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Thread: integration by trigonometric substitution.

  1. #1
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    integration by trigonometric substitution.

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

    $\displaystyle 3sec\theta = x$

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

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

    since secant is hypotenuse over adjacent sine theta must be

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

    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 $\displaystyle \tfrac{1}{9}$ along the way.


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

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

    Substitute: .$\displaystyle \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} \;=\; $

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