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Thread: Trig Substitution

  1. #1
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    Trig Substitution

    Evaluate the integral


    so I did x=atanΘ. which is x=3tanΘ and dx is $\displaystyle 3sec^2\Theta$. Then it is

    $\displaystyle \sqrt[]{9tan^2\Theta+9}$*$\displaystyle 3sec^2\Theta$ which evaluates after factoring to $\displaystyle \sqrt[]{9sec^2\Theta}$*$\displaystyle 3sec^2\Theta$ which is then $\displaystyle 3sec\Theta*3sec^2\Theta$ If i take the 9 out of the integral, that leaves $\displaystyle sec^3\Theta$. And i'm stuck
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  2. #2
    Ted
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    Quote Originally Posted by bfpri View Post
    Evaluate the integral


    so I did x=atanΘ. which is x=3tanΘ and dx is $\displaystyle 3sec^2\Theta$. Then it is

    $\displaystyle \sqrt[]{9tan^2\Theta+9}$*$\displaystyle 3sec^2\Theta$ which evaluates after factoring to $\displaystyle \sqrt[]{9sec^2\Theta}$*$\displaystyle 3sec^2\Theta$ which is then $\displaystyle 3sec\Theta*3sec^2\Theta$ If i take the 9 out of the integral, that leaves $\displaystyle sec^3\Theta$. And i'm stuck
    1- $\displaystyle dx=3sec^2(\theta){\color{red}d\theta}$.
    2- $\displaystyle \int sec^3(\theta) d\theta=\int sec(\theta) sec^2(\theta) d\theta$ , use integration by parts with: $\displaystyle u=sec(\theta)$ and $\displaystyle dv=sec^2(\theta)d\theta$.
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  3. #3
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    Ah, Thanks.
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