Posted by **Integration by Parts** on Saturday, August 30, 2008 at 6:05pm.

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?

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