Thursday
July 31, 2014

Homework Help: calculus

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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