Integrating e^-4tsint by Parts

[KateyLou]

Homework Statement

Integrating -e^-4tsint

Homework Equations

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

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!]

 

[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.

 

[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?

 

[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