Trig Substitution: Solving Integrals with sec^3Θ

[bfpri]

http://img708.imageshack.us/img708/8897/symimage.gif

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

 

[Mark44]

See the wikipedia article - http://en.wikipedia.org/wiki/Integral_of_secant_cubed