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Math Help - Integration.

  1. #1
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    Integration.

    Hello i have a question pplz help me it is
    how the:
    a is a constant number

    after integration became


    I really appreciate your help
    Thanks
    Last edited by mr fantastic; July 10th 2011 at 04:24 PM. Reason: Title.
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  2. #2
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    Because your are integrating with respect to the variable p, as that variable is changing not a, thats the way I see it
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  3. #3
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    Hello

    Thanks For your reply well i have tried to replace
    <br />
P^2+1=U<br />
    and finding:
    <br />
p=sqrt(U-1)<br />
    and findinf dp and subsitituing them into the original equation etc....
    if there is easier method plz give it to me
    Thnx
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  4. #4
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    I am working on your problem as well, I am still a learner compared to everyone here, I have tried you substitution, before you mention it as it gets tad confusing im re-working it, hope I can help you
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  5. #5
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    ok

    i dunno i have tried many other method and didnt work
    Thanks Anyway
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  6. #6
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    This problem is very drawn out for me, I will find out how to do it though! It will be good for both of us, but our methods seem to complicate thing while integrating and simplifying as I told by my professor, mathematicians aim to make hard things very simple when working out a problem, so we will get it! I am sure there a few around here that this is easy for, I was close to the solution but made a careless everyone and wound up with a few missing parts, like the numerator coming to p^2 instead of p^3

    Lets see if we can make sense of things together, first we know -3a is a constant so we can pull that to the outside and that the numerator p^2 when integrated as if it was just alone as such  p^2 becomes \frac{p^3}{3} and remember we pulled out the -3a so now the the integrated form we get -3a*\frac{p^3}{3} we get -a * p^3 so somehow we need to make a relation to the denominator and decide the right method of integrating to where we can relate to this

    Edit: have you tried integration by parts, I feel I am close using this method
    Last edited by RockHard; November 21st 2009 at 08:58 PM.
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  7. #7
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    Have you tried the substitution  p = \frac{1}{x} ?

    since

     dp  =  \frac{-1}{x^2} dx

    the integral ( let's omit the constant -3a ) becomes


     \int \frac{x^5}{\sqrt{x^2+1}^5} \cdot \frac{1}{x^2} \cdot \frac{-1}{x^2} dx

     = - \int \frac{x dx}{\sqrt{x^2+1}^5}

     =  - (1 + x^2)^{\frac{-3}{2}} (\frac{-2}{3}) \frac{1}{2} + C

     = \frac{1}{3} \frac{ 1}{ \sqrt{ 1 + \frac{1}{p^2} }^3 } + C

     = \frac{ p^3}{3 \sqrt{ 1 + p^2}^3  } + C
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  8. #8
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    Jeepers, the one I never thought of trying, how did you know what to correctly substitute?
    Also where x^5 comes from?
    Last edited by mr fantastic; July 10th 2011 at 04:24 PM.
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  9. #9
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    Quote Originally Posted by RockHard View Post
    Jeepers, the one I never thought of trying, how did you know what to correctly substitute?

    If thec0o0lest didn't give us the answer , i won't think of this substitution .

    I observed that the integral is algebraic function , not transcendental ( usually  \sinh^{-1}(x) , \cosh^{-1}(x) ) .


    so i attempt to change the variable ( algebraic ) . It is just a trial !

    Another case we can use this sub. is

     \int \frac{dx}{\sqrt{x^2+1}^3}

    to

     - \int \frac{u ~du}{\sqrt{u^2+1}^3}
    Last edited by mr fantastic; July 10th 2011 at 04:25 PM.
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  10. #10
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    simplependulum Thankssssssssssssssssssssssssss
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  11. #11
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    Quote Originally Posted by RockHard View Post
    Also where x^5 comes from?
    x^5 comes from the denominator ( after changing  p to  \frac{1}{x} )

     \frac{1}{\sqrt{p^2+1}^5} = \frac{1}{\sqrt{1 + \frac{1}{x^2}}^5 }
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  12. #12
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    Quote Originally Posted by RockHard View Post
    Jeepers, the one I never thought of trying, how did you know what to correctly substitute?
    Also where x^5 comes from?
    the x^5 come from
    <br />
sqrt{1/x^2+1}^5
    when you subtitute p=1/x
    you need to make some balance in the denominator you multiply by

    x^(5/2)
    Last edited by mr fantastic; July 10th 2011 at 04:25 PM.
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  13. #13
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    Man, this is a confusing integral, I was stuck on it, If I did not see the answer I wouldn't even came to close to guessing a right answer
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