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stripes
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Homework Statement
Gosh I've been asking a lot of questions lately...anyways...
I tried this question two separate times and couldn't manage to figure out where i went wrong...
[tex]\int e^{-x}cos2x dx[/tex]
Homework Equations
[tex] uv - \int v du = \int u dvdx
[/tex]
The Attempt at a Solution
let [itex]u = e^{-x}[/itex] and [itex]dv = cos2x dx[/itex]
so [itex]du = -e^{-x}[/itex] and [itex]v = \frac{1}{2}sin2x[/itex]
[tex]\int e^{-x}cos2x dx = \frac{1}{2}e^{-x}sin2x + \frac{1}{2}\int e^{-x}sin2x dx[/tex]
use integration by parts again, this time let [itex]u = e^{-x}[/itex] and [itex]dv = sin2x dx[/itex] so [itex]du = -e^{-x}[/itex] and [itex]v = \frac{-1}{2}cos2x[/itex]
so now we have:
[tex]\int e^{-x}cos2x dx = \frac{1}{2}e^{-x}sin2x - \frac{1}{2}e^{-x}cos2x - \frac{1}{2}\int e^{-x}cos2x dx[/tex]
bring the [itex]\frac{1}{2}\int e^{-x}cos2x dx[/itex] to the LHS and solve for the integral.
[tex]\frac{3}{2}\int e^{-x}cos2x dx = \frac{1}{2}e^{-x}sin2x - \frac{1}{2}e^{-x}cos2x[/tex]
so finally:
[tex]\int e^{-x}cos2x dx = \frac{1}{3}e^{-x}(sin2x - cos2x)[/tex]
which is incorrect... at which step did I go wrong?
thank you all in advance!
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