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Anti-derivatives! help!

  1. May 14, 2005 #1
    was just wondering if this was the right answer..
    take the anti-derivative of:

    cosx dx / sin^3x

    i got
    -1/2(sinx)^-2 + C
    is that right?
  2. jcsd
  3. May 14, 2005 #2
    Thats what maple says, good job.
  4. May 14, 2005 #3
    Just a tip, if your not sure, take the derivative and check if you get the original function.
  5. May 14, 2005 #4


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    Science Advisor

    Maple?? Maple?? !!! Let u= sin x. Then du= cos du and the integrand becomes
    [tex]\frac{cos x}{sin^3 x}dx= \frac{1}{u^3}du= u^{-3}du[/tex]

    The anti-derivative of that is [tex]\frac{1}{-3+1}u^{-3+1}+C= \frac{1}{2}u^{-2}+C= \frac{1}{2}\frac{1}{sin^2 t}+C[/tex]

    Good job, laker_gurl3
  6. May 14, 2005 #5
    Nothing wrong with being lazy!
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