calculus
posted by Integration by Parts on .
Use integration by parts to evaluate the integral of x*sec^2(3x).
My answer is
([x*tan(3x)]/3)[ln(sec(3x))/9]
but it's incorrect.
u=x dv=sec^2(3x)dx
du=dx v=(1/3)tan(3x)
[xtan(3x)]/3  integral of(1/3)tan(3x)dx
 (1/3)[ln(sec(3x))/3]
 [ln(sec(3x))/9]
What am I doing wrong?

integral of(1/3)tan(3x)dx
= (1/3)integral of(sin3x/cos 3x) dx
= (1/3)lnsin 3x + C
then the final answer is
(x tan3x)/3  (1/3)lnsin 3x + C