# Integrating e^-4tsint

1. Dec 27, 2008

### KateyLou

1. The problem statement, all variables and given/known data

Integrating -e-4tsint

2. Relevant equations

Our tutor suggested we did this by integrating by parts in a cyclic fashion and things would cancel out

3. The attempt at a solution

Taking u = sint t; thus u'=cost
And v'=-e-4t; thus (1/4)e-4t

$$\int-e$$4tsint = (1/4)e-4tsint-$$\int$$(1/4)e-4tcost

Taking u=cost thus u'= -sint
And v'=(1/4)e-4t; thus (-1/16)e-4t

$$\int-e$$4tsint = (1/4)e-4tsint+(1/16)e-4tcost+(1/16)$$\int$$e-4tsint

Nothing appears to cancel!]

2. Dec 27, 2008

### Dick

Yes it does. You have the integral of e^(-4t)*sin(t) on both sides. Put them both on the same side and solve for it.

3. Dec 27, 2008

### KateyLou

$$\int$$-e-4tsint=[(1/4)sinte-4t+[(1/16)e-4tcost+(1/16)$$\int$$e-4t

(-17/16)$$\int$$e-4tsint=e-4t((1/4)sint+(1/16)cos))

Not sure what to do now?

4. Dec 27, 2008

### KateyLou

oooh

$$\int$$-e-4tsint=(16/17)e-4t((1/4)sint+(1/16)cos))

or

$$\int$$-e-4tsin=e-4t(4sint+cost)

YAY that what the answer sheet says