On page one with title: Elapsed time on accelerating clock. by morrobay.
    In particular reply # 4 by JesseM .
    I seem to be missing something with The Integrator reference.
    And one semester calculus , (Thomas ) is not enough.
  2. jcsd
  3. Mar 2, 2008 #2

    Gib Z

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    Try a simple substitution, [tex]x=\frac{\cos u}{\sqrt{a}}[/tex]
  4. Mar 3, 2008 #3


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    Integral help from S.R. and G.R physicsforums

    If that substitution is simple for you or anyone else I would like to see it
  5. Mar 3, 2008 #4


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    Then look in any introductory calculus book!

    Since cos2(u)= 1- sin2(u), setting x= sin(u) is a "natural" choice for any integrand of the form [itex]\sqrt{1- x^2}dx[/itex]. With x= sin(u), that [itex]\sqrt{1- x^2}= \sqrt{1- sin^2(u)}= cos(u)[/itex] and dx= cos(u)du. [itex]\int \sqrt{1- x^2} dx= \int cos^2(u)du[/itex] which can be integrated using the trig identity "cos2(u)= (1/2)(1+ cos(2u))".

    If you have integrand [itex]\sqrt{a^2- x^2}[/itex] just factor out the "a2": [itex]a\sqrt{1- x^2/a^2}[/itex] and obvious substitution is x/a= sin(u).
  6. Mar 3, 2008 #5

    Gib Z

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    [tex] \int \sqrt{ 1- ax^2} dx[/tex]

    Make the substitution from the previous post.

    Then [tex]dx = - \frac{\sin u}{\sqrt{a}}[/tex].

    The integral is then transformed into;
    [tex]\int \sqrt{ 1- a \cdot \frac{\cos^2 u}{a} } ( - \frac{\sin u}{\sqrt{a}}) du[/tex].

    Constants may be taken out of an integral, and [tex] 1 - \cos^2 u = \sin^2 u[/tex] by the Pythagorean Identity.

    [tex]-\frac{1}{\sqrt{a}} \int \sin^2 u du[/tex]

    [tex]\sin^2 u = \frac{1 - \cos (2u)}{2} [/tex] which is verifiable by the double angle identity; [itex]\cos (2t) = \cos^2 t - \sin^2 t[/itex].

    [tex]-\frac{1}{\sqrt{a}} \int \left( \frac{1}{2} - \frac{2 \cos (2u)}{4} du \right)[/tex]

    Split the integral and in the second one, let w= 2u, so dw= 2 du.
    [tex] = -\frac{1}{\sqrt{a}} \left( \frac{u}{2} - \frac{\sin (2u)}{4} \right)[/tex]

    where [tex]x= \frac{\cos u}{\sqrt{a}}[/tex], or [tex] u = \arccos \left( \sqrt{a}u\right)[/tex]

    EDIT: Just saw halls post, should have chosen his substitution because the signs work out more nicely, but this still works out fine.
    Last edited: Mar 3, 2008
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