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Integral of x^2*sin(ax)dx

  1. Nov 25, 2007 #1
    [SOLVED] Integral of x^2*sin(ax)dx

    So, I'm a newb at Integration, especially by parts.

    This is what I have
    [tex]\int x^{2}sin(ax)dx[/tex]

    I set:

    u=x[tex]^{2}[/tex] and dv=sin(ax)dx

    so I got:

    du=2xdx and v=-cos(ax)/a

    So then I put everything together

    [tex]u*v- \int v*du[/tex]
    [tex]-x^{2}*\frac{-cos(ax)}{a} - \int \frac{-cos(ax)}{a}*2xdx[/tex]
    [tex]-x^{2}*\frac{-cos(ax)}{a} + 2\int \frac{-cos(ax)}{a}*xdx[/tex]

    So that's pretty much where I get lost. I don't know what do do about the second integral.

    I tried setting u = 2x and dv = cos(ax)/a * dx and I got

    -x^2*cos(ax)/a - 2xsin(ax) - 2cos(ax)
    but WebAssign (website where I submit homework) says that's wrong.

    This is my first post here. I had some trouble with some other problems that I have so I googled them and most of the search results pointed me here.

    So, any suggestions are appreciated.
  2. jcsd
  3. Nov 25, 2007 #2
    Hello Psyc, welcome to the forum.

    The method you use is indeed the correct one. However you must pay attention to the minus signs. (It might be a typo error). Anyway the intermediate result should read:

    [tex]-\frac{x^2}{a}cos(ax)+\frac{2}{a}\int {xcos(ax)dx}[/tex]

    The second integral is obtained with the same method as the first one, which you have attempted, but was wrong due to the errors mentioned. You will get the right answer if you write everything down with a bit more care. Give it another try, you are almost at the correct answer.
  4. Nov 25, 2007 #3
    Thanks for noticing that, I think signs are my number one weakness. I tend to look over stuff like that.

    I fixed the sign but I don't know if once again I missed something or I'm entering it into the answer box incorrectly.
    I've attached a screenshot.


    Attached Files:

  5. Nov 25, 2007 #4

    D H

    Staff: Mentor

    A good habit to get into is to always double-check your work. A partial checklist:
    • Do you have a well-formed formula? (Yes, you do)
    • Are the units correct? (not applicable here)
    • Does reverting the answer return the original problem? (No)

    In this case, "reverting the answer" means taking the derivative of your answer. The derivative of
    \frac 2 a\left(\frac x a \sin(ax)-\,\frac 1 a \cos(ax)\right)[/tex]
    with respect to x is
    [tex]\left(x^2+\frac 2 a\left(\frac 1 a +1\right)\right)\sin(ax)[/tex]
    which is not [itex]x^2\sin(ax)[/itex].
  6. Nov 26, 2007 #5
    You've made the same mistake. Be carefull with the signs,

    [tex]\int sin(ax) dx = - \frac{cos(ax)}{a}+C[/tex]

    and take care of the brackets as you can see in D H's reply.
    Believe me, you're almost there
  7. Nov 26, 2007 #6
    I got someone else to look at it and I was apparently distributing the 2/a incorrectly and some other errors.

    I got it all worked out and my answer was

    Thanks for all of your help! :)
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