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Math Help - Integrate Sqrt[x^2-a]?

  1. #1
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    Integrate Sqrt[x^2-a]?

    I believe the answer is 0.5(x(x^2-a)^0.5)-0.5aln(x+(x^2-a)^0.5)

    does anyone know how to get this/derive this i.e. not take from tables !!!!

    Thanks
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by coverband View Post
    I believe the answer is 0.5(x(x^2-a)^0.5)-0.5aln(x+(x^2-a)^0.5)

    does anyone know how to get this/derive this i.e. not take from tables !!!!

    Thanks
    that is correct.

    to get the derivative of the answer, you need to use the product rule and the chain rule. are you familiar with these.

    moreover, note that \frac d{dx} \ln u = \frac {u'}u (by the chain rule)

    where u is some function of x
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  3. #3
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    Quote Originally Posted by Jhevon View Post
    that is correct.

    to get the derivative of the answer, you need to use the product rule and the chain rule. are you familiar with these.

    moreover, note that \frac d{dx} \ln u = \frac {u'}u (by the chain rule)

    where u is some function of x
    I could be wrong, but I think s/he was asking how the result was reached.
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  4. #4
    MHF Contributor Mathstud28's Avatar
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    try letting x=\sqrt{a}\cosh(\theta)
    Last edited by mr fantastic; December 4th 2008 at 07:21 PM.
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    mathstud what if a is negative - then we can't consider sqrt(a) ?
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  6. #6
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by coverband View Post
    mathstud what if a is negative - then we can't consider sqrt(a) ?
    Ok then make it \sqrt{x^2+(-a)} and make the sub x=\sqrt{-a}\tan(\theta) or x=\sqrt{-a}\sinh(\theta)
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  7. #7
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    but we want to integrate Sqrt[x^2-a] and i forsee difficulty for if a<0 we can not use sqrt(a) in our substitutions
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  8. #8
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by coverband View Post
    but we want to integrate Sqrt[x^2-a] and i forsee difficulty for if a<0 we can not use sqrt(a) in our substitutions
    I edited my post
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  9. #9
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    Quote Originally Posted by Mathstud28 View Post
    Ok then make it \sqrt{x^2+(-a)} and make the sub x=\sqrt{-a}\tan(\theta) or x=\sqrt{-a}\sinh(\theta)

    but you took the square root of a neg number there. Is that allowed ?
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  10. #10
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by coverband View Post
    but you took the square root of a neg number there. Is that allowed ?
    But you said what if a<0\implies{-a>0} so yes
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  11. #11
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    Quote Originally Posted by coverband View Post
    you're a very stupid man jhevon
    And you're banned for 2 weeks to cool off. None of that.
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