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Math Help - just an integral :)

  1. #1
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    just an integral :)

    Can anyone integrate this?
    Integral ((Cos[x])^n)*Cos[(n+1)*x] dx
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  2. #2
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    Quote Originally Posted by grozot
    Can anyone integrate this?
    Integral ((Cos[x])^n)*Cos[(n+1)*x] dx
    For what its worth QuickMath gives the result in the attached image.

    This uses Mathematica as its processing engine (and I beleive that _2F_1 denotes the hypergeometric function).
    Attached Thumbnails Attached Thumbnails just an integral :)-quickint.jpg  
    Last edited by CaptainBlack; January 25th 2006 at 02:01 PM.
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  3. #3
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    thanks, mathematica gave that result also, but i still have no idea how to do that on my exam
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  4. #4
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    Quote Originally Posted by grozot
    thanks, mathematica gave that result also, but i still have no idea how to do that on my exam
    Mathematica does not always give the most compact/simple form
    for integrals, but the fact that there has been no substantive replys
    to your post suggests that this is not a well known or simple integral

    RonL
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  5. #5
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    Quote Originally Posted by grozot
    thanks, mathematica gave that result also, but i still have no idea how to do that on my exam
    Having thought about this integral and Mathematica, I think that M. might
    just be a bit to clever for its own good. It does not know that we want
    n to be a positive integer.

    If we make M. assume the domain of n is \mathbb{Z}_+ it
    might well give a simpler answer.

    RonL
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  6. #6
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    i realized i didnt tell you that integral is from 0 to Pi/2.
    and i dont know how to tell m that n is positive integer but that shoud be easy so ill figure it out.
    and i have no idea why but the result is 0
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  7. #7
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    Quote Originally Posted by grozot
    i realized i didnt tell you that integral is from 0 to Pi/2.
    and i dont know how to tell m that n is positive integer but that shoud be easy so ill figure it out.
    and i have no idea why but the result is 0
    Now you tell us! Look at the (anti)symmetry of the integrand about the
    mid point of the interval.

    I was looking at this earlier today, but I don't have my notes with me now,
    though I thought it was over an interval of length pi or 2pi that the integral
    was zero. However memory can be misleading.

    RonL
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  8. #8
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    Quote Originally Posted by CaptainBlack
    Now you tell us! Look at the (anti)symmetry of the integrand about the
    mid point of the interval.

    I was looking at this earlier today, but I don't have my notes with me now,
    though I thought it was over an interval of length pi or 2pi that the integral
    was zero. However memory can be misleading.

    RonL
    Just checked some special cases (n=2 and 3), and the integral is
    non-zero over 0-pi/2, but zero over 0-pi.

    RonL
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  9. #9
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    this is what mathematica has to say
    In[18]:=
    n\[Element]Naturals
    Integrate[(Cos[x]^n)*Cos[(n+2)x],x]

    Out[18]=
    n\[Element]Naturals

    Out[19]=
    \!\(\(Cos[x]\^\(1 + n\)\ Sin[\((1 + n)\)\ x]\)\/\(1 + n\)\)
    Out[19] is 0 over 0=pi/2

    this is just trying to understand how all that is true:
    In[20]:=
    \!\(D[\(Cos[x]\^\(1 + n\)\ Sin[\((1 + n)\)\ x]\)\/\(1 + n\), x]\)
    Out[20]=
    \!\(Cos[x]\^\(1 + n\)\ Cos[\((1 + n)\)\ x] -
    Cos[x]\^n\ Sin[x]\ Sin[\((1 + n)\)\ x]\)

    In[21]:=
    \!\(Simplify[
    Cos[x]\^\(1 + n\)\ Cos[\((1 + n)\)\ x] -
    Cos[x]\^n\ Sin[x]\ Sin[\((1 + n)\)\ x]]\)
    Out[21]=
    \!\(Cos[x]\^n\ Cos[\((2 + n)\)\ x]\)

    so i guess i should somehow integrate this
    Cos[x]\^\(1 + n\)\ Cos[\((1 + n)\)\ x] -
    Cos[x]\^n\ Sin[x]\ Sin[\((1 + n)\)\ x]]
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  10. #10
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    Quote Originally Posted by grozot
    this is what mathematica has to say
    In[18]:=
    n\[Element]Naturals
    Integrate[(Cos[x]^n)*Cos[(n+2)x],x]
    Cos[(n+2)x] ???? I thought it was Cos[(n+1)x].

    RonL
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  11. #11
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    sorry, i was in a hurry every time i posted something here
    so i made mistakes...what can i say...something happens...
    Last edited by ThePerfectHacker; January 26th 2006 at 11:41 AM.
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