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Thread: question about integral

  1. #1
    Newbie micheymichmich's Avatar
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    question about integral

    how do i do: $\displaystyle \int sqrt{x^2-1}\;dx$
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by micheymichmich View Post
    how do i do: $\displaystyle \int sqrt{x^2-1}\;dx$
    I suppose you mean $\displaystyle \int \sqrt{x^2 - 1}~dx$

    we can use trigonometric substitution here. try $\displaystyle x = \sec \theta$

    can you continue?
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  3. #3
    Newbie micheymichmich's Avatar
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    Quote Originally Posted by Jhevon View Post
    I suppose you mean $\displaystyle \int \sqrt{x^2 - 1}~dx$

    we can use trigonometric substitution here. try $\displaystyle x = \sec \theta$

    can you continue?
    yes. thank you
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