1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    Regards,

    Nenad
     
  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]

    Regards,

    Nenad
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Simple Integral w/ Trigonometric Substitution
  1. Trigonometric integral (Replies: 4)

Loading...