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Tricky Integral

  1. Dec 4, 2006 #1
    Hey guys,

    By what rules is the integral of

    [tex]\frac{cos(t)}{1+9sin^2(t)}=\frac{1}{3}tan^{-1}(sin(3t))[/tex]

    I know this is right, but I have no idea why and can't find any trig identities to help. Thanks.
     
  2. jcsd
  3. Dec 4, 2006 #2
    Let [tex] u = 3\sin t [/tex]

    and use the fact that [tex]\int \frac{dx}{x^{2}+a^{2}} = \frac{1}{a}\tan^{-1}\left(\frac{x}{a}\right) + C [/tex]
     
    Last edited: Dec 4, 2006
  4. Dec 4, 2006 #3
    gotcha, thanks
     
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