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Integrating cos(2x)*cos(6x)

  1. Oct 7, 2013 #1
    1. The problem statement, all variables and given/known data

    ∫cos(2x)cos(6x)dx


    2. Relevant equations



    3. The attempt at a solution

    When I do this one, I seem to get a different answer than my book. The book uses a product to summarize formula, but I hate memorizing formulas and want to do this without it. Here's what I did:

    Let u=6x then du=6dx

    (1/6)∫cos(u/3)cos(u)du

    Integration by parts: 1/6[cos(u/3)sin(u)+(1/3)∫sin(u/3)sin(u)du]

    Integrating (1/3)sin(u/3)sin(u) gave: (-1/3)sin(u/3)cos(u)+(1/9)∫cos(u/3)cos(u)du

    Call ∫cos(u/3)cos(u)du “I”

    (I)=cos(u/3)sin(u)-(1/3)sin(u/3)cosu+(I/9)

    (8/9)I=cos(u/3)sin(u)-(1/3)sin(u/3)cosu

    I=(9/8)cos(u/3)sin(u)-(3/8)sin(u/3)cosu

    I= (9/8)cos(2x)sin(6x)-(3/8)sin(2x)cos6x +C = ∫cos(u/3)cos(u)du=∫cos(2x)cos(6x)

    The answer key has: (1/8)sin(4x)+(1/16)sin(8x)+ C

    Are these equivalent or have I messed up?

    Thanks!
     
  2. jcsd
  3. Oct 7, 2013 #2
    cos(A+B)=cosAcosB-sinAsinB
    cos(A-B)=cosAcosB+sinASinB

    Add these two equations together and see what you get.
     
  4. Oct 8, 2013 #3
    This turned out to be better than I thought. Much better than the mess above. I'm still kind of frustrated that the long way didn't work :(
     
  5. Oct 8, 2013 #4

    arildno

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    It can perfectly well be that your expressions are equivalent!

    To find THAT out, simply use on your own expression the formula sin(a)cos(b)=1/2(sin(a+b)+sin(a-b)), and compare.
    If your answer deviates from the answer key's by merely a constant of integration, your original expression was correct as well.

    If they do NOT equate, you have made some trivial integration error somewhere.
     
  6. Oct 8, 2013 #5

    D H

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    You messed up. You dropped that factor of 1/6.

    Incorporating that factor of 1/6 into your answer yields ##\frac 3{16} \cos(2x)\sin(6x) - \frac 1{16} \sin(2x)\cos(6x) + C##, and that is the same as the answer key.
     
  7. Oct 8, 2013 #6
    lol this usually happens when I do long things like this. Thanks!
     
  8. Jan 8, 2015 #7
    Here is what I have done :

    First of all I would like to remind you that cos(s)cos(t) = [cos(s+t) + cos(s-t)]/2

    ∫cos(8x)+cos(4x) / 2 dx = 1/2 ∫cos(8x)+cos(4x) = sin(8x)/16 + sin(4x)/8 + C

    Best Regards
     
  9. Jan 8, 2015 #8
    Isn't that what I said in post #2?

    By the way, you need to resist the temptation to provide the complete solution, since that is a violation of Physics Forums rules.

    Chet
     
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