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