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August 29, 2014

August 29, 2014

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

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?

- calculus -
**Mohamed**, Saturday, August 30, 2008 at 11:08pmintegral of(1/3)tan(3x)dx

= (1/3)integral of(sin3x/cos 3x) dx

= (1/3)ln|sin 3x| + C

then the final answer is

(x tan3x)/3 - (1/3)ln|sin 3x| + C

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