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Math Help - Trig substitution

  1. #1
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    Trig substitution

    \int_1^2 \frac{\sqrt{x^2-1}}{x}dx

    x = sec\theta

    \sqrt{x^2-1} = tan\theta

    dx = sec\theta*tan\theta

     x = 2 ; \theta = \pi/3

     x = 1 ; \theta = 2\pi

    \int_{2\pi}^{\pi/3} \frac{tan\theta}{sec\theta}sec\theta*tan\theta

    This isnt yielding me the correct result, where am I going wrong?

    Thanks
    Last edited by silencecloak; February 26th 2009 at 06:15 PM. Reason: typos
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  2. #2
    Senior Member mollymcf2009's Avatar
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    Quote Originally Posted by silencecloak View Post
    \int_1^2 \frac{\sqrt{x^2-1}}{x}dx

    x = sec\theta

    \sqrt{x^2-1} = tan\theta

    dx = sec\theta*tan\theta

     x = 2 ; \theta = \pi/3

     x = 1 ; \theta = 2\pi

    \int_{2\pi}^{\pi/3} \frac{tan\theta}{sec\theta}sec\theta*tan\theta

    This isnt yielding me the correct result, where am I going wrong?

    Thanks

    I think your new limits should be 0 & \frac{\pi}{3} not 2\pi
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  3. #3
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    Quote Originally Posted by mollymcf2009 View Post
    I think your new limits should be 0 & \frac{\pi}{3} not 2\pi
     1 = sec\theta ; \theta = 0 ?

    Im probably gonna get slammed for this, but that's not what my calculator says
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  4. #4
    Senior Member mollymcf2009's Avatar
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    Quote Originally Posted by silencecloak View Post
     1 = sec\theta ; \theta = 0 ?

    Im probably gonna get slammed for this, but that's not what my calculator says
    arcsec (1) = \Theta = 0
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