1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Integral substituting

  1. Apr 19, 2014 #1

    wel

    User Avatar
    Gold Member

    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 [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

    CAF123

    User Avatar
    Gold Member

    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:
    $$I(x)=I(0)+I'(0)x+I''(0)\frac{x^2}{2!}+O(x^3)$$
    Obviously ##I(0)=0##.

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

    wel

    User Avatar
    Gold Member

    [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

    SammyS

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    Which part(s) of this procedure don't you understand ?
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Integral substituting
  1. Integral substitution? (Replies: 3)

Loading...