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Math Help - Ingtegral

  1. #1
    MHF Contributor Mathstud28's Avatar
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    Ingtegral

    ok so where did I go wrong.. \int{cos(mx^2)^2dx}...

    so I first did this \int{cos(mx^2)^2dx}=\int{\frac{1+cos(2mx^2)}{2}dx}...so then I seperated it to get



    \int{\bigg[\frac{1}{2}+\frac{cos(2mx^2)}{2}dx\bigg]} \Rightarrow{\int{\frac{1}{2}dx}+\int{\frac{cos(2mx  ^2)}{2}dx}\Rightarrow{\frac{x}{2}+\int{\frac{cos(2  mx^2)}{2}dx}}}



    Then using substitution by power series I got



    \frac{x}{2}+\int\frac{\sum_{n=0}^{\infty}\frac{(-1)^n\cdot(2mx^2)^{2n}}{(2n)!}}{2}dx\Rightarrow\fra  c{x}{2}+\int\sum_{n=0}^{\infty}\frac{(-1)^{n}\cdot{m^{2n}}\cdot{2^{2n-1}}\cdot{x^{4n}}}{(2n)!}dx


    ...then I went



    \frac{x}{2}+\sum_{n=0}^{\infty}\frac{(-1)^{n}\cdot{m^{2n}}\cdot{2^{2n-1}}\cdot{x^{4n+1}}}{(4n+1)(2n)!}

    ...could someone point out my mistake...thanks
    Last edited by Mathstud28; April 17th 2008 at 02:34 AM.
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  2. #2
    MHF Contributor Mathstud28's Avatar
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    O and by the way this is for leisurely purposes so no rush
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  3. #3
    Moo
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    Hello,

    \frac{x}{2}+\int\frac{\sum_{n=0}^{\infty}\frac{(-1)^n\cdot(2mx^2)^{\color{red}2n}}{(2n)!}}{2}dx

    Huh ?
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  4. #4
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Moo View Post
    Hello,

    \frac{x}{2}+\int\frac{\sum_{n=0}^{\infty}\frac{(-1)^n\cdot(2mx^2)^{\color{red}2n}}{(2n)!}}{2}dx

    Huh ?
    Thank you...I had the correct way on my paper but I just mistyped it....and I looked off of my typing...so I just kept copying my mistake...so the way I have it now is the way it is ony my paper...so that isnt my mistake if there is one....is there any other mistakes?
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  5. #5
    Moo
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    Why should there be an error ?

    Try factoring by x so that you have x^(4n)

    x^(4n)=(x)^(2n)

    Try to put everything in that power, putting away constants that are not useful..
    You got (2n)! on the denominator, so maybe it'll be something like cosine...
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  6. #6
    Super Member angel.white's Avatar
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    I don't think that
    \int{cos(mx^2)^2dx} = \int cos^2(mx^2) ~dx

    I think it it should be
    \int{cos(mx^2)^2dx} = \int cos(m^2x^4) ~dx

    Because I've gotten demerited on a test before for assuming the former (though that was with the natural log function, not a trig function).
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  7. #7
    Moo
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    It wouldn't be very logical... I mean why would they have written (mx) instead of mx^4 ?
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  8. #8
    Super Member angel.white's Avatar
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    Quote Originally Posted by Moo View Post
    It wouldn't be very logical... I mean why would they have written (mx) instead of mx^4 ?
    I argued it too ^_^ (didn't get the points back, though)

    Best I can figure is that the function applies to the funciton directly to it's right, In this case, that means f(x) = x^2, and g(x) = mx^2 so this is cos f(g(x))

    This may not be what the problem is, IDK, but since the OP doesn't include the answer, I figured it could be, since I've experienced similar frustration in the past.
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  9. #9
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by angel.white View Post
    I don't think that
    \int{cos(mx^2)^2dx} = \int cos^2(mx^2) ~dx

    I think it it should be
    \int{cos(mx^2)^2dx} = \int cos(m^2x^4) ~dx

    Because I've gotten demerited on a test before for assuming the former (though that was with the natural log function, not a trig function).
    You know what I think I am right...and it was \int{cos^2(mx^2)}....you want to know how? haha I made up the problem...it was just for fun
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  10. #10
    Super Member angel.white's Avatar
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    Quote Originally Posted by Mathstud28 View Post
    You know what I think I am right...and it was \int{cos^2(mx^2)}....you want to know how? haha I made up the problem...it was just for fun
    Fair enough.
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