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Integral of cos^3 theta

  1. Oct 20, 2004 #1
    Would very much appreciate a hint on how to start this off.
    So far I have cos theta = (1-t^2)/(1+t^2). After squaring this and doing various substitutions, I get (cos theta)^3, which can't be right.
     
  2. jcsd
  3. Oct 20, 2004 #2

    Tide

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    I'm not sure what you're asking but if you want the integral indicated in the title of your post this may help:

    [tex]\cos^3 \theta = \cos^2 \theta \cos \theta = \left(1-\sin^2 \theta \right) \frac {d \sin \theta}{d\theta}[/tex]
     
  4. Oct 20, 2004 #3
    Many thanks. I'll give it a shot.
     
  5. Oct 20, 2004 #4
    I got cos theta - (1/3)cos^3 theta + C
    Does this look right?
     
  6. Oct 20, 2004 #5

    Fredrik

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    Almost. It should be sin, not cos.
     
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