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Integration by parts

  1. Mar 26, 2008 #1
    [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).



    [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: Mar 26, 2008
  2. jcsd
  3. Mar 26, 2008 #2
  4. Mar 26, 2008 #3
    Ok, now show some work!!! You obviously know it's redundant, so show your steps up to when you figured that out.
  5. Mar 26, 2008 #4
    sorry about that. I'm just getting use to the latex sourcing
  6. Mar 26, 2008 #5


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    Homework Helper

    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...)
  7. Mar 26, 2008 #6
    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: Mar 26, 2008
  8. Mar 26, 2008 #7
    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!
  9. Mar 26, 2008 #8
    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: Mar 26, 2008
  10. Mar 26, 2008 #9
    Just add a +C at the end. No fuss, no hassle.
  11. Mar 26, 2008 #10
  12. Mar 26, 2008 #11
    Ok thanks =). So the C came from previous integrations?
  13. Mar 26, 2008 #12
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