Integrating e^-4tsint

  • Thread starter KateyLou
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  • #1
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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

[tex]\int-e[/tex]4tsint = (1/4)e-4tsint-[tex]\int[/tex](1/4)e-4tcost

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

[tex]\int-e[/tex]4tsint = (1/4)e-4tsint+(1/16)e-4tcost+(1/16)[tex]\int[/tex]e-4tsint

Nothing appears to cancel!]
 

Answers and Replies

  • #2
Dick
Science Advisor
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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
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[tex]\int[/tex]-e-4tsint=[(1/4)sinte-4t+[(1/16)e-4tcost+(1/16)[tex]\int[/tex]e-4t

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

Not sure what to do now?
 
  • #4
17
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oooh

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

or

[tex]\int[/tex]-e-4tsin=e-4t(4sint+cost)

YAY that what the answer sheet says
 

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