Integrate by Parts: Solve \int e^{x}sinxdx

[LadiesMan]

[SOLVED] Integration by parts

1. Evaluate

\int e^{x}sinxdx

[Hint: Integrate by parts twice.]

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

Thanks

Work:

\int e^{x}sinxdx

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

Using Integration by parts in Differential Notation

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

What am I doing wrong?

 

[rocomath]

Re-type!

 

[rocomath]

Ok, now show some work! You obviously know it's redundant, so show your steps up to when you figured that out.

 

[LadiesMan]

sorry about that. I'm just getting use to the latex sourcing

 

[dynamicsolo]

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...)

 

[LadiesMan]

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)

 

[rocomath]

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!

 

[LadiesMan]

what do you mean divide by the constant?

\int e^{x}sinxdx = e^{x}sinx - e^{x}cosxdx - \int e^{x}sinxdx

Oh wait, I get it! =P

\int e^{x}sinxdx = 1/2 (e^{x}sinx - e^{x}cosxdx) But how do we get a C (Constant)?

 

[Snazzy]

Just add a +C at the end. No fuss, no hassle.

 

[rocomath]

Congrats!

 

[LadiesMan]

Ok thanks =). So the C came from previous integrations?

 

[rocomath]

Ok thanks =). So the C came from previous integrations?

Yes.