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Integration by parts

  1. Oct 7, 2009 #1

    [tex]

    \int x^3cos(x^2)dx
    [/tex]


    [tex]
    -\frac{1}{2}x^2sin(x^2)+\frac{3}{2}\int xsin(x^2)dx
    [/tex]


    [tex]
    -\frac{1}{2}x^2sin(x^2)+\frac{3}{4}cos(x^2)-\frac{3}{4}\int \frac{cos(x^2)}{x}
    [/tex]
    the last integral
     
    Last edited: Oct 7, 2009
  2. jcsd
  3. Oct 7, 2009 #2

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    Hi nameVoid! :smile:

    (i think that the minus is wrong, and where did the 3/2 come from? but anyway …)

    Where did the last integral come from? What did you think you were differentiating? :confused:
     
  4. Oct 7, 2009 #3

    Mark44

    Staff: Mentor

    In your integration by parts, you don't show u and dv, etc., but you seem to be mostly on the right track.

    For the first integral, u = x2 and dv = x cos(x2) dx. So du = 2xdx and v = -(1/2)sin(x2).
    So
    [tex]\int x^3cos(x^2)dx[/tex]
    [tex]=~-(1/2)x^2sin(x^2) + \int x sin(x^2)dx[/tex]
    So far, the only difference between your work and mine is the multiplier in front of the last integral. This one can be done with an ordinary substitution (u = x2), instead of the integration by parts that it looks like you tried.
     
  5. Oct 7, 2009 #4
    right..thank you.
     
  6. Oct 7, 2009 #5
    Just a suggestion, make life a little easier on yourself by doing a substitution before int by parts:

    [tex]\int x^3cos(x^2)dx[/tex]

    [tex]let p = x^2[/tex]

    [tex]dp = 2xdx[/tex]

    [tex]\frac{1}{2}*\int 2x^3cos(x^2)dx[/tex]

    [tex]\frac{1}{2}*\int pcos(p)dp[/tex]

    From there, the integration by parts is soo much simpler.

    Here's the variable breakdown for integration by parts:
    u = p dv = cos(p) dp
    du = dp v = sin(p)

    Sincerely,

    NastyAccident
     
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