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Simple Integral w/ Trigonometric Substitution

  1. May 9, 2005 #1
    Hello everyone, Im having some trouble with an integral.
    [tex] \int \sqrt{x^2 - 1} dx [/tex]

    so far:
    [tex] x = sec \theta [/tex]
    [tex] \frac{dx}{d \theta} = sec \theta tan \theta [/tex]
    [tex] dx = sec \theta tan \theta d\theta [/tex]

    now we substitute:

    [tex] \int \sqrt{x^2 - 1} dx [/tex]
    [tex]= \int \sqrt{sec^2 \theta - 1} sec \theta tan \theta d \theta [/tex]

    since [tex] sec^2 \theta - 1 = tan^2 \theta [/tex]

    [tex]= \int \sqrt{sec^2 \theta - 1} sec \theta tan \theta d \theta = \int \sqrt{tan^2 \theta} sec \theta tan \theta d \theta [/tex]

    [tex]= \int tan^2 \theta sec \theta d \theta [/tex]

    this is where Im stuck. A hint would be appreciated. Thanks in advance


  2. jcsd
  3. May 9, 2005 #2
    Express tangent in terms of secant, separate the two integrals. One will be trivial, and If i remember correctly the other can be done with a substitution.
  4. May 9, 2005 #3
    thanks, I got it, but I had to use Integration by parts like 5 times to get [tex] \int sec^3 \theta d\theta[/tex]


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