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tangibleLime
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Homework Statement
Integrate.
[tex]\int e^{2\theta}sin(9\theta) d\theta[/tex]
[tex]\int e^{2\theta}sin(9\theta) d\theta[/tex]
Homework Equations
Integration by Parts
[tex]\int udv = uv - \int vdu[/tex]
[tex]\int udv = uv - \int vdu[/tex]
The Attempt at a Solution
I started out with integration by parts (IPB).
This was marked wrong, even after I simplified the answer (which I did not do here, but I checked the simplification with WolframAlpha). Any suggestions?
[tex]u = e^{2\theta}[/tex]
[tex]du = 2e^{2\theta} d\theta[/tex]
[tex]v = \frac{-1}{9}cos(9\theta)[/tex]
[tex]dv = sin(9\theta) d\theta[/tex]
[tex]uv - \int vdu = -\frac{1}{9}cos(9\theta)*e^{2\theta}+\frac{2}{9}\int cos(9\theta) d\theta[/tex]
Noticing that the new integral on the right is just about the same as how I started (sin just turned into cos), I did IPB once again.[tex]du = 2e^{2\theta} d\theta[/tex]
[tex]v = \frac{-1}{9}cos(9\theta)[/tex]
[tex]dv = sin(9\theta) d\theta[/tex]
[tex]uv - \int vdu = -\frac{1}{9}cos(9\theta)*e^{2\theta}+\frac{2}{9}\int cos(9\theta) d\theta[/tex]
[tex]u = e^{2\theta}[/tex]
[tex]du = 2e^{2\theta} d\theta[/tex]
[tex]v = \frac{1}{9}sin(9\theta}[/tex]
[tex]dv = cos(9\theta) d\theta[/tex]
[tex]uv - \int vdu = \frac{1}{9}sin(9\theta)*e^{2\theta}-\frac{2}{9}\int sin(9\theta)*e^{2\theta} d\theta[/tex]
Combining all of this into the equation so far...[tex]du = 2e^{2\theta} d\theta[/tex]
[tex]v = \frac{1}{9}sin(9\theta}[/tex]
[tex]dv = cos(9\theta) d\theta[/tex]
[tex]uv - \int vdu = \frac{1}{9}sin(9\theta)*e^{2\theta}-\frac{2}{9}\int sin(9\theta)*e^{2\theta} d\theta[/tex]
[tex]-\frac{1}{9}cos(9\theta)*e^{2\theta}+\frac{2}{81}sin(9\theta)*e^{2\theta}-\frac{2}{9}\int sin(9\theta)*e^{2\theta} d\theta[/tex]
Noticing that this new integral on the right is the exactly equal to what I started with, I added it to both sides in order to remove it from the right side. I mentally noted that the left version could have a coefficient of [tex]\frac{9}{9}[/tex], and added the right side's coefficient of [tex]\frac{2}{9}[/tex] to give me the final coefficient for the left version of [tex]\frac{11}{9}[/tex].[tex]\frac{11}{9}\int e^{2\theta}sin(9\theta) d\theta = -\frac{1}{9}cos(9\theta)e^{2\theta}+\frac{2}{81}sin(9\theta)e^{2\theta}[/tex]
To get that [tex]\frac{11}{9}\[/tex] out of the left side of the equation to get my original integral back, I multiplied both sides by it's reciprocal, [tex]\frac{9}{11}[/tex]. Then I tacked on the constant of integration to arrive at my final answer.[tex]\int e^{2\theta}sin(9\theta) d\theta = \frac{9}{11}(-\frac{1}{9}cos(9\theta)e^{2\theta}+\frac{2}{81}sin(9\theta)e^{2\theta}) + C[/tex]
This was marked wrong, even after I simplified the answer (which I did not do here, but I checked the simplification with WolframAlpha). Any suggestions?
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