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Integration with k as constant

  1. Aug 5, 2008 #1
    1. The problem statement, all variables and given/known data


    PLease HELP ME INTEGRATE THIS>
    [tex]\int\frac{dh}{\sqrt{h^{2}-k^{2}}}[/tex]
    k is constant
     
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  3. Aug 5, 2008 #2

    Kurdt

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    Re: Integration:

    You will have to show some attempt first.
     
  4. Aug 5, 2008 #3
    Re: Integration:

    If I take h^2 common from the integral, it will be:
    [tex]\int\frac{dh}{h\sqrt{1-\frac{k^{2}}{h^{2}}}}[/tex]
    which finally gives:
    [tex]\frac{\sqrt{1-\frac{k^{2}}{h^{2}}}}{k^{2}}[/tex]
     
  5. Aug 5, 2008 #4

    Defennder

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    Re: Integration:

    I don't see how that follows. Your first step is right, next you should use a trigo substitution.
     
  6. Aug 9, 2008 #5
    Re: Integration:

    I am redefining the question:
    [tex]\int\frac{dx}{\sqrt{x^{2}-a^{2}}}[/tex]


    taking x^2 common

    [tex]\int\frac{dx}{x\sqrt{1-\frac{a^{2}}{x^{2}}}}[/tex]

    let [tex]1-\frac{a^{2}}{x^{2}}=t[/tex]
    [tex]\frac{1}{2}\int\frac{dt}{(1-t)(\sqrt{t})}[/tex]
    let[tex]\sqrt{t}=j[/tex]
    [tex]\int\frac{dj}{1-j^{2}}[/tex]

    how shall i proceed
     
  7. Aug 9, 2008 #6

    Kurdt

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    Re: Integration:

    As mentioned before trig substitution is the way to go from the very start.
     
  8. Aug 9, 2008 #7
    Re: Integration:

    I also agree with others, but assuming that you have the last step right:
    [tex]\int\frac{dj}{1-j^{2}}[/tex]

    You should do decomposition and partial fraction ...
     
  9. Aug 9, 2008 #8

    Defennder

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    Re: Integration:

    You've made it a lot more tedious than it could have been with a simple trigo substitution.
     
  10. Aug 9, 2008 #9
    Re: Integration:

    please be more explicit. Actually there is no sign of trigo ratio in the answer :S
     
  11. Aug 9, 2008 #10

    Kurdt

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  12. Aug 9, 2008 #11

    HallsofIvy

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    Re: Integration:

    Pretty much every step is wrong. You seem to be consistly forgetting to replace the "dx" or "dt" term.
    For example, if you let [tex]1-\frac{a^2}{x^2}=t[/tex] then [tex]\frac{2a^2}{x^3}dx= dt[/tex]. How are you going to substitute for that?

    Even if that were correct when you say "let [tex]\sqrt{t}= j[/tex]", you should have [tex]\frac{1}{2\sqrt{t}}dt= dj[/tex].

    As everyone has been telling you from the start, use a trig substitution. [tex]sin^2 \theta+ cos^2 \theta= 1[/tex] so [tex]tan^2 \theta+ 1= sec^2 \theta[/tex].
     
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