Integral: e^{(ax)} cos(bx) - Solve with Integration by Parts

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Homework Help Overview

The original poster is attempting to solve the integral of the form \(\int e^{(ax)} \cos(bx) \, dx\). They have reached an intermediate step involving integration by parts but feel stuck and are seeking assistance.

Discussion Character

  • Exploratory, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the steps taken in integration by parts and the resulting expressions. There is mention of the challenges faced when trying to isolate the original integral and the confusion surrounding the signs in the process.

Discussion Status

Some participants have offered guidance on how to proceed with the integral, suggesting a reevaluation of the integration steps. There is an ongoing exploration of the implications of the signs in the integration process, but no consensus has been reached on a definitive method.

Contextual Notes

The original poster expresses frustration with the LaTeX formatting, indicating potential communication barriers in conveying mathematical expressions clearly. This may affect the clarity of the discussion.

The Bob
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Hi all,

I am having problems with the integral:

\int e^{(ax)} cos(bx) dx

I have got to \frac{e^{ax} sin(bx)}{b} - \int \frac{a e^{ax} sin(bx)}{b} dx

After this I can only see myself going around in circles.

Any help would be appreciated. :smile:

Cheers,

The Bob (2004 ©)
 
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The Bob said:
\int e^{(ax)} cos(bx) dx

I have got to \frac{e^{ax} sin(bx)}{b} - \int \frac{a e^{ax} sin(bx)}{b} dx

The integrals should be:

\int e^{(ax)} cos(bx) dx

and

\frac{e^{ax} sin(bx)}{b} - \int \frac{a e^{ax} sin(bx)}{b} dx

LaTex doesn't seem to be editable anymore. :frown:

The Bob (2004 ©)
 
Go again, taking the integral on the RHS Since you get a -cos(x), where you would normally have subtraction from your integration by parts you get addition, except the whole thing is negative anyway, so your new integral (which is a bunch of constants times eaxcos(bx) ) turns out negative. Add that to both sides, and multiply/divide by constants to isolate your original integral
 
Office_Shredder said:
Go again, taking the integral on the RHS Since you get a -cos(x), where you would normally have subtraction from your integration by parts you get addition, except the whole thing is negative anyway, so your new integral (which is a bunch of constants times eaxcos(bx) ) turns out negative. Add that to both sides, and multiply/divide by constants to isolate your original integral
Cheers Office_Shredder. Sorry for the late reply, I have been very busy recently.

Thanks so much again :biggrin:,

The Bob (2004 ©)
 

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