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?

## Answer This Question

## Related Questions

- Integration - Intergrate ¡ì sec^3(x) dx could anybody please check this answer. ...
- calculus (check my work please) - Not sure if it is right, I have check with the...
- calculus - find dy/dx y=ln (secx + tanx) Let u= secx + tan x dy/dx= 1/u * du/dx ...
- Calculus 12th grade (double check my work please) - 2- given the curve is ...
- Calculus - Evaluate the indefinite integral integral sec(t/2) dt= a)ln |sec t +...
- math - How do I derive the secant reduction rule? Integral (sec x)^n dx = ...
- calculus - Integrate: dx/sqrt(x^2-9) Answer: ln(x + sqrt(x^2 - 9)) + C I'm ...
- Calculus AP - I'm doing trigonometric integrals i wanted to know im doing step ...
- Calculus - Integration - Hello! I really don't think I am understanding my calc ...
- calculus - So I am suppose to evaulate this problem y=tan^4(2x) and I am ...

More Related Questions