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Fun Integral

  1. Oct 20, 2007 #1
    1. The problem statement, all variables and given/known data

    [tex]
    \int {\frac{{dx}}{{\sqrt {x^2 - 2} \left( {x^2 - 1} \right)}}}
    [/tex]


    2. Relevant equations

    [tex]


    \frac{d}{{dx}}\arctan (x)


    [/tex]


    3. The attempt at a solution

    [tex]

    \frac{d}{{dx}}\arctan (x)
    [/tex] seems to be part of it, I can't quite get much farther...
     
    Last edited: Oct 20, 2007
  2. jcsd
  3. Oct 20, 2007 #2
    very interesting, fun time indeed!
     
  4. Oct 20, 2007 #3

    Gib Z

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    Thats not a fun integral you bullrat :P Try letting u=x^2-1, du = 2x dx. Solve for dx and express 2x in terms of u, the resultant integral is a great candidate for a trig substitution no?
     
    Last edited: Oct 20, 2007
  5. Oct 21, 2007 #4
    My head just exploded.

    But I end up with an arcsec, which becomes decidedly unhappy in terms of range below the x=0 point... Is that just a limitation of the method?

    Thanks for your input, though. :)
     
  6. Oct 21, 2007 #5
    did you get it? lol
     
  7. Oct 21, 2007 #6

    Gib Z

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    You ended up with [tex]\frac{1}{2} \arcsec (x^2-1) [/tex]?

    Label a right angled triangle with one of the other angles being G. Now label the opposite and hypotenuse sides, in accordance with the fact that sec G = (x^2-1)/1. So cos G 1/(x^2-1).

    Eg arcsec (x^2-1) = arccos (1/ (x^2-1) ).

    The smallest value x^2 can have is 0, in which case arccos -1 is defined quite well, and the largest value it can have it positive infinity, in which case arccos 0 is also well defined. =] I dont see a problem?
     
  8. Oct 21, 2007 #7
    yes ... lol, wow i actually did it right up to that point, let me keep working it.

    *not peeking at your sol'n* :-x
     
  9. Oct 21, 2007 #8
    When I differentiate the solution I get, [tex]
    \frac{1}{2}arc\sec \left( {x^2 - 1} \right)
    [/tex], it only applies well for x>0, where defined.. Am I doing something funny?
     
  10. Oct 21, 2007 #9
    woohoo! i solved it, lol

    man i learned trig subst. like 3 weeks ago, i'm already rusty :D
     
  11. Oct 21, 2007 #10

    Gib Z

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    Well when I differentiate [tex]\frac{1}{2} \arcsec (x^2-1)[/tex] i get [tex] \frac{1}{\sqrt{x^2-2} |x^2-1| }[/tex].

    Since x^2-1 and |x^2-1| are only different for |x| < 1 , this means the anti derivative is valid for all real x except when |x|< 1. Thats only fair, because the integrand is not even defined for those values =]

    For some very odd reason the tex isnt showing the arcsec, but you know what i mean
     
    Last edited: Oct 21, 2007
  12. Oct 21, 2007 #11
    ok i screwed up somewhere ... uh
    edit nvm

    ok hmm ... i'm taking the derivative and i'm not getting the original integral ... i think i need some sleep, lol.

    i shall check it tomorrow! gnite
     
    Last edited: Oct 21, 2007
  13. Oct 21, 2007 #12

    Gib Z

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    Remember that there is a constant of 1/2 to be considered, remember to use the chain rule as well, and remember that the derivative of arcsec is [tex]\frac{1}{|x|\sqrt{x^2-1}}[/tex].
     
  14. Oct 21, 2007 #13
    works like a charm!

    if i have x^2-1

    should it be and is it okay as

    |x^2-1| or does it have to be in parenthesis? |(x^2-1)|
     
    Last edited: Oct 21, 2007
  15. Oct 21, 2007 #14

    Gib Z

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    I wouldn't have written it |x^2-1| myself (post 10) if it wasn't ok =]
     
  16. Oct 21, 2007 #15
    The way to solve this is by using a hyperbolic substitution, not a goniometric one. Put in x=sqrt(2)cosh(t).

    You should get 1/2*integral[du/cosh(u)] after letting 2t=u.

    Then the substitution w=tanh(u/2). We have therefore cosh(u)=((1+w^2)/(1-w^2)) and
    du=(2dw)/(1-w^2).

    This gives the arctan function as a solution.
    Remembering that tanh(cosh^(-1)(x))=sqrt(x^2-1)/x should give you the solution.
    cosh^(-1) is the inverse function of cosh, it's not a power.

    It is not a difficult integral, but a tedious one, specially if you have to derive the
    hyperbolic relations like I had to :-s
     
  17. Oct 21, 2007 #16

    Gib Z

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    Many mistakes :( Firstly, a different method is not an incorrect one. Secondly, the answer is in terms of Arg Tanh, not arctan. And also, you will find that arcsec and arg tanh can be expressed in terms of each other...same answer mate, except your one is longer.
     
  18. Oct 21, 2007 #17
    Hmm, many mistakes....
    On the first point, you are right, a different method is not an incorrect one. I didn't want it to sound like that. My mistake if it did.

    Secondly, the answer I got is in fact a solution in terms of arctan. I was under the impression that it was required, looking back to the original post. F(x)=arctan(sqrt(x^2-2)/x)+C is the one I got and it can be transformed (haven't done it though) into you're solution if necessary.
     
  19. Oct 21, 2007 #18

    Gib Z

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    But if your substitution is w=tanh(u/2), how did you get the answer in terms of arctan?

    ps sorry if i sounded touchy, hard day. another method is never unwelcome =]
     
  20. Oct 21, 2007 #19
    No problem.

    It is a very interesting substitution based on the following equalities, if
    w=tanh(u/2), then cosh(u)=((1+w^2)/(1-w^2)) and sinh(u)=((2w)/(1-w^2)) and tanh(u)=((2w)/(1+w^2)), also the following can be found dw=1/2*(1-tanh^2(u/2))du or
    du=(2dw)/(1-w^2). Putting this into the integral makes it the standard integral for arctan.

    P.S. these exist also for goniometric properties. Here you have w=tan(u/2) and thus
    sin(u)=(2w)/(1+w^2) and cos(u)=((1-w^2)/(1+w^2)) and tan(u)=((2w)/(1-w^2)).

    I started from this to derive the hyperbolic ones because I didn't have those yet.

    Edit: Using goniometric and hyperbolic substitutions is a good way of eliminating the square root. After this you end up with an integral in these functions. Using then the substitutions I explained gets rid of them and transforms it back into something polynome like, which is hopefully easier than the original equation.
     
    Last edited: Oct 21, 2007
  21. Oct 21, 2007 #20
    Many thanks to all of you.

    The answer wasn't necessary supposed to be in arctan, but I found the solution that is more similar to coomast's through some type of teacher-prodded guess-and-check (magic and really lucky).

    Of course, reversing the differentiation was very difficult, so I thought I'd start anew.

    On the other hand, we didn't quite study hyperbolic things yet, so ... I'm not quite sure how I would transform it into arctan while it was in this state.

    Thanks very much, though, coomast, Gib Z, and Rocophysics. :)
     
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