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Indefinite integral

  1. Feb 1, 2005 #1
    [tex]\int sin(9x)sin(16x)dx[/tex]

    is there another way of solving the problem above besides using the multiple angles formula?
  2. jcsd
  3. Feb 1, 2005 #2
    use this identity:

    sin u = \frac{e^{iu}-e^{-iu}}{2i}

    expand the sine in term of exponential, multiply them and regroup them into 2 cosine, then do the integral
    Last edited: Feb 1, 2005
  4. Feb 1, 2005 #3
    wow looks more difficult than using multiple angles formula. i just thought there's an easier way to do it so i wont have to remeber crazy amount of formulas when it's test day.
  5. Feb 1, 2005 #4


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    [tex] \sin{\alpha}\sin{\beta} = \frac{\cos(\alpha-\beta) - \cos(\alpha + \beta)}{2}[/tex]

  6. Feb 1, 2005 #5
    actually, my way is much much much more easier than remember you formulas.....
    I can eye ball the answer using my way......
    the answer is....
    1/14 sin7x - 1/50 sin25x
    the expansion of the complex number is easy.... there are only 4 terms
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