Saturday

November 1, 2014

November 1, 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

**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 ...

math - How do I derive the secant reduction rule? Integral (sec x)^n dx = ...

Calculus - Evaluate the indefinite integral integral sec(t/2) dt= a)ln |sec t +...

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 ...

Calculus - could anybody please explain how sec x tan x - ¡ì sec x tan^2(x) dx...