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Integral of x*sin(ax)

  1. Nov 29, 2004 #1
    What is the integral of x*sin(x) and x*sin(ax)?
    I have no idea since I have neveer integrated something to get a product...
    Ohh, it's supposed to be integrated from 0 to 1 for the sin(ax)
     
  2. jcsd
  3. Nov 29, 2004 #2

    HallsofIvy

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    Use "integration by parts".

    From the product rule for derivatives, d(uv)/dx= u(dv/dx)+ v(du/dx). We can write that in "differential" form as d(uv)= u dv+ vdu and then rewrite it as

    u dv= d(uv)- vdu.

    Integrating both sides gives the integral formula
    [tex]\int u dv= uv- \int vdu[/tex].

    In particular, to integrate x sin(ax), let u= x, dv= sin(ax) dx. Then du= dx and
    v= -(1/a)cos(ax) so
    [tex]\int x sin(ax)dx= -(\frac{1}{a}x cos(ax)+ \frac{1}{a}\int cos(ax) dx[/tex]

    [tex]= -\frac{1}{a}(x cos(x)+ \frac{1}{a}sin(ax))[/tex].
     
  4. Nov 29, 2004 #3

    dextercioby

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    Sorry,there's a minus,a paranthesis too much and an "a" missing:
    [tex]\int x sin(ax)dx= -\frac{1}{a}x cos(ax)+ \frac{1}{a}\int cos(ax) dx[/tex]

    [tex]= -\frac{1}{a}[x cos(ax)- \frac{1}{a}sin(ax)][/tex]
     
  5. Nov 29, 2004 #4
    and if its / ? =) as in sin(ax)/x
    Or is it so easy that I can do it by myself, don't have time right now...
     
  6. Nov 29, 2004 #5
    This is what I seem to get, very annoying
    [tex]\int(sin(ax)\frac{1}{x})=sin(ax) ln(x)+\int(ln(x) cos(ax))[/tex]
    or
    [tex]\int(\frac{1}{x}sin(ax))=-\frac{1}{x}cos(ax)+\int(\frac{1}{x^2}cos(ax))[/tex]

    Any ideas? Are any of the following integrals easy to do?
     
  7. Nov 29, 2004 #6
    did u try setting u=x and dv=sin(ax)dx ?
    this is what i got
    [tex] \int xsin(ax) dx=-\frac{x}{a}cos(ax) +\frac{1}{a}\int cos(ax) dx[/tex]
    [tex] = -\frac{x}{a}cos(ax)+\frac{1}{a^2}sin(ax)[/tex]
     
  8. Nov 30, 2004 #7
    heh vladimir, I understand the xsin(ax) integral but now I am trying to do 1/x*sin(ax) is this possible? If you look at my previous post you'll see me trying to integrate 1/x*sin(ax)
     
  9. Nov 30, 2004 #8

    dextercioby

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    That's because there is no primitive of the function sinx/x.
    I assume you know that ordinary functions can be differentiated and the result be another "familiar" function.But this does not apply for primitives.There are functions like sinx/x,cos/x,exp(x^2),etc. which do not have primitives.That is,u cannot find a function which to differentiate to get the function you wish to integrate.
    However,numerical methods based on Taylor/Mac Laurin formula(s) can be used to obtain results.For example,to find the primitive of sinx/x,u need to expand sinx and devide each term of the expansion term by x and integrate the results.You'll have then a new infinite series,which could be seen as the Taylor/Mac Laurin exapansion of the function u are looking for.
    This thing works for functions which "behave" pretty well as to apply Taylor/Mac Laurin formula(s) to them.The 3 examples i have stated prove this assertion.
    To find definite integral values for the 3 functions mentioned above,try to get a hand on 2 books:M.Abramowitz,I.Segun:"Mathematical functions and tables" and Rytzhik and Gradstein:"Tables of integrals" and search for sine integral function,cosine integral function and erf(error) function.

    P.S.I'm not at the library anymore,so from now on,when i give indications to certain books always doubt the veridicity of the names and titles stated,as i give them from my memory to which i have no recollection of having ever been treated with glucosis.So it cold fail me someday.Hopefully not soon.
     
    Last edited: Nov 30, 2004
  10. Dec 1, 2004 #9
    Thanks man, I was expecting the likes...
     
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