Here's the question you clicked on:
Mertsj
Integral Question Using integration by parts
\[\int\limits_{?}^{?}e ^{2\theta}\sin 3\theta d \theta\]
e^(2x) sin(3x) 2e^(2x) -1/3*cos(3x) 4e^(2x) -1/9*sin(3x) ..... \[=e^{2x} \cdot \frac{-1}{3} \cos(3x)-\int\limits_{}^{}2 e^{2x} \cdot \frac{-1}{3}\cos(3x) dx\] \[=e^{2x} \cdot \frac{-1}{3} \cos(3x)-[2 e^{2x} \cdot \frac{-1}{9} \sin(3x)-\int\limits_{}^{} 4 e^{2x} \cdot \frac{-1}{9} \sin(3x) dx]\] So we have \[\int\limits_{}^{}e^{2x} \sin(3x) dx=\frac{-e^{2x}}{3} \cos(3x)+\frac{2e^{2x}}{9} \sin(3x)-\frac{4}{9}\int\limits_{}^{}e^{2x} \sin(3x) dx\]
Solve for \[\int e^{2x} \sin(3x) dx\]
\[\frac{13}{9} \int\limits_{}^{}e^{2x} \sin(3x) dx=\frac{-e^{2x}}{3} \cos(3x)+\frac{2e^{2x}}{9} \sin(3x) +C\]
You might want to check me for airthmetic errors.
Thank you very much, my dear. I did have the correct f and g' identified and got to the point where you said "and so we have" but I couldn't identify that as progress. Thanks again.
Np. :) I hope I put enough details as far as me showing my work. Let me know if you don't understand one of the steps.
I will. Hopefully I can figure out the algebra.
When we have: \[\int\limits_{?}^{?}2e ^{2x}(\frac{-1}{3}\cos (3x)dx)\] Why don't we just pull out the constant -2/3?
i was just showing how you can use a table
Perfect!! I got it. \[\frac{1}{13}e ^{2x}(2\sin 3x-3\cos 3x)\]
Yep. I never would have thought of that trick of solving for the integral after it appeared again.
That is a nice little trick that comes in handy with these trig guys!
Yes. I think I knew it 50 years ago but didn't remember it now.
You sound as old as amistre. hehe