Question

Integral Question Using integration by parts

Answer 1

\[\int\limits_{?}^{?}e ^{2\theta}\sin 3\theta d \theta\]

Answer 2

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
\[\int\limits_{}^{}e^{2x} \sin(3x) dx=\frac{-e^{2x}}{3} \cos(3x)+\frac{2e^{2x}}{9} \sin(3x)-\frac{4}{9}\int\limits_{}^{}e^{2x} \sin(3x) dx\]

Answer 3

Solve for \[\int e^{2x} \sin(3x) dx\]

Answer 4

\[\frac{13}{9} \int\limits_{}^{}e^{2x} \sin(3x) dx=\frac{-e^{2x}}{3} \cos(3x)+\frac{2e^{2x}}{9} \sin(3x) +C\]

Answer 5

You might want to check me for airthmetic errors.

Answer 6

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.

Answer 7

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.

Answer 8

I will. Hopefully I can figure out the algebra.

Answer 9

When we have:
\[\int\limits_{?}^{?}2e ^{2x}(\frac{-1}{3}\cos (3x)dx)\]
Why don't we just pull out the constant -2/3?

Answer 10

you can

Answer 11

Thanks

Answer 12

i was just showing how you can use a table

Answer 13

Perfect!! I got it.
\[\frac{1}{13}e ^{2x}(2\sin 3x-3\cos 3x)\]

Answer 14

+C of course

Answer 15

nice stuff! :)

Answer 16

Yep. I never would have thought of that trick of solving for the integral after it appeared again.

Answer 17

That is a nice little trick that comes in handy with these trig guys!

Answer 18

Yes. I think I knew it 50 years ago but didn't remember it now.

Answer 19

You sound as old as amistre. hehe

Answer 20

Probably older.