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Homework Help: Need some help with an integral.

  1. Sep 17, 2011 #1
    1. The problem statement, all variables and given/known data

    [itex]\int x^{3}sin(2x) dx[/itex]

    Relevant equations

    3. The attempt at a solution

    We are doing table integration - where I use a formula for integrals. I'm having trouble here because I keep getting new integrals through my work which don't seem to be leading anywhere.

    [itex]\int x^{3}sin(2x) dx[/itex]

    [itex]-\frac{1}{2}x^{3}cos2x + \frac{1}{2}\int3x^{2}cos2xdx[/itex]

    I can solve that integral with integration by parts or using that same formula, am I going down the correct path for this?
  2. jcsd
  3. Sep 17, 2011 #2
    This integral is solvable with integration by parts. Notice that you started with x^3 and after your first integration you have x^2. So just keep integrating!
  4. Sep 17, 2011 #3


    Staff: Mentor

    Actually, you are getting somewhere. Notice that the new integral has a smaller power of x (x2). At your next step, you should get that down to x, and then one more step will result in an integral that involves only sin(2x) or cos(2x).
  5. Sep 17, 2011 #4


    User Avatar
    Science Advisor

    Since this integral involves [itex]x^3[/itex] and integration by parts (taking [itex]u= x^n[/itex] each time) reduces the power of x by 1, you need to do three consecutive integrations by parts to get to a integral of a trig function only.
  6. Sep 17, 2011 #5
    Alright guys, I see what you're saying. However, while the process does eventually lead me to a final expression without integrals, it doesn't seem to be correct.


    That's a link to the work I did.. I didn't think itexing all of that would be feasible. Hopefully it's legible.
  7. Sep 17, 2011 #6
    I got the same answer you did. Perhaps it is a case of equivalent answers, where your answer could be re-arranged to match the given answer. If you post what the answer should be I can check for equivalence on my calculator very easily.
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