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Math Help - Integration Help (using trig identities)

  1. #1
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    Integration Help (using trig identities)

    please help! any tips or help would be awesome. ive gotten through part of the problem, but get lost.

    \int\frac{\sqrt{x^2-9}}{x}dx\\

    x=3\sec\Theta\\

    dx=3\sec\Theta\tan\Theta\ d\Theta

    \int\frac{\sqrt{3\sec^2\Theta-9}}{3\sec\Theta}\cdot3\sec\Theta\tan\Theta\ d\Theta

    -clay
    Last edited by claymation14; January 26th 2009 at 08:43 AM.
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  2. #2
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    You're looking good, except you did not square your 3 in the radical after you made the sub.

    \frac{\sqrt{9sec^{2}(t)-9}}{3sec(t)}\cdot 3sec(t)tan(t)dt

    Then, it simplifies to:

    \frac{\sqrt{9(sec^{2}(t)-1)}}{3sec(t)}\cdot 3sec(t)tan(t)dt

    \int 3tan^{2}(t) dt
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  3. #3
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    Quote Originally Posted by galactus View Post
    Then, it simplifies to:

    1 \frac{\sqrt{9(sec^{2}(t)-1)}}{3sec(t)}\cdot 3sec(t)tan(t)dt

    2 \int 3tan^{2}(t) dt
    One question, how did you get from 1 to 2, I understand that sec^2(t)-1=tan^2(t), but what happens to the radical and the denominator?
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  4. #4
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    Obviously, the square root of tan^2(t) gives tan(t).

    The denominator cancels with the 3sec(t) in the dt
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  5. #5
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    Well, you don't actually need any sort of trig. identity, so put u^2=x^2-9 and you'll turn your integral into an arctangent.
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  6. #6
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    Yes, of course, but perhaps the poster is required to use trig sub. That was my impression anyway.
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