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Integrating product of trig functs with different args

  1. Feb 17, 2007 #1
    1. The problem statement, all variables and given/known data

    How do you integrate a product of two trigonometric functions of x, when the argument is different, ie:

    what is the integral of

    sin x cos 3x

    I ought to know this, but don't seem to be able to do it.

    Thanks in adv!

    2. Relevant equations

    3. The attempt at a solution
  2. jcsd
  3. Feb 17, 2007 #2


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    Use the formulas for sin(a+b) and sin(a-b):

    sin(a+b) = sin a cos b + cos a sin b
    sin(a-b) = sin a cos b - cos a sin b


    sin(a+b) + sin(a-b) = 2 sin a cos b
  4. Feb 17, 2007 #3
    you can solve this by the formula
    choose one part as v
    choose the other as du

    {-integral sigh

    {vdu=v*u -{dv*u
  5. Feb 17, 2007 #4


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    Dearly Missed

    Note that transgalactic's proposal will enable you to find the answer by entering an integrand loop of finite length.

    To see what I mean with an "integrand loop", I'll take an example:

    We are to compute:
    we use partial integration to write this as:
    Therefore, we have gained the equation for I:
    which is easily solved for I.
  6. Sep 5, 2007 #5
    Thanks for all the help and advice, everyone. I got an A in maths, btw.
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