# Integration problem

• Mar 3rd 2007, 10:43 AM
clockingly
Integration problem
i have to integrate e^(2x) * cos(3x) dx by parts.

so far, I have u = e^(2x), u' = 2e^(2x), v' = cos(3x), and v = 1/3 sin (3x)

This means that the integral of e^(2x) * cos(3x) dx is:

e^(2x) * 1/3 sin (3x) - the integral of 2e^(2x) * 1/3 sin(x)

I'm not exactly sure where to go from here, so any help would be appreciated!
• Mar 3rd 2007, 11:22 AM
Jhevon
Quote:

Originally Posted by clockingly
i have to integrate e^(2x) * cos(3x) dx by parts.

so far, I have u = e^(2x), u' = 2e^(2x), v' = cos(3x), and v = 1/3 sin (3x)

This means that the integral of e^(2x) * cos(3x) dx is:

e^(2x) * 1/3 sin (3x) - the integral of 2e^(2x) * 1/3 sin(x)

I'm not exactly sure where to go from here, so any help would be appreciated!

Personally, I would have chosen e^(2x) as the function to integrate, but let's do it your way.

int{e^(2x)cos(3x)}dx
= (1/3)sin(3x)e^(2x) - int{2(1/3)sin(3x)e^(2x)}dx
= (1/3)sin(3x)e^(2x) - (2/3)int{sin(3x)e^(2x)}dx .........do this last integral by parts again
= (1/3)sin(3x)e^(2x) - (2/3)[(-1/3)cos(3x)e^(2x) - int{2(-1/3)cos(3x)e^(2x)}dx]
= (1/3)sin(3x)e^(2x) - (2/3)[(-1/3)cos(3x)e^(2x) + (2/3)int{cos(3x)e^(2x)}dx]
= (1/3)sin(3x)e^(2x) - [(-2/9)cos(3x)e^(2x) + (4/9)int{cos(3x)e^(2x)}dx]
= (1/3)sin(3x)e^(2x) + (2/9)cos(3x)e^(2x) - (4/9)int{cos(3x)e^(2x)}dx

So now we have:
int{e^(2x)cos(3x)}dx = (1/3)sin(3x)e^(2x) + (2/9)cos(3x)e^(2x) - (4/9)int{cos(3x)e^(2x)}dx

So we solve this equation for int{e^(2x)cos(3x)}dx

=> int{e^(2x)cos(3x)}dx + (4/9)int{cos(3x)e^(2x)}dx = (1/3)sin(3x)e^(2x) + (2/9)cos(3x)e^(2x)
=> (13/9)int{e^(2x)cos(3x)}dx = (1/3)sin(3x)e^(2x) + (2/9)cos(3x)e^(2x)
=> int{e^(2x)cos(3x)}dx = (9/13)[(1/3)sin(3x)e^(2x) + (2/9)cos(3x)e^(2x)]
=> int{e^(2x)cos(3x)}dx = (3/13)sin(3x)e^(2x) + (2/13)cos(3x)e^(2x) + C

Also, see the last note of this post: http://www.mathhelpforum.com/math-he...l-problem.html