## Mertsj 2 years ago Integral Question Using integration by parts

1. Mertsj

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

2. myininaya

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

3. myininaya

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

4. myininaya

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

5. myininaya

You might want to check me for airthmetic errors.

6. Mertsj

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.

7. myininaya

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.

8. Mertsj

I will. Hopefully I can figure out the algebra.

9. Mertsj

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

10. myininaya

you can

11. Mertsj

Thanks

12. myininaya

i was just showing how you can use a table

13. Mertsj

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

14. Mertsj

+C of course

15. myininaya

nice stuff! :)

16. Mertsj

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

17. myininaya

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

18. Mertsj

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

19. myininaya

You sound as old as amistre. hehe

20. Mertsj

Probably older.