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Integral substituting

  1. Apr 19, 2014 #1


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    Consider the integral
    I(x)= \frac{1}{\pi} \int^{\pi}_{0} sin(xsint) dt
    show that
    I(x)= \frac{2x}{\pi} +O(x^{3})
    as [itex]x\rightarrow0[/itex].

    => [itex] sin(x.sint)= x.sint - \frac{(x-sint)^3}{3!}+...[/itex]

    and integrate term by term should give
    [itex] - x.sint - \frac{1}{12}(cos3t-9cost)+...[/itex]
    when substituting $t=\pi$ and $t=0$ something else comes up.
    please help me.
  2. jcsd
  3. Apr 19, 2014 #2


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    As x → 0, the linear approximation of sin is enough and indeed the question wants you to ignore further terms in the expansion (the next term being cubic).
  4. Apr 19, 2014 #3
    Hi wel!

    As suggested in your previous thread, solve it in the following way:
    Obviously ##I(0)=0##.

    Can you find ##I'(x)## and then ##I'(0)##?
  5. Apr 19, 2014 #4


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    [itex]sin(x\cdot sint) = x\cdot sint - \dfrac{(x\cdot sint)^3}{3!} + ...[/itex],
    then I guess I have to integrate term by term.
    [itex] I(0)=0[/itex]
    integrate [itex]I(x)= x sint[/itex] at [itex]t =\pi[/itex] and [itex]t=0[/itex] gives the answer but I really don't know why and how?
  6. Apr 19, 2014 #5


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    Which part(s) of this procedure don't you understand ?
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