(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

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]

2. Relevant equations

[tex] uv - \int v du = \int u dvdx

[/tex]

3. 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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# Homework Help: Integration by parts question

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