How can integration by parts be used twice to solve ∫ e^at. sinωt dt?

[alex282]

∫ e^at. sinωt dt

This is the second part of an electrical circuit DE problem from our notes (first part not required to solve the above integral) however in-between this integral and the answer our professor only told us that we would get to the answer by using integration by parts twice. I am curious to understand this missed step and have tried integration by parts twice but can't seem to get the same answer as our professor gave us which should work out as:

e^at (asinωt - ωcosωt) / (ω^2 + a^2)

I would really appreciate it if someone could help me here.. I basically tried integrating by parts twice and taking the original negative integral from the second integration by parts to the LHS and then using that to half the RHS and ended up with e^at(sintωt - cosωt)/2, which is different to our lecturers result.

Thanks to anyone for their time!

 

[micromass]

I basically tried integrating by parts twice and taking the original negative integral from the second integration by parts to the LHS and then using that to half the RHS and ended up with e^at(sintωt - cosωt)/2, which is different to our lecturers result.

Could you show us step-by-step what you did??

 

[alex282]

∫ e^at. sinωt dt

u = e^at, u' = ae^at

v' = sinωt, v = -cosωt

∫ uv' = uv - ∫ u'v

-e^at cosωt - ∫ - ae^at cosωt

-e^at cosωt + ∫ ae^at cosωtnow the second integration by parts,

a = ae^at; a' = a^2 e^at;
b' = cosωt; b = sinωt;

ab - ∫ a'b

ae^at sinωt - ∫ a^2 e^at sinωt

divide everything by a^2

e^at sinωt/a - ∫ e^at sinωt

let original integral = y now - ∫ e^at sinωt = -y

take that to RHS, 2y = e^at sinωt/a

y = e^atsinωt/2aThere is my solution, I can't find where I went wrong so please feel free to point out where I screwed up or if anyone can find the solution and post it I should be able to work it out from there

 

[alex282]

Could you show us step-by-step what you did??

Forgot to quote

 

[micromass]

∫ e^at. sinωt dt

u = e^at, u' = ae^at

v' = sinωt, v = -cosωt

It seems to me that you're forgetting a factor \omega here.
If v=-\cos\omega t, then v^\prime= \omega \sin\omega t, no??

You made the same mistake in your second integration by parts.

 

[alex282]

It seems to me that you're forgetting a factor \omega here.
If v=-\cos\omega t, then v^\prime= \omega \sin\omega t, no??

You made the same mistake in your second integration by parts.

Ahh I get it now. Thanks for pointing that out I should be okay from here (hopefully)

 

[ross1219]

Hi, how about this:

∫e^at sinωt dt

u=sinωt du=ωcosωt dt
v=∫e^at dt = 1/a e^at

∫e^at sinωt dt = 1/a e^at sinωt - ∫1/a e^at ωcosωt dt

∫e^at sinωt dt = 1/a e^at sinωt - ω/a ∫e^at cosωt dt

u=cosωt du= -ωsinωt dt
v=∫e^at dt = 1/a e^at

∫e^at sinωt dt = 1/a e^at sinωt - ω/a[ 1/a e^at cosωt - ∫-ω/a e^at sinωt dt ]

∫e^at sinωt dt = 1/a e^at sinωt - ω/a[ 1/a e^at cosωt + ω/a∫e^at sinωt dt ]

∫e^at sinωt dt = 1/a e^at sinωt - ω/a² e^at cosωt - ω²/a² ∫e^at sinωt dt

∫e^at sinωt dt + ω²/a² ∫e^at sinωt dt = 1/a e^at sinωt - ω/a² e^at cosωt

a²∫e^at sinωt dt + ω²∫e^at sinωt dt = ae^at sinωt - ωe^at cosωt

(ω² + a²)∫e^at sinωt dt = e^at(asinωt - ωcosωt)

=> ∫e^at sinωt dt = e^at(asinωt - ωcosωt) / (ω² + a²) + C

Ross :)

 

[Dick]

Yes, that's just great!

 

[alex282]

Hi, how about this:

∫e^at sinωt dt

u=sinωt du=ωcosωt dt
v=∫e^at dt = 1/a e^at

∫e^at sinωt dt = 1/a e^at sinωt - ∫1/a e^at ωcosωt dt

∫e^at sinωt dt = 1/a e^at sinωt - ω/a ∫e^at cosωt dt

u=cosωt du= -ωsinωt dt
v=∫e^at dt = 1/a e^at

∫e^at sinωt dt = 1/a e^at sinωt - ω/a[ 1/a e^at cosωt - ∫-ω/a e^at sinωt dt ]

∫e^at sinωt dt = 1/a e^at sinωt - ω/a[ 1/a e^at cosωt + ω/a∫e^at sinωt dt ]

∫e^at sinωt dt = 1/a e^at sinωt - ω/a² e^at cosωt - ω²/a² ∫e^at sinωt dt

∫e^at sinωt dt + ω²/a² ∫e^at sinωt dt = 1/a e^at sinωt - ω/a² e^at cosωt

a²∫e^at sinωt dt + ω²∫e^at sinωt dt = ae^at sinωt - ωe^at cosωt

(ω² + a²)∫e^at sinωt dt = e^at(asinωt - ωcosωt)

=> ∫e^at sinωt dt = e^at(asinωt - ωcosωt) / (ω² + a²) + C

Ross :)

Thanks for your solution, I couldn't manage to see where the omega squared came from until I read yours

 

[ross1219]

Great! Glad to help. That was fun. :)