C4 Integration by parts questio Watch

Youk
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Hi, could someone show me the workings for this integration


e^(2x)sin(2x)


I think it's integrated by parts twice, but I just find myself going in circles!


Thanks
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Mr M
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(Original post by Youk)
Hi, could someone show me the workings for this integration


e^(2x)sin(2x)


I think it's integrated by parts twice, but I just find myself going in circles!


Thanks
There's a trick to it. Call the integral I. After you have integrated by parts twice you obtain I again so it appears on both sides of your equation. Now simplify.
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L'Evil Fish
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You'll get your original integral again.

Let that equal I.

Solve for I.

Voila!
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CTArsenal
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(Original post by Youk)
Hi, could someone show me the workings for this integration


e^(2x)sin(2x)


I think it's integrated by parts twice, but I just find myself going in circles!


Thanks
Let I = \int e^{2x}\sin2x\ dx

Your first IBP should look something like:

- \frac {1}{2} e^{2x}\cos2x + \int e^{2x}\cos2x\ dx

Now, integrate again, and you should get back to your first integral, or I

I've put my answer in the spoiler, check it once you've attempted it

Spoiler:
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I = \frac {1}{4} e^{2x}(\sin2x - \cos2x)
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Youk
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(Original post by CTArsenal)
Let I = \int e^{2x}\sin2x\ dx

Your first IBP should look something like:

- \frac {1}{2} e^{2x}\cos2x + \int e^{2x}\cos2x\ dx

Now, integrate again, and you should get back to your first integral, or I

I've put my answer in the spoiler, check it once you've attempted it

Spoiler:
Show
I = \frac {1}{4} e^{2x}(\sin2x - \cos2x)
Thank you all for the help, just a correction for anyone looking at this in the future, the first IBP looks like

 \frac {1}{2} e^{2x}\sin2x - \int e^{2x}\cos2x\ dx
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lazy_fish
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Neither is wrong. You just chose to differentiate and integrate the reverse parts.

Good luck.
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