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C4 Integration by parts questio

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 :smile:
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 :smile:


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.
You'll get your original integral again.

Let that equal I.

Solve for I.

Voila!
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 :smile:


Let I=∫e2xsin⁡2x dxI = \int e^{2x}\sin2x\ dx

Your first IBP should look something like:

−12e2xcos⁡2x+∫e2xcos⁡2x dx- \frac {1}{2} e^{2x}\cos2x + \int e^{2x}\cos2x\ dx

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

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

Spoiler

(edited 9 years ago)
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Youk
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Original post by CTArsenal
Let I=∫e2xsin⁡2x dxI = \int e^{2x}\sin2x\ dx

Your first IBP should look something like:

−12e2xcos⁡2x+∫e2xcos⁡2x dx- \frac {1}{2} e^{2x}\cos2x + \int e^{2x}\cos2x\ dx

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

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

Spoiler



Thank you all for the help, just a correction for anyone looking at this in the future, the first IBP looks like

12e2xsin⁡2x−∫e2xcos⁡2x dx \frac {1}{2} e^{2x}\sin2x - \int e^{2x}\cos2x\ dx
Neither is wrong. You just chose to differentiate and integrate the reverse parts.

Good luck. :smile:

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