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Primitive funktion help

  1. Aug 30, 2012 #1
    1. The problem statement, all variables and given/known data
    I have y'(x)= xcos(x2)
    How do I get the primitive function for this?

    2. Relevant equations



    3. The attempt at a solution
    I know that f'(x)=2xcos(x2) is f(x)=sin(x2)
    How will removing that 2 in front of xcos affect the primitive function?
     
  2. jcsd
  3. Aug 30, 2012 #2
    y(x)=1/2f(x)
     
  4. Aug 30, 2012 #3

    HallsofIvy

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    A more "formal" way to do this would be to let [itex]u= x^2[/itex] so that [itex]du= 2xdx[/itex] or [itex](1/2)du= xdx[/itex]. Then [itex]\int x sin(x^2)dx[/itex] becomes [itex](1/2)\int sin(u)du[/itex]. Alternatively, write [itex]\int x sin(x^2)dx[/itex] as [itex](1/2)(2)\int x sin(x^2)dx= (1/2)\int sin(x^2)(2x)dx[/itex]

    Note that you can only move constants into and out of the integral like that. If that first "x" were not there, you could not do the integral.
     
    Last edited by a moderator: Aug 30, 2012
  5. Aug 30, 2012 #4
    So, the answer should be f(x)=1/2 sin(x) ?
     
  6. Aug 30, 2012 #5
    I'm not completely understand everything you wrote there, and I feel a bit stupid now.
     
  7. Aug 30, 2012 #6
    @jakobs
    well you actually put f(x)=sin(x²). So I meant y(x)= 1/2 sin(x²). Then indeed y'(x)=xcos(x²).

    @Halls of Ivy
    Agreed this would be the 'follow the recepy' way of doing it, but with a problem this straightforward it really isn't necessary.
     
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