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Homework Help: Can't figure out this integral

  1. May 19, 2006 #1
    Hello,

    I am having trouble with this integral, I don't know how to solve it.

    integral(x * cos(x))dx
    I tried it in the calculator and it gave me the integral back, and I don't know what method of integration to use to figure it out.
    Well if anyone has a solution that would be great, thanks.
     
  2. jcsd
  3. May 19, 2006 #2

    berkeman

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    Staff: Mentor

    What do you get when you differentiate x * sin(x)?

    How about x^2 * sin(x)?

    Do these give you some ideas about what you could differentiate to get to x * cos(x)?
     
  4. May 19, 2006 #3

    nrqed

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    I would suggest integration by parts. There is on choice of "u" and "dv" that will lead to a very simple form after a single integration by part.s
     
  5. May 19, 2006 #4
    Okay, I had to do integration by parts twice and I got:
    ((x^2 * cos(x))/6) + ((x^2 * cos(x))/3), I don't know if that is right though but it's what I came up with.
     
  6. May 19, 2006 #5
    dark_omen, that answer is not correct.

    Here's my solution:

    u = x
    dv = cos x dx

    So:
    du = dx
    v = sin x

    uv - I(v * du) = x sin x - I(sin x dx) = x sin x + cos x.

    I've verified this answer in Mathematica as well.
     
  7. May 19, 2006 #6
    Okay, so I guess I made the wrong choice in u and dv, and that's why my answer didn't come out right. Thanks.
     
  8. May 20, 2006 #7

    VietDao29

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    :grumpy: :grumpy: :grumpy:
    Why can't people just stop to give out COMPLETE solutions?!?!?!
    ???
    Why?!?!?! :confused: :confused: :confused: What's so tempting about posting complete solutions like that?
    Noone bothered to read the rules, eh???
    And also, where has the Constant of Integration gone? Vanished?
     
    Last edited: May 20, 2006
  9. May 20, 2006 #8

    arildno

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    Well, I find it excusable in that OP had shown quite a bit of work already.
     
  10. May 20, 2006 #9
    First you chide me for giving out the complete solution. Then you lecture me about not giving the complete solution... :) I thought that since he knew how to take an indefinite integral in the first place, he'd be smart enough to remember the +C.

    Anyhow, no, I guess I missed that sentence in the rules. My apologies. Though I don't quite understand why it isn't left to the discretion of the person helping as to how much they want to help.
     
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