# calculus

Use integration by parts to evaluate the integral of x*sec^2(3x).

([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?

1. 0
1. integral of(1/3)tan(3x)dx
= (1/3)integral of(sin3x/cos 3x) dx
= (1/3)ln|sin 3x| + C

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

posted by Mohamed

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