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Math Help - Integrations

  1. #1
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    Integrations

    Consider the integral
    \begin{equation}
    I(x)= \frac{1}{\pi} \int^{\pi}_{0} sin(xsint) dt
    \end{equation}
    show that
    \begin{equation}
    I(x)= \frac{2x}{\pi} +O(x^{3})
    \end{equation}
    as $x\rightarrow0$.
    => I Have used the expansion of McLaurin series of $I(x)$ but did not work.
    please help me.
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  2. #2
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    Re: Integrations

    it does work.

    What is your Maclaurin expansion of $\sin(x \sin(t))$ ?
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  3. #3
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    Re: Integrations

    Quote Originally Posted by romsek View Post
    it does work.

    What is your Maclaurin expansion of $\sin(x \sin(t))$ ?
    $sin(x\cdot sint) = x\cdot sint - \dfrac{(x\cdot sint)^3}{3!} + ...$,
    then I guess I have to integrate term by term.
    $ I(0)=0$
    integrate $I(x)= x sint$ at $t =\pi$ and $t=0$ gives the answer but I really don't know why and how?
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  4. #4
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    Re: Integrations

    Quote Originally Posted by grandy View Post
    $sin(x\cdot sint) = x\cdot sint - \dfrac{(x\cdot sint)^3}{3!} + ...$,
    then I guess I have to integrate term by term.
    $ I(0)=0$
    integrate $I(x)= x sint$ at $t =\pi$ and $t=0$ gives the answer but I really don't know why and how?
    basically as $x \to 0, \sin(x \sin(t))$ can be closely approximated by it's Maclaurin series in x

    which in this case is simply $\sin(t) x+O(x^3)$

    we drop the $O(x^3)$ part and integrate

    $\dfrac 1 \pi \displaystyle{\int_0^{\pi}} \sin(t)x~dt=\dfrac x \pi \cos(t)|_{\pi}^0=\dfrac {2x}{\pi}$
    Thanks from prasum
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