Question

Integration by parts q.

e^2x sin(3x) dx

Answer 1

I know how to do it, but in this situation when I get to e^2x sin(3x) dx = something integral e^2x sin(3x) dx do I divide or do I - out one from the other..?

Answer 2

There's some special rule about this, I forgot how to do it :).

Answer 3

running thru twice tends to get it you to a point where you can algebra it into submission

Answer 4

\[\huge \int e^{2x} \sin (3x) dx?\]

Answer 5

yeah iggy.

Answer 6

since e^2x is not integratable; use it as the u parts

Answer 7

that's only because we don't have a du if we were to say u = 2x right?

Answer 8

I guess I'll just do the problem here then.

Answer 9

correct

Answer 10

but according to LIATE you should use sin (3x) right @amistre64 ?

Answer 11

according to common sense .... LIATE is jsut suggestions that are common, but not always applicable

Answer 12

u = 2x
1/2 du = dx

dv = sin(3x)

v = 3cos(3x)?

Answer 13

u = 2x?

Answer 14

u = e^2x
du = 2e^2x dx

dv = sin(3x)
v = -cos(3x)/3

Answer 15

actually I guess I gotta say u = e^2x.

Answer 16

shouldnt it be e^2x?

Answer 17

Yeah sorry I messed up stupidly.

Answer 18

couple times.

Answer 19

youll have to run it thru a second time, and you should get the same integral as you started with, or some multiplied version of it

Answer 20

yuh... Do I divide that?

Answer 21

I'm going to do this on paper, I make more mistakes when I type everything fast on here.

Answer 22

\[\int udv = u_1v_1-\int v_1du_1\]
\[\int udv = u_1v_1-(v_2u_2 -\int u_2dv_2)\]
\[\int udv = u_1v_1-v_2u_2 +\int u_2dv_2\]
\[A = u_1v_1-v_2u_2 +A\]

Answer 23

so you could subtract it then?

Answer 24

yes, you would subtract it; combine it with the other side; and divide off the "coefficient"

Answer 25

itll look better after you work it to that step and go ... ohhh!!

Answer 26

if it's exactly the same wouldn't it then be = 0?

Answer 27

IF, then yes; but its not going to be exactly the same, it was just easier to notate it inthe latex like that for simplicity

Answer 28

alright so fa rI have

Answer 29

\[A = u_1v_1-v_2u_2 +nA\]
\[A-nA = u_1v_1-v_2u_2\]
\[(1-n)A = u_1v_1-v_2u_2\]
\[A = \frac{u_1v_1-v_2u_2}{1-n}\]

Answer 30

\[\frac{e^{2x} \cos(3x)}{3} + 1/3 \int\limits \cos(3x) 2 e^{2x}dx\]

Answer 31

I could throw the 2 out in front right?

Answer 32

\[\int e^{2x} sin(3x) dx= -\frac{e^{2x} \cos(3x)}{3} +\frac 23 \int\limits e^{2x} cos(3x) dx\]

the 2s a constant that can be pulled out yes

Answer 33

u = e^2x
du = 2e^2x dx

Answer 34

dv = cos(3x)dx
v = 1/3 sin(3x)

Answer 35

i cant see the preview so i hope this is right :)

\[\int e^{2x} sin(3x) dx= -\frac{e^{2x} \cos(3x)}{3} +\frac 23 \left(\frac{e^{2x}sin(3x)}{3}-\frac 23 \int\limits e^{2x} sin(3x) dx\right)\]

Answer 36

yes

Answer 37

so it's multiplied when you do it a second time? I tohught you just add or subtract?

Answer 38

no, your simply doing the int by parts again but its all multiplied by the outer constant, -2/3 in this case

Answer 39

OH IC....

Answer 40

yeah the original 2/3 gotcha... I thought it was multiplying by the first integration by parts we did I was like wuh... :)

Answer 41

so now we are stuck with 2/3 sin(3x)e^2xdx

Answer 42

\[\int A dx= -\frac{e^{2x} \cos(3x)}{3} +\frac 23 \left(\frac{e^{2x}sin(3x)}{3}-\frac 23 \int A dx\right)\]

stuck with? not quite

\[\int A dx= -\frac{e^{2x} \cos(3x)}{3} +\frac 23 \left(-\frac 29 {e^{2x}sin(3x)}-\frac 49 \int A dx\right)\]

Answer 43

oh woops I did 4/6 for some reason LOL.

Answer 44

\[\int A dx= -\frac{e^{2x} \cos(3x)}{3} +-\frac 29 {e^{2x}sin(3x)}-\frac 49 \int A dx\]

Answer 45

which would go back to 2/3... Man I need to get something to eat soon no brain food...

Answer 46

i got an extra "-" lol :)

Answer 47

as you can see, hopefully; we can algebra the rest

Answer 48

\[\int A dx= -\frac 13 {e^{2x} \cos(3x)} +\frac 29 {e^{2x}sin(3x)}-\frac 49 \int A dx\]
\[\int A dx+\frac 49 \int A dx= -\frac 13 {e^{2x} \cos(3x)} +\frac 29 {e^{2x}sin(3x)}\]
\[\int A dx(1+\frac 49)= -\frac 13 {e^{2x} \cos(3x)} +\frac 29 {e^{2x}sin(3x)}\]
\[\int A dx= \frac{-\frac 13 {e^{2x} \cos(3x)} +\frac 29 {e^{2x}sin(3x)}}{1+\frac 49}\]

Answer 49

sorry I had to help my parents with the groceries and such...

Answer 50

so could we combine the 9/9 + 4/9 to get 13/9 then so a flip?

Answer 51

@amistre64

Answer 52

what do you mean a "flip" ?

Answer 53

oh yeah, I see what you mean
yeah, flip it

Answer 54

Alrighty, gotta feed my dog's I'll be back, sorry.

Answer 55

Alrighty, gotta feed my dog's I'll be back, sorry.

Answer 56

@amistre64 I was curious how it turned into +2/9sin(3x)?

Answer 57

\[\int\limits A dx= -\frac{e^{2x} \cos(3x)}{3} +\frac 23 \left(-\frac 29 {e^{2x}\sin(3x)}-\frac 49 \int\limits A dx\right)\]

Answer 58

Oh I think this was accidentally added, the one before wasn't -, and the one after isn't... That is where I was confused okay....

Answer 59

I got 3/13 e^2xcos(3x) + 3/26 e^2x sin(3x) = Adx

Answer 60

http://www.wolframalpha.com/input/?i=integration+e%5E%282x%29+sin%283x%29+dx ?? :(?

Answer 61

@TuringTest