Question

how to solve this?

Answer 1

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

Answer 2

Use integration by parts twice. Watch:
\[ I = \int e^{2\theta}\sin(3\theta) d\theta = \frac{1}{2}e^{2\theta}\sin(3\theta) - \frac{3}{2} \int e^{2\theta}\cos(3\theta) d\theta \]
\[ = \frac{1}{2}e^{2\theta}\sin(3\theta) - \frac{3}{4} e^{2\theta}\cos(3\theta) - \frac{9}{4}\int e^{2\theta}\sin(3\theta)\]
\[ =\frac{1}{2}e^{2\theta}\sin(3\theta) - \frac{3}{4} e^{2\theta}\cos(3\theta) -\frac{9}{4} I \]
so adding the last term to the other side, you get that
\[ \frac{13}{4}I = \frac{1}{2}e^{2\theta} \sin(3\theta) - \frac{3}{4}e^{2\theta}\cos(3\theta) \]
so
\[ I = \frac{2}{13} e^{2\theta}\sin(3\theta) - \frac{3}{13}e^{2\theta} \cos(3\theta) \]
+ C, of course...

Answer 3

thank you,im trying to digest this :)

Answer 4

\[1\div 2 e?\] where did you get that ?

Answer 5

Are you familiar with integration by parts?

Answer 6

I would think so given the nature of the question.

Answer 7

oh!now i got it, just now im a lil bit confuse with integration and differentiation, sorry :)

Answer 8

Okay, sounds good. If you ever need a refresher on it, this is a good resource:

http://tutorial.math.lamar.edu/Classes/CalcII/IntegrationByParts.aspx

Example 8 is relevant to this question.

Answer 9

thanks.you helping me lot! :)