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Integral of y=sqrt.(x^2+a^2) or y^2=x^2+a^2

  1. Jul 28, 2009 #1
    how do you integrate this...
    y=sqrt.(x^2+a^2)

    would greatly appreciate any help. thanks
     
  2. jcsd
  3. Jul 28, 2009 #2

    Cyosis

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    Easiest way to do this is to make the the substitution [itex]x=a \sinh u[/itex]. Note that [itex]\cosh^2 u-\sinh^2 u=1[/itex]. Try it out! If you're totally unfamiliar with hyperbolic functions use [itex]x=\tan u[/itex] instead.
     
    Last edited: Jul 28, 2009
  4. Jul 29, 2009 #3
    would i be right in thinking that cosh(x)=cos(ix) and sinh(x)=-isin(ix) and therefore
    cosh(x)+sinh(x)=e^x?
     
  5. Jul 29, 2009 #4

    Cyosis

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    Yes that's correct, although I don't see how this is relevant to your problem.
     
  6. Jul 29, 2009 #5

    HallsofIvy

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    Since [itex]sin^2(u)+ cos^2(u)= 1[/itex] leads to [itex]tan^2(u)+ 1= sec^2(u)[/itex] (divide both sides by [itex]cos^2(u)[/itex]), you could also make the substitution ax= tan(u).
     
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