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## Main Question or Discussion Point

does any one know how to integrate e^(-2x)(cos(3x)dx

using eulers formula e^(ix)= i(sinx)+cosx

using eulers formula e^(ix)= i(sinx)+cosx

- Thread starter cragar
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- #1

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does any one know how to integrate e^(-2x)(cos(3x)dx

using eulers formula e^(ix)= i(sinx)+cosx

using eulers formula e^(ix)= i(sinx)+cosx

- #2

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Use eular's identity to express the above equation as an exponential function.does any one know how to integrate e^(-2x)(cos(3x)dx

using eulers formula e^(ix)= i(sinx)+cosx

- #3

HallsofIvy

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[tex]cos(3x)= \frac{e^{3ix}+ e^{-3ix}}{2}[/tex]

[tex]e^{2x}cos(3x)= \frac{e^{(2+3i)x}+ e^{(2-3i)}}{2}[/tex]

- #4

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ok i get it now thanks

- #5

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k i think i get it now but how would we integrate e^(2x)*(sin(-2x))dx

- #6

HallsofIvy

Science Advisor

Homework Helper

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[tex]e^{2x}sin(-2x)= -e^{2x}sin(2x)[tex]

[tex]sin(2x)= \frac{e^{2x}- e^{-2x}}{2i}[/tex]

[tex]sin(2x)= \frac{e^{2x}- e^{-2x}}{2i}[/tex]

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