Question

The integral of (e^-x) cos(4x) dx.

Answer 1

Use by parts twice.. keeping the original question on the left.
The answer will come through recursion... does that make sense?

Answer 2

Oh, so i think the reason i didn't get it is after the first "by parts" i stopped cuz it made no sense.. so you're saying if i do "by parts" again on that i should get it?

Answer 3

yeah, your original "question" will show up again on the right, then you add/subract it to the other side and solve for what is left.

Answer 4

Oh, true i always leave that integral out and just have the equals sign.. i should be more careful to always write what i have. similar to when i did the integral of sec^3 :P

Answer 5

yes... it is very much like sec^3

Answer 6

@nj1202 Are you going to show your work?

Answer 7

I might, i'm just working it out on paper... so far i'm at this step:

\[\int\limits_{}^{}e ^{-x}\cos(4x)dx=-e ^{-x}\cos(4x)-\int\limits_{}^{}4e ^{-x}\sin(4x)dx\]

Answer 8

V = (1/4 ) sin4x

Answer 9

oh... i switched my u and dv xD

Answer 10

uv - ∫ vdu

Answer 11

I chose the trig function to be u each time and e^-x to be dv each time... otherwise it undoes itself

Answer 12

well, then that's what i did:
dv = e^-x dx
v = -e^-x
u = cos(4x)
du = -4sin(4x) dx

right?

Answer 13

meaning.. stay consistent... I think that you can go the other way... I just like integrating e's

Answer 14

yes... that's how I did it

Answer 15

minus signs... suck... watch them

Answer 16

Yea i'm being careful, but i think i screwed one up cuz i have this now:

\[\int\limits_{}^{}e ^{-x}\cos(4x)dx=-e ^{-x}\cos(4x)+e ^{-x}\cos(4x)+\int\limits_{}^{}e ^{-x}\cos(4x) dx\]

Where's my mix up hahahahaha xD

Answer 17

I know i shoud be able to add my integral to the other side, so that should be -, but does that mean the other part should be - as well?

Answer 18

you are missing a 4 in the second term, which compounds to a 16 in the third term

Answer 19

as well as the third term should be negative

Answer 20

would you mind typing that out? it'd be of great use just to see the wrong

Answer 21

oh boy... see you in a bit... :)

Answer 22

hahaha xD

Answer 23

\[\int\limits_{}^{}e ^{-x}\cos(4x) dx=-e ^{-x}\cos(4x)-4\int\limits_{}^{}e ^{-x}\sin(4x) dx\]for the first round... 3 negatives make the second term negative.

Answer 24

that i did have :) so i was alright so far!

Answer 25

u=sin(4x)
du=4cos(4x)
dv=e^(-x)
v=-e^(-x)

Answer 26

perfect oh.... forgot to multiply 4 for du... that's the 16 now i see it xD

Answer 27

\[\int\limits_{}^{}e ^{-x}\cos(4x)dx=-e ^{-x}\cos(4x)+4e ^{-x}\sin(4x)-16\int\limits_{}^{}e ^{-x}\cos(4x)dx\]

Answer 28

ahh now it works beautifully :)

Answer 29

negative from the 4 with negative from v with negative from by parts makes a negative in the third term.

Answer 30

cool... so you have divide by 17 in your answer?

Answer 31

17 what? xD

Answer 32

i should work this out first xD

Answer 33

k

Answer 34

= \[\frac{ 3e ^{-x}\cos(4x) }{ 17 }\]

this it?

Answer 35

\[17\int\limits_{}^{}e ^{-x}\cos(4x)dx=-e ^{-x}\cos(4x)+4e ^{-x}\sin(4x)\]

Answer 36

combine like-terms and divide by 17, so i think i'm right... i'll type it in online and see.. my hw site thingy tells me if it's right or wrong.

Answer 37

there is a cos term and a sin term

Answer 38

oh... where'd i get the two cos xD

Answer 39

got it now :D says i'm right for my hw problem!

Answer 40

cool! check your signs @Chlorophyll if you factor the negative... it changes both signs inside.

Answer 41

I'm getting better with this by parts stuff, just learned it thursday in lecture, so i've been doing tons of problems... this one was the first one by myself that does it twice, so it was strange. we did the sec^3 in class. but thanks euler!

Answer 42

You're welcome... oddly enough I did sec^3 with a student today.

Answer 43

haha xD next one is hyperbolic cosine... bring it math! :P

Answer 44

@EulerGroupie Thanks, the result doesn't have negative side in front of e^-x

Answer 45

I = (e^-x/ 17) ( 4sin4x - cos4x)

Answer 46

yeee haaa! hyperbolics are a challenge for me because my instructors never pushed it very hard. @Chlorophyll I agree.

Answer 47

haha i have never learned anything applicable with them... they just pop up to integrate xD

Answer 48

I've heard that civil engineering uses them... hanging cables are apparently modeled very well by them.

Answer 49

hmm that's interesting.. i almost think they can be easier than the trig functions sometimes... cuz usually it's only sinh and cosh i see, and it's always positive either der. or int. so it's less to remember xD

Answer 50

Some things do seem simpler with them, and I have played with them in parametric functions along with complex numbers... what makes a hyperbola in real space makes a circle in the complex plane... its very wierd. Anyway, you may be on your own with the next one... it is after midnight here.

Answer 51

haha yea it's 2 am where i'm at haha xD i work more efficiently in late hours... and i can sleep in xD no classes on sunday :D

Answer 52

Good luck! See ya' around.

Answer 53

Thanks buddy :D