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Substitution for indef. integral

  1. Mar 19, 2008 #1
    1. The problem statement, all variables and given/known data
    Hey folks, I think I know how to solve this by parts but I need a substitution to get there. I've been staring at examples for a while but I still don't understand how to apply the substitution rule. Anyway, here's the integral:

    [tex]\int x^9cos(x^5)[/tex]

    2. Relevant equations
    integration by parts: [tex]\int f(x)g\prime (x)dx=f(x)g(x)-\int g(x)f\prime (x)dx[/tex]
    substitution rule: [tex]\int f(g(x))g\prime (x)dx=\int f(u)du[/tex]

    3. The attempt at a solution

    By applying integration by parts:
    [tex]f\prime (x)=9x^{8}[/tex]
    [tex]g\prime (x)=cos(x^5)[/tex]

    Now I need to apply the substitution rule to find [tex]g(x)[/tex] by integrating [tex]cos(x^5)[/tex]:
    Then maybe [tex]sin(u)du[/tex]? I'm not sure how to proceed. Thanks!
    I've got another post that's still unanswered just in case you've got some more time to kill :wink:
    Last edited: Mar 19, 2008
  2. jcsd
  3. Mar 19, 2008 #2


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    Nooo … if you're going to use substitution, do the substitution first!

    (If you do the integrating by parts first, then split it differently - the object is to make it easier!)

    As you said, u=x^5 is a good idea … but it's du=5x^4dx.

    Then integrate by parts!

    What do you get? :smile:
  4. Mar 19, 2008 #3
    It;s simple , just break x^9 into x^5*x^4 , then take X^5 = t 5x^4dx=dt , therefore integral becomes 5 t cos(t) dt , solve using by parts
  5. Mar 19, 2008 #4


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    Hint: [tex]x^4 \cos(x^5) = \frac{1}{5} \frac{\mathrm d}{\mathrm dx} \sin(x^5)[/tex].
  6. Mar 19, 2008 #5
    Okaaay, here goes:

    [tex]x^9sin(u)5x^4dx-\int sin(u)5x^4dx9x^8dx[/tex]
  7. Mar 19, 2008 #6
    No, you misunderstood the suggestions.

    [tex]\int x^9\cos{x^5}dx=\int x^4x^5\cos{x^5}dx[/tex]


    [tex]\frac 1 5\int u\cos udu[/tex]
    Last edited: Mar 20, 2008
  8. Mar 19, 2008 #7
    You're doing a substitution first!!! Then you will apply Parts.

    u=x^5, so replace all x^5 with u
    du=5x^4dx, so x^4dx appears in your original Integral, so that goes away.
  9. Mar 19, 2008 #8
    ahhha, ok. I'll be back!
  10. Mar 19, 2008 #9
    Got it? You need to review if you're not getting it. This should be beyond you!!! First step is Calc 1!
  11. Mar 19, 2008 #10
    Yeah, I did cal 1 years ago so I'm sort of reviewing points from it as I slam up against them. I have been looking at substitution examples for a while but I need to grind through a few problems to shake the rust off. I think this one is coming along now though. I'll be back with an answer.
  12. Mar 20, 2008 #11
    [tex]\frac{1}{5}[u(sin(u))-(cos(u)+C)][/tex] getting warmer?
  13. Mar 20, 2008 #12
    Almost!!! Don't forget that ...

    [tex]\int sin xdx=-\cos x+C[/tex]

    [tex]\frac 1 5 (x^5\sin x^5+\cos x^5)+C[/tex]
  14. Mar 23, 2008 #13
    Does it matter whether the C is inside or outside the brackets?
  15. Mar 24, 2008 #14

    Gib Z

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    No it doesn't. One fifth of an arbitrary real constant is still an arbitrary real constant isn't it.
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