Question

integration by parts: e^ax sin bx dx

Answer 1

\[\int\limits_{}^{}e ^{ax}) \sin bx dx\]

Answer 2

is that sin(bx) or sin(b)x

Answer 3

forget teh parenthessi..thats a typo and its teh first one

Answer 4

\[\int\limits_{}^{}e ^{ax} \sin(bx) dx\]

Answer 5

its is cyclic.. you see that right?

Answer 6

is it b.c its a sin

Answer 7

and derivative of taht is -cos n der of taht is -sin and so on so forth

Answer 8

its is because of both functions.. either integrating or differentiating it will eventually produce the same problem you started with.

Answer 9

how do i integrate it by parts?

Answer 10

well.. that depends on how your teacher wants you to learn

Answer 11

wat do u mean?

Answer 12

do the by parts twice

Answer 13

wat shud i set as my u and my dv?

Answer 14

or use euler's thing \[e^{i\theta} = i\sin\theta + \cos \theta \]

Answer 15

whoah i havent learned that yet!

Answer 16

can we do it with u, du, v , and dv?

Answer 17

oh sorry hmm let me show you the by parts way

Answer 18

\[\int e^{ax}\sin{bx}dx = \frac{e^{ax}}{a}\times \sin{bx} - \frac{b}{a}\int e^{ax}\cos{bx} \times dx \]
Let \(\int e^{ax}\sin{bx}dx = I\)
\[\implies -\frac{b}{a}\left(\frac{e^{ax}}{a}\cos{bx} + \frac{b}{a}\int e^{ax}\times \sin{bx}\right)\]

\[\implies -\frac{b}{a}\left(\frac{e^{ax}}{a}\cos{bx} + \frac{b}{a}I\right)\]

\[\implies -\frac{be^{ax}}{a^2}\cos{bx} - \frac{b^2}{a^2}I\]

Answer 19

Now the whole equation becomes

\[ I = \frac{e^{ax}}{a}\sin{bx} - \frac{b}{a^2}\cos{bx} - \frac{b^2}{a^2}I\]
\[ I \frac{(a^2 + b^2)}{b^2} = \frac{e^{ax}}{a}\sin{bx} - \frac{b}{a^2}\cos{bx}\]

Answer 20

Now Important thing is you follow up the whole thing I have typed!

Answer 21

how did u get the very first equation

Answer 22

sorry this looks new to me so im kinda confused what ur doing

Answer 23

it is really hard to explain all that like i said you need to follow it up, it's hard to type that out and i am kinda lazy :/


http://www.youtube.com/watch?v=_MuKOm8IHBA <<try this

Answer 24

ok ill watch thanks