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Please Help Integration!

  1. Feb 6, 2008 #1
    Please Help!! Integration!

    can anyone help me solve the following integration? thanks a lot.

    [tex]\int x^5 cos(x^3) dx[/tex]
  2. jcsd
  3. Feb 6, 2008 #2
    try to solve it by parts
  4. Feb 6, 2008 #3
    I have tried it, but it doesn't work, at least, for me.

    If I make [tex] u=cos(x^3), dv=x^5 dx [/tex], the grade of x, which is in the [tex] cos(x^3) [/tex], won't be reduce.

    If I make [tex] u=x^5, dv=cos(x^3) dx [/tex], I can't solve the [tex] \int cos(x^3) dx[/tex].

    Could you give me some more advice? thanks
  5. Feb 6, 2008 #4
    What would you like to have inside the integral in order to have

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

    be solvable? Can you choose slightly different u and dv to accomplish that?
  6. Feb 6, 2008 #5
    From [itex](\sin x^3)'=3\,x^2\,\cos x^3[/itex] deduce that [itex]\cos x^3=\frac{1}{3\,x^2}\,(\sin x^3)'[/itex] and use that to integrate by parts.
  7. Feb 6, 2008 #6


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    Oh, for goodness sake! Make the substitution u= x3, then integrate by parts!

    (I notice now that that is essentially what Rainbow Child said.)
  8. Feb 6, 2008 #7
    How it can help?

    How this can help?
  9. Feb 6, 2008 #8


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    I don't think there's any harm in showing the substitution,

    [tex]u = x^3 \Rightarrow \frac{du}{dx} = 3x^2 \Rightarrow dx = \frac{du}{3x^2}[/tex]

    Hence when we make the substitution the integral becomes,

    [tex]\int \frac{x^5}{3x^2}\cos(u)du = \frac{1}{3}\int u\cos(u)du[/tex]

    Which is a simple integral to solve.
    Last edited: Feb 6, 2008
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