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Thread: trig substitution

  1. #1
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    trig substitution

    What have I done wrong

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

    $\displaystyle sec\theta = \frac{x}{3}$

    $\displaystyle x = 3sec\theta$

    $\displaystyle dx = 3sec\theta tan\theta d\theta$

    $\displaystyle \sqrt{x^2-9} = 3tan\theta$

    Therefore...

    $\displaystyle \int \frac{3tan\theta}{3sec\theta} 3sec\theta tan\theta d\theta$

    $\displaystyle \int 3tan^2\theta d\theta$

    $\displaystyle 3 \int sec^2\theta - 1 d\theta$

    $\displaystyle 3tan\theta - 3x + C$
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  2. #2
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    Re: trig substitution

    And I know that equates to

    $\displaystyle \sqrt{x^2-9} - 3x + C$

    But mine book sayth the answer my soul so desireth is:

    $\displaystyle \sqrt{x^2-9} - arctan (\frac {\sqrt {x^2-9}}{3}) + C$
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  3. #3
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    Re: trig substitution

    Quote Originally Posted by Nervous View Post
    What have I done wrong ...

    ...

    $\displaystyle 3 \int sec^2\theta - 1 d\theta$

    $\displaystyle 3tan\theta - {\color{red}3x} + C$
    last line should be

    $\displaystyle 3\tan{\theta} - {\color{red}3\theta} + C$
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