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Integration by parts

  • Thread starter LadiesMan
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96
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[SOLVED] Integration by parts

1. Evaluate

[tex]\int e^{x}sinxdx[/tex]

[Hint: Integrate by parts twice.]


I can't seem to get an answer, but by integrating, the process is redundant (repeats itself).

Thanks

Work:

[tex]\int e^{x}sinxdx[/tex]

Let u = sin x, therefore du = cosxdx
Let [tex]dv = e^{x}dx[/tex], therefore v = [tex]e^{x}[/tex]

Using Integration by parts in Differential Notation

[tex]\int e^{x}sinxdx = e^{x}sinx - \int e^{x}cosxdx[/tex] <--- See how [tex]\int e^{x}cosxdx[/tex] The process of integration will repeat over and over again.

What am I doing wrong?
 
Last edited:

Answers and Replies

1,750
1
Re-type!!!
 
1,750
1
Ok, now show some work!!! You obviously know it's redundant, so show your steps up to when you figured that out.
 
96
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sorry about that. I'm just getting use to the latex sourcing
 
dynamicsolo
Homework Helper
1,648
4
One other thing -- have faith! Apply integration by parts on your new integral, then look at the full equation you end up with for your integral; there is something you'll notice. (The process repeats, but not endlessly...)
 
96
0
yes but it ends back to another integral and then another... however using a different trigonometric function (i.e. sin instead of cos, or vice-versa)

Throughout it makes a process of e^x sinx - e^x cosx.....(This process repeats over and over again)
 
Last edited:
1,750
1
Do it once again, and you will end up with your original Integral. From here it's only Algebra, just bring your original Integral to the left side and divide by the constant, and you're done!
 
96
0
what do you mean divide by the constant?

[tex]\int e^{x}sinxdx = e^{x}sinx - e^{x}cosxdx - \int e^{x}sinxdx[/tex]

Oh wait, I get it! =P

[tex]\int e^{x}sinxdx = 1/2 (e^{x}sinx - e^{x}cosxdx)[/tex] But how do we get a C (Constant)?
 
Last edited:
458
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Just add a +C at the end. No fuss, no hassle.
 
1,750
1
Congrats!
 
96
0
Ok thanks =). So the C came from previous integrations?
 
1,750
1

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