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Math Help - Integrate cos(at)(cos(bt) ... Can't figure it out

  1. #1
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    Integrate cos(at)(cos(bt) ... Can't figure it out

    Well, I've been looking at this problem for a while, and need some help. I'm sure some smart folks on this forum can help me, perhaps? This is somewhat urgent unfortunately...

    { cos(at)cos(bt) where a and b are positive integers and not equal to zero.

    There is a little more to the problem, but I just need to know how to integrate these. I can probably do the rest myself.
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  2. #2
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    Quote Originally Posted by taichu View Post
    Well, I've been looking at this problem for a while, and need some help. I'm sure some smart folks on this forum can help me, perhaps? This is somewhat urgent unfortunately...

    { cos(at)cos(bt) where a and b are positive integers and not equal to zero.

    There is a little more to the problem, but I just need to know how to integrate these. I can probably do the rest myself.
    May I assume your whole integrand is cos(at)*cos(bt) *dt?
    Because that last part, the "dt", is very important.

    INT.[cos(at)cos(bt)]dt ---------------(1)

    As it is, the integrand is the product of two functions. To integrate that, we can do it by parts. INT[u]dv = uv -INT.[v]du. But that won't do here because there is no sin(at) or sin(bt) involved.
    So we have to separate/decompose the product into addition or subtraction.

    There is this trig identity
    cosAcosB = (1/2)[cos(A-B) +cos(A+B)] ---------(i)
    that we can use to decompose the original integrand.

    cos(at)cos(bt)
    = (1/2)[cos(at -bt) +cos(at +bt)]
    = (1/2)[cos(t(a-b)) +cos(t(a+b))]

    So, (1) becomes
    = (1/2)INT.[cos((a-b)t) +cos((a+b)t)]dt -----------(2)

    It is easy now.

    = (1/2)INT.[cos((a-b)t)]dt +1/2INT.[cos((a+b)t)]dt

    = (1/2)INT.[cos((a-b)t)]*[(a-b)/(a-b)]dt +(1/2)INT.[cos((a+bt))]*[(a+b)/(a+b)]dt

    = [(1/2)(1/(a-b))]INT.[cos((a-b)t)]*(a-b)dt +[(1/2)(1/(a+b))]INT.[cos((a+b)t)]*(a+b)dt

    = [1 / (2(a-b))]sin((a-b)t) +[1 / (2(a+b))]sin((a+b)t) +C

    That is it.
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  3. #3
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    ticbol forget to mention one thing, the equation fails when a=b.

    In that case the formula is not true.
    Rather you have,
    INT sin (at) cos (bt) dt = (1/2)*INT 2sin(at)*cos(at) dt=(1/2)*INT sin(2at) dt=-(1/4a) cos(2at)+C
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  4. #4
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    Quote Originally Posted by ThePerfectHacker View Post
    ticbol forget to mention one thing, the equation fails when a=b.

    In that case the formula is not true.
    Rather you have,
    INT sin (at) cos (bt) dt = (1/2)*INT 2sin(at)*cos(at) dt=(1/2)*INT sin(2at) dt=-(1/4a) cos(2at)+C
    Yeah?

    If a=b,
    then cos(at) cos(bt) becomes cos^2(at) or cos^(bt).

    So what's the use of asking for Integral of cos(at)cos(bt) dt then? )-:
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  5. #5
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    So what's the use of asking for Integral of cos(at)cos(bt) dt then? )-:
    There is none.
    But as a mathematician you need to consider all cases, even the simpler ones. Otherwise that leads to errors.
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  6. #6
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    Quote Originally Posted by ThePerfectHacker View Post
    There is none.
    But as a mathematician you need to consider all cases, even the simpler ones. Otherwise that leads to errors.
    In this case it is useless. Again, if a=b, .....blah, blah, blah.
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  7. #7
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    Quote Originally Posted by ticbol View Post
    In this case it is useless. Again, if a=b, .....blah, blah, blah.
    But your solution is wrong unless your write a not = b.

    For example, solve the differencial equation,
    y'=2xy
    Divide by "y",
    y'/y=2x
    Variables seperable,
    ln |y| = x^2 +C
    Thus,
    y=C*e^(x^2) where C>0.
    But that cannot be right!
    Because y=0 is also a solution.
    The error occured after the division, that case was not considered.
    Like here.
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