Integration by Parts

  • Thread starter The Bob
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  • #1
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Hi all,

I am having problems with the integral:

[tex]\int e^{(ax)} cos(bx) dx [/tex]

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

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

Any help would be appreciated. :smile:

Cheers,

The Bob (2004 ©)
 

Answers and Replies

  • #2
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The Bob said:
[tex]\int e^{(ax)} cos(bx) dx [/tex]

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

The integrals should be:

[tex]\int e^{(ax)} cos(bx) dx [/tex]

and

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

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

The Bob (2004 ©)
 
  • #3
Office_Shredder
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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
 
  • #4
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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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