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Indefinite Integration Derivation ?

  1. Feb 22, 2007 #1
    1. The problem statement, all variables and given/known data
    I want to prove the standard result for indefinite integrals that [tex] \int \frac{1}{a^2-x^2} .dx\ =\ \frac{1}{2a}\ln{(\frac{a+x}{a-x})} \\ \mbox{But the problem is that i am unable to do so. I get this--}
    \int {\frac{1}{a^2-x^2} .dx } \\ [/tex]
    [tex]
    = \frac{1}{2a}\int {\frac{2a}{a^2-x^2} .dx} \\
    = \frac{1}{2a}\int{ \frac{a+x+a-x}{a^2-x^2}} \\ [/tex]
    [tex]

    = \frac{1}{2a}\int {\frac{1}{a-x}+\frac{1}{a+x} .dx} \\
    [/tex]
    [tex]
    = \frac{1}{2a}(\int {\frac{1}{a-x}}+\int{\frac{1}{a+x}} ) \\
    [/tex]

    [tex]
    = \frac{1}{2a}\ln{(a-x)}+ln{(a+x)} .dx [/tex]
    I used this from an earlier derivation of [tex]
    \int{\frac{1}{x^2-a^2} } \mbox{where } \frac{1}{2a}(\int{ \frac{1}{x-a}}+\int{\frac{1}{x+a}} ) [/tex]
    was converted to [tex]
    = \frac{1}{2a}\ln{(x-a)}-ln{(x-a)} .dx [/tex] similarly


    Plz suggest where i went wrong and suggest a way to get the desired proof
    Thx

    EDIT: There's some trouble with latex. The code i entered was -: (space to avoid repetition)
    [i tex] \int \frac{1}{a^2-x^2} .dx\ =\ \frac{1}{2a}\ln{(\frac{a+x}{a-x})} \\ \mbox{But the problem is that i am unable to do so. I get this--}
    \int {\frac{1}{a^2-x^2} .dx } \\
    = \frac{1}{2a}\int {\frac{2a}{a^2-x^2} .dx} \\
    = \frac{1}{2a}\int{ \frac{a+x+a-x}{a^2-x^2}} \\
    = \frac{1}{2a}\int {\frac{1}{a-x}+\frac{1}{a+x} .dx} \\
    = \frac{1}{2a}(\int {\frac{1}{a-x}}+\int{\frac{1}{a+x}} ) \\
    = \frac{1}{2a}\ln{a-x}+ln{a+x} .dx mbox{I used this from an earlier derivation of }
    \int{\frac{1}{x^2-a^2} } \ mbox{where } \frac{1}{2a}(\int{ \frac{1}{x-a}}+\int{\frac{1}{x+a}} )
    mbox{ was converted to ln similarly } [/i tex]
    I am unable to get the complete msg?
     
    Last edited: Feb 22, 2007
  2. jcsd
  3. Feb 22, 2007 #2
    Have you tried trig. substitutions? If so, which one?

    And, btw, asking others where you have gone wrong without showing your work will not work.
     
  4. Feb 22, 2007 #3
    I HAVE shown my work...its only that i cant get it working with latex. Anyways its more or less working..so plz check again :)
     
  5. Feb 22, 2007 #4

    Dick

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    If your final answer is (ln(a-x)+ln(a+x))/(2*a), there is a sign error on the first log. It should be negative. (Why?). Then if you combine the logs using ln(a)-ln(b)=ln(a/b) you get exactly what you are looking for.
     
  6. Feb 22, 2007 #5

    Ahh fine...i got it. Its s'posed to be negatibe because in 1/(a-x) , x is -ve
    Thx for the help
     
  7. Feb 22, 2007 #6
    Oh, sorry about that. I guess Dick has provided you the answer. :smile:
     
  8. Feb 22, 2007 #7
    As said: there is a trig substitution that you can use that will help out a lot =). Simply, draw a triangle and you will understand what you can substitute it with.

    You should end up with a trig function, sin, cos, csc, sec, tan, cot (x). From there, you can go back to the original equation based on your choice heh.
     
  9. Feb 22, 2007 #8

    D H

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    If you had read the previous posts, you would have seen that this problem has been solved without trig substitutions. He simply multiplied the original integrand by one and added zero (rather creatively), and voila, out came the result. The only problem in the original post was in computing [itex]\int \frac 1{a-x} dx = \ln(a-x)[/itex].
     
  10. Feb 22, 2007 #9

    Dick

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    If you actually do a trig substitution on this, you'll see that it is really the long way around.
     
  11. Feb 23, 2007 #10
    Yeah I thought about trig substitution....IMO i think it has to be either [tex] x=acos\Theta [/tex] or [tex] x=asin \Theta [/tex] But on solving further it still leaves [tex] sec\Theta or cosec\Theta [/tex] to be integrated, which gives something in form of [tex] \frac{1}{a} \ln{\mid cosec\Theta-cot\Theta \mid} +c [/tex]
     
    Last edited: Feb 23, 2007
  12. Feb 23, 2007 #11

    dextercioby

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    You can also use the change of variable [itex] x=a\tanh t [/itex]. The result is immediate.
     
  13. Feb 23, 2007 #12
    I m yet unfamiliar with hyperbolic functions...:shy:
     
  14. Feb 23, 2007 #13

    Gib Z

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    What is it with dexter and his hyperbolic substitutions :P

    Could you do us a favour and show us how it works? I havent worked with hyperbolics much either.
     
  15. Feb 23, 2007 #14

    dextercioby

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    [tex] \int \frac{dx}{a^2 -x^2}\rightarrow \frac{1}{a}\int \frac{dt}{\cosh^2 t} \ \frac{1}{1-\tanh^2 t}=\frac{t}{a}+C\rightarrow \frac{1}{a} \mbox{argtanh}\frac{x}{a} +C [/tex]
     
  16. Feb 23, 2007 #15

    Gib Z

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    assuming thats an "arc" instead of "arg", i see it now. thanks
     
  17. Feb 23, 2007 #16

    dextercioby

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    It's "arg", not "arc" for good reasons.
     
  18. Feb 23, 2007 #17

    Gib Z

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    O right that is so huh? Argument :) Forgot about my imaginary stuff for a while :)
     
  19. Feb 23, 2007 #18

    dextercioby

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    Why do you think it's "arc" for "arctangent", "arcsine",... ?
     
  20. Feb 23, 2007 #19

    Gib Z

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    Arcument :)
     
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