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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

    AlephZero

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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

    so

    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

    arildno

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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:
    [tex]I=\int_{0}^{1}e^{x}\cos(x)dx[/tex]
    we use partial integration to write this as:
    [tex]I=\int_{0}^{1}e^{x}\cos(x)dx=e^{x}\cos(x)\mid^{x=1}_{x=0}+\int_{0}^{1}e^{x}\sin(x)dx=(e^{x}\cos(x)+e^{x}\sin(x))\mid_{x=0}^{x=1}-\int_{0}^{1}e^{x}\cos(x)dx=(e^{x}\cos(x)+e^{x}\sin(x))\mid_{x=0}^{x=1}-I[/tex]
    Therefore, we have gained the equation for I:
    [tex]I=(e^{x}\cos(x)+e^{x}\sin(x))\mid_{x=0}^{x=1}-I[/tex]
    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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