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Tricky trig substitution

  1. Apr 24, 2005 #1
    im hoping i worked this out right; its long:
    [tex]\int x(81-x^2)^{5/2}dx[/tex]
    the integral contains [itex]a^2-x^2[/itex], so i set [itex]x=asin\theta[/itex]. that would make [itex]x=9sin\theta[/itex] and [itex]dx=9cos\theta d\theta[/itex]:
    [tex]\int 9sin\theta(81-81sin^2\theta)^{5/2}9cos\theta d\theta = \int 9sin\theta[81(1-sin^2\theta)]^{5/2}9cos\theta d\theta[/tex]
    the integral now contains [itex]1-sin^2\theta =cos^2\theta[/itex]
    [tex]\int 9sin\theta(9cos\theta)^59cos\theta d\theta[/tex]
    i used u-sub by setting [itex]u=9cos\theta[/itex] and [itex]du=-9sin\theta d\theta[/itex].
    [tex]-\int u^6 du=-\frac{u^7}{7}+C[/tex]
    i plugged my u back in:
    then, i drew my little triangle.
    i plugged that into the [itex]cos\theta[/itex] and simplified and came out with:
    that's my answer
  2. jcsd
  3. Apr 24, 2005 #2
    You could also try the substitution u = 81 - x^2, which should be significantly easier.
  4. Apr 24, 2005 #3
    I punched it into mathematica and it produced the same as yours, save for a 7/2 exponent rather than 7.
  5. Apr 24, 2005 #4
    7/2 exponent is like my square root to the 7th
    also, we had to use trig substitution, so that left out the easy 81-x^2 sub
  6. Apr 24, 2005 #5
    It doesn't have to be.

    You shouldn't have to use a trig or u substitution. Inspection isn't too hard with something like this.

    [tex]\frac{(2)}{(-2)(7)}\int (-2x)(\frac{7}{2})(81-x^2)^{5/2}dx[/tex]
    [tex]= -\frac{1}{7} (\sqrt {81-x^2})^7[/tex]
    Last edited by a moderator: Apr 24, 2005
  7. Apr 24, 2005 #6
    *Cough* Table *Cough*

    Excuse me... something in my throat--- tables are great to have around so you barely have to think.
  8. Apr 24, 2005 #7
    Guess having someone else do it is the easiest way. :smile:
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