Integrate sqrt (x^2 -4) using the substitution x = 2sec@ I had: x = 2sec@ dx = 2sec@tan@ sqrt (4sec^2 @ -4) x 2sec@tan@ d@ = 4tan^2 @ sec@ and now I'm stuck. Thanks
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Originally Posted by differentiate Integrate sqrt (x^2 -4) using the substitution x = 2sec@ I had: x = 2sec@ dx = 2sec@tan@ sqrt (4sec^2 @ -4) x 2sec@tan@ d@ = 4tan^2 @ sec@ and now I'm stuck. Thanks Ted had written answer before I did. So I deleted mine. Anyway this integration could be easily done if you use x=2sin@. Hope this helps you.
Originally Posted by differentiate Integrate sqrt (x^2 -4) using the substitution x = 2sec@ I had: x = 2sec@ dx = 2sec@tan@ sqrt (4sec^2 @ -4) x 2sec@tan@ d@ = 4tan^2 @ sec@ and now I'm stuck. Thanks $\displaystyle \int tan^2(\theta) sec(\theta) \, d\theta=\int (sec^2(\theta)-1)sec(\theta) \, d\theta=\int sec^3(\theta) \, d\theta - \int sec(\theta) \, d\theta$ the first one is a well-known integral and you should know how to integrate the second one
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