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April 18, 2015

April 18, 2015

Posted by **Confused** on Friday, February 15, 2013 at 11:29pm.

I used integration by parts and I let u=e^2x and dv=sin(3x)

My final answer was (-3/8)(e^(2x))(cos(3x)) - (1/8)(e^(2x))(sin(3x)) but it's wrong. Please help!!

- Calculus Problem -
**Reiny**, Saturday, February 16, 2013 at 12:10amu = e^(2x)

du/dx = 2 e^(2x)

du = 2 e^(2x) dx

dv = sin(3x) dx

v = -(1/3) cos(3x)

∫e^(2x)sin(3x) dx

= (e^(2x) )(-1/3)cos(3x) - ∫(-2/3)(e^(2x))(cos(3x) ) dx

=(e^(2x) )(-1/3)cos(3x) + (2/3)∫( e^(2x) (cos(3x)) dx

do it again on that last part:

let u = e^2x

du/dx = 2e^(2x)

let dv = cos(3x) dx

v = (1/3)sin(3x)

and ∫( e^(2x) (cos(3x)) dx

= (1/3)e^(2x) (sin(3x)) - ∫(1/3)sin(3x) (2e^(2x)) dx

= (1/3)e^(2x) (sin(3x)) - (2/3) ∫e^(2x) sin(3x) dx

but look at that last part, isn't that what we started with on the original left side of our equation

let ∫e^(2x)sin(3x) dx = A

so we have

A = (e^(2x) )(-1/3)cos(3x) + (2/3)[(1/3)e^(2x) (sin(3x)) - (2/3) A ]

A = (-1/3)(e^(2x)) (cos(3x) + (2/9)e^(2x) sin(3x) - (4/9)A

times 9

9A = -3 e^(2x) cos(3x) + 2 e^(2x) sin(3x) - 4A

13A = -3 e^(2x) cos(3x) + 2 e^(2x) sin(3x)

A = (-1/13) ( 3 e^(2x)cos(3x) - 2 e^(2x) sin(3x)

∫e^(2x)sin(3x) dx =**(-1/13) ( 3 e^(2x)cos(3x) - 2 e^(2x) sin(3x) )**

or as Wolfram has it

(1/13) ( 2 e^(2x) sin(3x) - 3 e^(2x)cos(3x) )

http://integrals.wolfram.com/index.jsp?expr=e%5E%282x%29sin%283x%29+&random=false

- Calculus Problem -
**Confused**, Saturday, February 16, 2013 at 1:02amThank you so much for showing the steps, I really appreciate the help! (:

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