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Develop the argument:

  1. Apr 1, 2005 #1
    Develop the argument:

    Integral cos(3x)cos(2x)dx , with out using integral tables.

    I know,

    cos (A+B) = cosA.cosB-sinA.sinB

    Any ideas:
     
  2. jcsd
  3. Apr 1, 2005 #2

    dextercioby

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    Okay.

    [tex] \cos (A+B) =\cos A \cos B -\sin A \sin B [/tex]

    [tex] \cos (A-B) =\cos A \cos B +\sin A \sin B [/tex]

    Add the 2 expressions and c what u get...

    Daniel.
     
  4. Apr 1, 2005 #3
    Yes, I added them, and
    I came up with:

    cos(A+B) + cos (A-B) = 2cosAcosB
    = 2cos(3x)cos(2x) if we let A = 3x and B = 2x
    But how to get rid of that extra 2.

    if we write 3x = x + 2x, and then write the cos (A+B) of the LHS as:

    cos (x+2x) and expand, similarly expand cos (A-B)

    Equate the LHS and the RHS, and do some cancellations, but how to get that integral sign..... that's another thing to worry about.
     
  5. Apr 1, 2005 #4
    just put a 1/2 outside the integral and put a 2 inside. Same result.

    As for the integral, you ahvent posed an argument. You've just made a statement.
     
  6. Apr 1, 2005 #5

    dextercioby

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    [tex] \cos A \cos B =\frac{1}{2} \left[\cos (A+B) + \cos (A-B) \right] [/tex]

    Agree?

    Then set

    [tex] \left\{\begin{array}{c} A=3x\\B=2x \end{array}\right [/tex]

    And apply the formula.The integrations will be simple,involving almost trivial substitutions.

    Daniel.
     
  7. Apr 1, 2005 #6
    Ok, did this:

    Using cosA.cosB = 1/2[cos(A+B) + cos (A-B)]

    A = 3x, B = 2x

    cos3x.cos2x = 1/2 [ cos (5x) + cos (x) ]

    Integrating both sides now:

    Integral ( cos3x.cos2x) = Integral ( 1/2 [ cos (5x) + cos(x) ]

    = 1/2 Integral cos 5x + 1/2 Integral cos x
    = 1/2 sin5x /5 + 1/2 sinx

    = 1/10 ( sin 5x) + 1/2 sinx /2

    Integral cos 2x. dx = cos3x = 1/2 [ sin5x/5 +sinx ]

    Is the above correct!
     
  8. Apr 1, 2005 #7
    I mean:
    Integral cos 2x.cos 3x dx = 1/2 [ sin5x/5 +sinx ]
     
  9. Apr 1, 2005 #8

    dextercioby

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    Don't forget the integration constant.You might consider learning to edit formulas in Latex..:wink:

    Daniel.
     
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