1. ## 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

2. Originally Posted by bfpri
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 $\displaystyle dv=sec^2(\theta)d\theta$.

3. Ah, Thanks.