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Integration by Parts of x^5cos(x^3)

  1. Apr 30, 2009 #1
    1. The problem statement, all variables and given/known data

    [tex]\int x^5cos(x^3) dx[/tex]

    2. Relevant equations
    [tex]\int uv' = uv - \int u'v[/tex]


    3. The attempt at a solution
    [tex]\int x^5cos(x^3) dx[/tex]

    [tex]\frac{(x^5)*(sin(x^3)}{(3x^2)} - \int\frac{5x^4*sin(x^3)}{(3x^2)} dx[/tex]

    [tex]\frac{(x^3)*sin(x^3)}{3} - \int\frac{(5x^2)*sin(x^3)}{(3)} dx[/tex]

    [tex]\frac{(x^3)*sin(x^3)}{3} - \frac{5}{3} \int x^2*sin(x^3) dx[/tex]

    [tex]\int x^2*sin(u) *\frac{du}{3x^2}[/tex]

    [tex]\frac{1}{3} \int sin(u) du[/tex]

    [tex]\frac{-1}{3} cos(u)[/tex]

    [tex]\frac{-1}{3} cos(x^3)[/tex]

    Revisting

    [tex]\frac{(x^3)*sin(x^3)}{3} - \frac{5}{3} * \frac{-1}{3} cos(x^3)[/tex]

    My answer:

    [tex]\frac{(x^3)*sin(x^3)}{3} + \frac{5}{9} * cos(x^3) + C[/tex]

    Correct Answer:

    [tex]\frac{(x^3)*sin(x^3)}{3} + \frac{1}{3} * cos(x^3) + C[/tex]
     
  2. jcsd
  3. Apr 30, 2009 #2
    no, your first u-dv doesn't work that way. think compostion chainrule and u-substitution.
     
  4. May 1, 2009 #3

    HallsofIvy

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    The integral of [itex]cos(x^3)[/itex] is not [itex]sin(x^3)/3x^2[/itex]!

    First write the integral as
    [tex]\int x^3\left[x^2cos(x^3)\right]dx[/tex]
    and let [itex]u= x^3[/itex].
     
  5. May 2, 2009 #4
    May I ask what the derivative of sin(x^3)/3x^2 may be then?

    Thanks, rewriting the function got me the right answer.
     
    Last edited: May 2, 2009
  6. May 2, 2009 #5

    HallsofIvy

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    I don't know why you would ask that. It might be helpful to recognize that, by the chain rule, the derivative of [itex]sin(x^3)[/itex] is [itex](2x^2)cos(x^3)[/itex].
     
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