Calculus II question - integration by parts?

Question

integrate e^3x sin(2x) dx. a step by step process would help. I set u as sin 2x and dv as e^3x but I don't know what to do from there. Thanks!

Answer 1

RESPONSE TO EDIT:

The fraction comes from when I differentiate u and integrate v to get, respectively:

Then du = 2·cos(2·x) dx

Then v = e^(3·x)/3

You have to use the chain rule with differentiation, which means the derivative of 2x (which is 2) gets multiplied by it. Then you have to use the reverse of that to integrate dv into v, which means division by the derivative of 3x (which is 3). That gives 1/3.

They are then used in the next integral ∫ v du

Constants can be factored out of an integral, so I factored out the 2 from du and the 1/3. If you multiply (2)×(1/3), you get 2/3

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The integrals and derivatives of sine and cosine "loop" back and forth, and e^(u) is unchanged. So the goal here is to do parts a couple times until you get a multiple of the original integral again. Then add it to both sides of the equation and divide to find the integral.

∫ e^(3·x)·sin(2·x) dx

——————————————————————————————————————

Let u = sin(2·x)

Then du = 2·cos(2·x) dx

Let dv = e^(3·x) dx

Then v = e^(3·x)/3

——————————————————————————————————————

→ u·v - ∫ v du

= [ sin(2·x) ] · [ e^(3·x)/3 ] - ∫ [ e^(3·x)/3 ] · [ 2·cos(2·x) dx ]

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

——————————————————————————————————————

Let q = cos(2·x)

Then dq = -2·sin(2·x) dx

Let dr = e^(3·x) dx

Then r = e^(3·x)/3

——————————————————————————————————————

→  sin(2·x)·e^(3·x)/3 - (2/3) · { q·r - ∫ r dq }

= sin(2·x)·e^(3·x)/3 - (2/3) · { [ cos(2·x) ] · [ e^(3·x)/3 ] - ∫ [ e^(3·x)/3 ] [ -2·sin(2·x) dx ] }

= sin(2·x)·e^(3·x)/3 - (2/9)·cos(2·x)·e^(3·x) - (4/9)·∫ e^(3·x)·sin(2·x) dx

——————————————————————————————————————

Notice how you now have the original integral on the far right.

Set this whole line equal to the original integral:

∫e^(3·x)·sin(2·x) dx = sin(2·x)·e^(3·x)/3 - (2/9)·cos(2·x)·e^(3·x) - (4/9)·∫e^(3·x)·sin(2·x)dx

Move the "copy" of the original integral over by adding it to both sides. 1 + 4/9 = 13/9:

(13/9)·∫ e^(3·x)·sin(2·x) dx = sin(2·x)·e^(3·x)/3 - (2/9)·cos(2·x)·e^(3·x) + C,

Divide by (13/9) to solve for the original integral:

∫ e^(3·x)·sin(2·x) dx = (3/13)·sin(2·x)·e^(3·x) - (2/13)·cos(2·x)·e^(3·x) + C

You can factor this answer if you want:

= 1/13·[ 3·sin(2·x) - 2·cos(2·x) ]·e^(3·x) + C

Answer 2

these are pretty laborious,

start as you did, and the first integration leads you to:

1/3 e^(3x) sin(2x)-2/3Inegral(e^(3x) Cos(2x)

this gives you another integration by parts, where you set dv=exp^(3x) and u=cos(2x)

this will yield a third integral, and your first reaction will be to think this is an infinite recursion...but...the integral you get from the second integration by parts will be a multiple of the original integral,

Integral (e^(3x) Sin(2x)

once you get to this step, you can solve for the integral algebraically and you should get as your final answer:

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

hope this helps you get started