# Integrating e^-4tsint

## 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!]

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Dick
Homework Helper
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.

$$\int$$-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?

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