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Math Help - integration by trigonometric substitution.

  1. #1
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    integration by trigonometric substitution.

    \int \frac{\1}{(x^2*\sqrt{x^2-9}}dx

    3sec\theta = x

    \int \frac{3sec\theta tan\theta}{sec^2\theta\sqrt{9tan^2\theta}}dx

    \int sec\theta dx = sin\theta+c

    since secant is hypotenuse over adjacent sine theta must be

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

    the book has a nine in the denominator and I can't figure out where it comes from.
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  2. #2
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    Re: integration by trigonometric substitution.

    Hello, bkbowser!

    You dropped a \tfrac{1}{9} along the way.


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

    Let x = 3\sec\theta \quad\Rightarrow\quad dx = 3\sec\theta\tan\theta\,d\theta

    Substitute: . \int \frac{3\sec\theta\tan\theta\,d\theta}{(3\sec\theta  )^2\sqrt{9\tan^2\theta}} \;=\;\int\frac{3\sec\theta\tan\theta\,d\theta}{9 \sec^2\theta\cdot3\tan\theta} \;=\;

    . . . . . . . . . =\;{\color{red}\frac{1}{9}}\int\frac{d\theta}{\sec  \theta}
    Thanks from bkbowser
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