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Simplification of an integral please explain

  1. Apr 19, 2010 #1
    In doing math I try not to memorize shortcuts/simplifications, but instead understand what's happening. When studying integrals, my professor gave out a few "simplifications" for us to use on problems; most of which I have been able to figure out, except one. I just cannot seem to figure out the relationship given, please help clarify/explain. Thankyou.



    [tex]\int(\sqrt{a^2-u^2})du[/tex] = [tex]\left(\frac{u}{2}\right)[/tex][tex]\times[/tex][tex]\left(\sqrt{a^2-u^2}\right)[/tex] + [tex]\left(\frac{a^2}{2}\right)[/tex][tex]\times[/tex][tex]\left(sin^{-1}\left(\frac{u}{a}\right)\right)[/tex] + C



    I have tried:
    - u substitution
    - trig substitution
    - IBP

    *It could be my steps, maybe I'm just doing the intermediates wrong.
    **it took a really long time to put that formula in
     
  2. jcsd
  3. Apr 19, 2010 #2

    tiny-tim

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    Hi phys-lexic! :smile:

    (have a square-root: √ and try using the X2 tag just above the Reply box :wink:)

    Try substituting u = asinθ, then use a couple of the standard trigonometric identities :smile:
     
    Last edited: Apr 19, 2010
  4. Apr 19, 2010 #3
    What he said. An integral with a difference of squares suggests a certain right triangle, and that can be used to suggest a certain trigonometric substitution. In this case, [itex]u = a \sin\theta[/itex]
     
  5. Apr 19, 2010 #4
    Thankyou, the trig-substitution worked and I set up the correct proof. If you wouldn't mind, we were assigned homework for an exam review. I finished the regular problems, but he gave us three "critical thinking" problems that would be way harder than the exam. I can do them all, but one is giving me issue (I cannot seem to find the final value of x after doing the integral). Here is a link to the problem I posted in the homework section, thankyou again).

    https://www.physicsforums.com/showthread.php?p=2679179#post2679179
     
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