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Integration by trig substitution

  1. Mar 5, 2009 #1
    1. The problem statement, all variables and given/known data
    Integrate: [tex]\int[/tex][tex]\sqrt{1-9t^{2}}[/tex]dt

    2. Relevant equations

    3. The attempt at a solution
    t = 1/3 sin[tex]\Theta[/tex]
    dt/d[tex]\Theta[/tex] = 1/3 cos[tex]\Theta[/tex]
    dt = 1/3 cos[tex]\Theta[/tex]d[tex]\Theta[/tex]
    3t = sin[tex]\Theta[/tex]

    1/3[tex]\int\sqrt{1-sin^{2}}\Theta[/tex] cos[tex]\Theta[/tex]d[tex]\Theta[/tex]
    1/3[tex]\int cos^{2}\Theta[/tex]d[tex]\Theta[/tex]
    1/3[tex]\int[/tex](1 + cos 2[tex]\Theta[/tex]) / 2 d[tex]\Theta[/tex]
    1/6[tex]\int1 + cos2\Theta[/tex] d[tex]\Theta[/tex]
    1/6([tex]\Theta + 1/2 sin 2\Theta[/tex]) + C
    1/6([tex]\Theta + 1/2(sin\Theta cos\Theta[/tex]) + C
    1/6([tex]\Theta + 1/2(sin\Theta\sqrt{1-sin^{2}\Theta}[/tex])) + C
    1/6([tex]\Theta[/tex] + 1/2(3t[tex]\sqrt{1-9t^{2}}[/tex])) + C

    I can't figure out how to get rid of [tex]\Theta[/tex] in the result.
  2. jcsd
  3. Mar 5, 2009 #2
    Why not using arcsin? Also, there is a factor of two missing in the 3rd step from the end.

    The Bob
  4. Mar 5, 2009 #3
    Perfect! Thanks.
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