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How do you find the indefinite integral of xsinx

  1. Mar 3, 2005 #1
    I am just curious I was thinking about this and if anyone could explain I would appreciate it. I am curious to know how to find the indefinite integral of xsinx with respect to x.

    Thanks, David
     
  2. jcsd
  3. Mar 3, 2005 #2

    dextercioby

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    Are u familiar to the part integration method...?If so,then do it...

    Daniel.
     
  4. Mar 3, 2005 #3
    im not, i did a lot bit of research however i am having difficulty comprehending it, im in calculus ab now, we have yet to reach that if it is in our agenda
     
  5. Mar 3, 2005 #4

    dextercioby

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    Since part integration is the first method of integration one learns,i suspect you haven't reached indefinite integration at all.In which case,why bother...?Is it curiosity,or what...?

    Daniel.
     
  6. Mar 3, 2005 #5

    James R

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    The integration by parts formula is:

    [tex]\int u \frac{dv}{dx} dx = uv - \int \frac{du}{dx} v dx[/tex].

    In your problem, put

    [tex]u = x, \qquad \frac{dv}{dx} = \sin x[/tex].

    Now find [itex]du/dx[/itex] and [itex]v[/itex], then plug them into the formula above. Do the second integration, and you're done.
     
  7. Mar 3, 2005 #6
    ive reached integration, i can integrate like 2sin2x when using substitution as long as the constant cancels out from du, in my book, thomas/finney 9th edition calculus, it isn't introduced until the latter part of the book, and since my class is only for the ap test, its not like we will get that far anyway i dont think and im just curious, i looked in the book but its past me
     
  8. Mar 3, 2005 #7
    ok that helps i get it that way, thanks
     
  9. Mar 3, 2005 #8

    dextercioby

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    Part integration uses product rule of differentiation for proving.I think that's a little easier than substitution,which would require chain rule for proving...:wink:

    Daniel.

    P.S.I think someone else offered the solution.
     
  10. Mar 3, 2005 #9
    thats it, it slipped my mind i couldnt remember what its called, but yeah, chain rule is what i was getting at, we just havent done integration by parts yet
     
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