Q: ∫e^(2x) * sin(3x) dx
A: This is a more involved, integration by parts problem. So, I will assume you know the basics of integration. If you need more explanation on any step, let me know!
Integration by Parts (the gist): ∫udv = uv – ∫vdu
So, I always fill in the following chart:
u = v =
du = dv =
Remember, we get to “select” the u and the dv. The v and the du are found via differentiation or integration. So, I select the following
u = e^(2x) v =
du = dv = sin(3x) dx
Now, fill in the rest:
u = e^(2x) v = – 1/3 cos(3x)
du = 2e^(2x) dx dv = sin(3x) dx
So, we know that the solution to our original problem is:
∫udv = uv – ∫vdu
And, I plug in the parts to get (with a little house-keeping):
(1) ∫e^(2x) * sin(3x) dx = – 1/3 e^(2x) * cos(3x) + 2/3 ∫e^(2x) * cos(3x) dx
Hmmm. Now we have another integration to do (by parts even)! We will select the “e” part to be u again, and the “cos” part to be dv. This way, we do not undo what we just did. So, just removing the integration above for our analysis, we have:
(2) ∫e^(2x) * cos(3x) dx
u = e^(2x) v = 1/3 sin(3x)
du = 2e^(2x) dx dv = cos(3x) dx
∫e^(2x) * cos(3x) dx = 1 / 3 e^(2x) * sin(3x) – 2/3 ∫e^(2x) *sin(3x) dx
Notice that the answer above contains the original problem?!? This is great news. This is what I call a “cycling” problem. Watch below:
So, going back to equation (1):
∫e^(2x) * sin(3x) dx = – 1/3 e^(2x) * cos(3x) + 2/3 ∫e^(2x) * cos(3x) dx
Plug in our solution for (2):
∫e^(2x) * sin(3x) dx = – 1/3 e^(2x) * cos(3x) + 2/3 [1 / 3 e^(2x) * sin(3x) – 2/3 ∫e^(2x) *sin(3x)]
Clean house:
∫e^(2x) * sin(3x) dx = – 1/3 e^(2x) * cos(3x) + 2/9 e^(2x) * sin(3x) – 4/9 ∫e^(2x) *sin(3x) dx
Add [4/9 ∫e^(2x) *sin(3x) dx] to each side of the equation to get:
13/9 ∫e^(2x) *sin(3x) = – 1/3 e^(2x) * cos(3x) + 2/9 e^(2x) * sin(3x) dx
Multiply both sides by 9/13:
∫e^(2x) *sin(3x) dx = – 9/39 e^(2x) * cos(3x) + 2/13 e^(2x) * sin(3x)
Tada!
Thank you very much the explanation makes it so easy!
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Glad it was helpful! These cycling problems can be messy and require good “house keeping”.
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you aint kiddin they do! Had a couple of these for homework and now that you showed me this one I can get through the mess of the other ones lol
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