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Math Help - itnegral of sin x using the definition

  1. #1
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    integral of sin x using the definition

    Hi,

    I am trying to calculate:

    \int_0^{\pi}\sin x \,dx

    using the definition only, that is, that \int_0^{\pi}\sin x\,dx = \lim_{n \to \infty}\sum_{i = 1}^{\infty}\sin(x_{i}^{*})\Delta x, where \Delta x = \frac{\pi}{n}. If we take take x_i^* = i\Delta x = \frac{\pi}{n} we obtain:

    \int_0^{\pi}\sin x\,dx = \lim_{n \to \infty}\left(\frac{\pi}{n}\right)\sum_{i=1}^n \sin\left(\frac{i\pi}{n}\right).

    I am stuck at this point. Is there any kind of identity, perhaps trigonometric or power series related, to help me evaluate this limit by hand only?

    Thank in advanced or any help,

    James
    Last edited by james121515; July 10th 2011 at 09:53 PM. Reason: typo
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  2. #2
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    Re: integral of sin x using the definition

    Quote Originally Posted by james121515 View Post
    Hi,

    I am trying to calculate:

    \int_0^{\pi}\sin x \,dx

    using the definition only, that is, that \int_0^{\pi}\sin x\,dx = \lim_{n \to \infty}\sum_{i = 1}^{\infty}\sin(x_{i}^{*})\Delta x, where \Delta x = \frac{\pi}{n}. If we take take x_i^* = i\Delta x = \frac{\pi}{n} we obtain:

    \int_0^{\pi}\sin x\,dx = \lim_{n \to \infty}\left(\frac{\pi}{n}\right)\sum_{i=1}^n \sin\left(\frac{i\pi}{n}\right).

    I am stuck at this point. Is there any kind of identity, perhaps trigonometric or power series related, to help me evaluate this limit by hand only?

    Thank in advanced or any help,

    James
    Surely you should know that \displaystyle \int{\sin{x}\,dx} = -\cos{x} + C by definition...
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  3. #3
    MHF Contributor Also sprach Zarathustra's Avatar
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    Re: integral of sin x using the definition

    Quote Originally Posted by james121515 View Post
    Hi,

    I am trying to calculate:

    \int_0^{\pi}\sin x \,dx

    using the definition only, that is, that \int_0^{\pi}\sin x\,dx = \lim_{n \to \infty}\sum_{i = 1}^{\infty}\sin(x_{i}^{*})\Delta x, where \Delta x = \frac{\pi}{n}. If we take take x_i^* = i\Delta x = \frac{\pi}{n} we obtain:

    \int_0^{\pi}\sin x\,dx = \lim_{n \to \infty}\left(\frac{\pi}{n}\right)\sum_{i=1}^n \sin\left(\frac{i\pi}{n}\right).

    I am stuck at this point. Is there any kind of identity, perhaps trigonometric or power series related, to help me evaluate this limit by hand only?

    Thank in advanced or any help,

    James

    Hint:

    Use Euler's formula - Wikipedia, the free encyclopedia.


    Edit:

    An alternative:

    \sum_{i=1}^n \sin\left(\frac{i\pi}{n}\right)=S

    Now multiply both sides by  \frac{\cos{\frac{\pi}{2n}}}{\sin\frac{\pi}{2n}}.



    And third alternative:

    Proving by induction:

    \sum_{i=1}^n{\sin(i\alpha)}=\frac{\sin\frac{n}{2}{  \alpha}\cdot\sin\frac{n+1}{2}\alpha}{\sin\frac{ \alpha }{2}}
    Last edited by Also sprach Zarathustra; July 10th 2011 at 11:38 PM.
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  4. #4
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    Re: integral of sin x using the definition

    Quote Originally Posted by Prove It View Post
    Surely you should know that \displaystyle \int{\sin{x}\,dx} = -\cos{x} + C by definition...
    I'm not sure it's a definition. I suspect the OP knows the rule and wants to derive the rule from first principles.
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  5. #5
    MHF Contributor chisigma's Avatar
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    Re: integral of sin x using the definition

    Quote Originally Posted by mr fantastic View Post
    ... I suspect the OP knows the rule and wants to derive the rule from first principles...
    May be that that is not deriving 'from the first principles' but a possibility is to use the 'geometric sum'...

    \sum_{k=1}^{n} x^{n} = x\ \frac{1-x^{n}}{1-x} (1)

    Using the 'identity'...

    \sin (k\ \frac{\pi}{n}) = \frac{e^{i\ k\ \frac{\pi}{n}} -e^{- i\ k\ \frac{\pi}{n}}}{2\ i} (2)

    ... and (1) You arrive to write...

    \frac{\pi}{n}\ \sum_{k=1}^{n} \sin (k\ \frac{\pi}{n}) = \frac{\pi}{2\ i\ n}\ \{e^{i\ \frac{\pi}{n}}\ \frac{1 - e^{i\ k\ \frac{\pi}{n}} }{1- e^{i\ \frac{\pi}{n}}} - e^{- i\ \frac{\pi}{n}}\ \frac{1 - e^{-i\ k\ \frac{\pi}{n}} }{1- e^{-i\ \frac{\pi}{n}}} \} (3)

    The effective computation of (3) is however 'just a little hard' for me and is left to 'a young mind' ...

    Kind regards

    \chi \sigma
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  6. #6
    MHF Contributor FernandoRevilla's Avatar
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    Re: integral of sin x using the definition

    Following previous outlines and denoting

    C_a=\cos a+\cos 2a+\ldots+\cos na,\; S_a=\sin a+\sin 2a+\ldots+\sin na

    If e^{ia}\neq 1 , we have:

    C_a+iS_a=e^{ia}+e^{2ia}+\ldots+e^{nia}=e^{ia}\;\df  rac{(e^{nia}-1)}{e^{ia}-1}=

    e^{ia}\;\dfrac{e^{\frac{nia}{2}}}{{e^{\frac{ia}{2}  }}}\;\dfrac{e^{\frac{nia}{2}}-e^{\frac{-nia}{2}}}{e^{\frac{ia}{2}}-e^{\frac{-ia}{2}}}=e^{\frac{i(n+1)a}{2}}\;\dfrac{\sin \frac{na}{2}}{\sin \frac{a}{2}}=

    \frac{\sin \frac{na}{2}}{\sin \frac{a}{2}}\left(\cos \frac{(n+1)a}{2}+i\sin \frac{(n+1)a}{2} \right)

    Identifying imaginary parts

    S_a=\dfrac{\sin \frac{na}{2}}{\sin \frac{a}{2}}\;\sin\frac{(n+1)a}{2}

    So, S_{\pi/n}=\dfrac{1}{\sin \frac{\pi}{2n}}\;\sin \frac{(n+1)\pi}{2n}

    Then,

    \int_0^{\pi}\sin x\;dx=\lim_{n\to +\infty} \dfrac{\pi}{n}S_{\pi/n}=2
    Last edited by FernandoRevilla; July 11th 2011 at 08:06 AM.
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    Re: integral of sin x using the definition

    Quote Originally Posted by chisigma View Post
    Using the 'identity'...

    \sin (k\ \frac{\pi}{n}) = \frac{e^{i\ k\ \frac{\pi}{n}} -e^{- i\ k\ \frac{\pi}{n}}}{2\ i} (2)
    Thanks for your reply. I had no idea the solution to this involves complex analysis. This first 'identity' you mention here, is this derived from Euler's formula e^{i\theta} = \cos \theta + i\sin \theta or is it more involved?
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    Re: integral of sin x using the definition

    Quote Originally Posted by james121515 View Post
    Thanks for your reply. I had no idea the solution to this involves complex analysis. This first 'identity' you mention here, is this derived from Euler's formula e^{i\theta} = \cos \theta + i\sin \theta or is it more involved?
    Yes. And the sum of a geometric series is being used.
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