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Integral of root(-x^2+10x-16)dx

  1. Mar 4, 2009 #1
    1. The problem statement, all variables and given/known data

    integral of root(-x^2+10x-16)dx

    2. Relevant equations



    3. The attempt at a solution
    i can factor the thing inside the root to (x-8)(x-2) if i pull out the negative sign
    no idea what to do after that
     
  2. jcsd
  3. Mar 4, 2009 #2

    lanedance

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    Re: Integrals

    could you try completing the square, then a trig substitution?
     
  4. Mar 4, 2009 #3
    Re: Integrals

    which root the square root? that factorization won't help you get anywhere good, but if you complete the square I think you'll find yourself with something you can work with.
     
  5. Mar 4, 2009 #4

    lanedance

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    Re: Integrals

    scratch the trig sub, first step should help
     
  6. Mar 4, 2009 #5
    Re: Integrals

    Try rewriting the quadratic expression under the square root symbol in vertex form

    sqrt[a]*sqrt[(x-h)^2 + k^2)] , as follows:

    sqrt[-1]*sqrt[(x-5)^2 - 9], then let u = x - 5, and use the formula from a table of integrals for an integral of the form

    sqrt[u^2 - a^2], for a > 0, and then make the appropriate back substitutions.


    Note: the sqrt[-1] is just a constant, albeit a complex constant, so it can be pulled out side of the integral. The integral will turn out to be real-valued none-the-less.

    I hope this helps.
     
  7. Mar 4, 2009 #6

    lanedance

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    Re: Integrals

    no need or use to pull -1 outside the squareroot

    complete the square make the substitution, then integrate
     
  8. Mar 4, 2009 #7
    Re: Integrals

    hmm im having trouble integrating sqrt(u^2 - 9)
    would i need to use integration by parts?
     
  9. Mar 4, 2009 #8
    Re: Integrals

    That makes sense, then it would just be an integral of the form

    sqrt[a^2 - u^2], for a > 0.
     
  10. Mar 4, 2009 #9
    Re: Integrals

    no, trig substituion
     
  11. Mar 4, 2009 #10
    Re: Integrals

    Try the triq substitution of

    u = 3*sin(v) for

    sqrt[9 - u^2], here we just didn't factor out the -1.
     
  12. Mar 5, 2009 #11
    Re: Integrals

    im lost...this is as far as i get

    -(integral) sqrt(u^2 - 9)

    i dont understand how i can use a trig identity to simplify that
     
    Last edited: Mar 5, 2009
  13. Mar 5, 2009 #12
    Re: Integrals

    where the heck did you get a cube from?
     
  14. Mar 5, 2009 #13
    Re: Integrals

    typo lol
     
  15. Mar 5, 2009 #14

    lanedance

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    Re: Integrals

    Hmm, not sure how well trig sub works here

    have a look at the derivative of arcsin...
     
  16. Mar 5, 2009 #15

    HallsofIvy

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    Re: Integrals

    The trig substitution x= 3 sin t will give the same thing as the arcsine integral.
     
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