how do i do: $\displaystyle \int sqrt{x^2-1}\;dx$
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Originally Posted by micheymichmich 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?
Originally Posted by Jhevon 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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