Posted by Confused on Friday, February 15, 2013 at 11:29pm.
u = 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
Thank you so much for showing the steps, I really appreciate the help! (:
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