Trig Substitution

**bfpri** · February 23rd 2010, 09:50 AM

Evaluate the integral

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

$\sqrt[]{9tan^2\Theta+9}$ * $3sec^2\Theta$ which evaluates after factoring to $\sqrt[]{9sec^2\Theta}$ * $3sec^2\Theta$ which is then $3sec\Theta*3sec^2\Theta$ If i take the 9 out of the integral, that leaves $sec^3\Theta$ . And i'm stuck

**Ted** · February 23rd 2010, 10:52 AM

Originally Posted by bfpri

Evaluate the integral

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

$\sqrt[]{9tan^2\Theta+9}$ * $3sec^2\Theta$ which evaluates after factoring to $\sqrt[]{9sec^2\Theta}$ * $3sec^2\Theta$ which is then $3sec\Theta*3sec^2\Theta$ If i take the 9 out of the integral, that leaves $sec^3\Theta$ . And i'm stuck

1- $dx=3sec^2(\theta){\color{red}d\theta}$ .
2- $\int sec^3(\theta) d\theta=\int sec(\theta) sec^2(\theta) d\theta$ , use integration by parts with: $u=sec(\theta)$ and $dv=sec^2(\theta)d\theta$ .

**bfpri** · February 23rd 2010, 01:12 PM

Ah, Thanks.

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