Question

int e^-theta cos(5theta) d theta

Answer 1

I have u= cos(5theta) du= -sin(5theta) dv= e^-theta and v= e^-theta

Answer 2

woops!
du=-5sin(5theta)

Answer 3

v=-e^-theta

Answer 4

Let's make sure we start this one off right cause it's a doozy! :)

Answer 5

\[\Large \int\limits e^{-\theta}\cos(5\theta)\;d \theta\]

Answer 6

ill try to show you what i have so far

Answer 7

So you've got the right idea, approaching this with `By Parts`.
In this case it doesn't matter which function you choose for u and dv.
So looks good so far.

Careful with your du and v though! :(

Answer 8

\[uv- intvdu = (\cos5\theta)(e^-\theta) - \int\limits (e^-\theta) (-sinx)\]

Answer 9

then

Answer 10

\[e^-\theta \cos5\theta - [ e^-\theta -\sin5\theta - \int\limits e^-\theta \cos5\theta]\]

Answer 11

thats it. i'm lost! lol

Answer 12

those e's are e to the neg theta

Answer 13

\[\Large u=\cos(5\theta) \qquad\qquad\qquad\qquad dv=e^{-\theta}\;d \theta\]\[\Large du=-5\sin(5\theta) \qquad\qquad\qquad v=-e^{-\theta}\]

Answer 14

Your du and v are not correct :( might wanna clean that up before you get too deep into this problem.

Answer 15

ok. I see what you did with the 5

Answer 16

i forgot about that

Answer 17

Your setup looks good though.
If we make those corrections:\[\Large =(\cos5\theta)(-e^{-\theta})-\int\limits\limits(-e^{-\theta}) (-5\sin5\theta)d \theta\]

Answer 18

will everything after the integral sign become positive now?

Answer 19

i see how this is set up. working through it confuses me though

Answer 20

ya it's a bit tricky :c

Answer 21

Cancelling out the negatives? yah that's a good idea.\[\Large =-e^{-\theta}\cos5\theta-5\int\limits e^{-\theta}\sin5\theta\;d \theta\]

Answer 22

\[ e ^{-\theta}(\cos \theta) + 5 \int\limits e ^{-\theta} (\sin 5\theta) d\]

Answer 23

ok, so the 5 isn't a pos then.

Answer 24

sorry, the equation helper takes me forever

Answer 25

Ya it takes some getting used to.
You can use the `draw` function also.
It's pretty tough to write on if you don't have a pen tablet though.

Answer 26

and i forgot the 5 with the cos theta

Answer 27

should i pull the e to the - theta out front?

Answer 28

Do you know how to solve this integral?\[\Large \int\limits e^x \cos x\;dx\]

Answer 29

The problem we're working on would be solved very similarly to the one I wrote above.
The problem is, if you don't know how to solve that one it's going to be a wild ride because of all these negatives and 5's :P

Answer 30

i think so. yes. i would make u cosx and e to the x dv

Answer 31

I saw a similar problem to the one we are doing now and this seems much more complicated

Answer 32

Lemme just ummmm...

Answer 33

Want me to burn through this one really quick?\[\Large \int\limits\limits e^x \cos x\;dx\]
So you can get a `general` idea of where we're trying to go with this?

Answer 34

If it's not necessary, that's ok.

Answer 35

\[e ^{-\theta} (\cos 5\theta) - [ e ^{-\theta} (\sin 5\theta) - \int\limits \sin 5\theta(e ^{-\theta})] d \theta \]

Answer 36

no not really. I still don't know where to go from here

Answer 37

really confused still

Answer 38

You're still missing a negative on the first term. and the second term.. hmm

Answer 39

Did you see my \(\Large v\) above? :\

Answer 40

oh yea, - e

Answer 41

all i see is e theta sin and cos over and over

Answer 42

I'm seeing a lot of mistakes D: you gotta slow down!!

Answer 43

When you integrated a second time from here,\[\large =-e^{-\theta}\cos5\theta-5\int\limits\limits e^{-\theta}\sin5\theta\;d \theta\]
Where did the factor of 5 go? D:

Answer 44

my first answer to this problem was \[-5/2 (e ^{-\theta} ) (\cos 5\theta - \sin 5\theta) + C\] and it said i was wrong

Answer 45

i keep having typos

Answer 46

From this setup,\[\large =-e^{-\theta}\cos5\theta-5\int\limits\limits\limits e^{-\theta}\sin5\theta\;d \theta\]
Making the same by parts as before, `trig as u`, `exponential as dv`:\[\Large u=\sin(5\theta) \qquad\qquad\qquad\qquad dv=e^{-\theta}\;d \theta\]\[\Large du=5\cos(5\theta)\qquad\qquad\qquad v=-e^{-\theta}\;d \theta\]

Answer 47

Understand what 5 im talking about? :o\[\large -e^{-\theta}\cos5\theta-5\left[-e^{-\theta}\sin5\theta-\int\limits (-e^{-\theta})(5\cos(5\theta)\;d \theta\right]\]

Answer 48

I know that equation editor can be rough :3 lol
it's not nice like paper

Answer 49

the one in front of the square bracket?

Answer 50

ya :o

Answer 51

So it looks like that simplifies to:\[\large =-e^{-\theta}\cos5\theta+5e^{-\theta}\sin5\theta-25\int\limits\limits e^{-\theta}\cos5\theta\;d \theta\]

Answer 52

ok. this is where i made the biggest mistake. I never had a 25

Answer 53

Yah integrating again gave us an extra factor of 5 :o

Answer 54

so confusing

Answer 55

now do we add the integrals on one side and the e's on the other?

Answer 56

Personally, I like to give a variable name to our initial integral, it's easier to keep track of that way:\[\Large \color{#3366CF}{\int\limits e^{-\theta}\cos(5\theta) \quad=\quad I}\]

Answer 57

So we currently have:\[\Large \color{royalblue}{I}=-e^{-\theta}\cos5\theta+5e^{-\theta}\sin5\theta-25 \color{royalblue}{I}\]

Answer 58

Yah we want to solve for I :)

Answer 59

i like the variable idea (a lot)

Answer 60

Hmm I think that last term should be +25I.. lemme see if we messed up somewhere...

Answer 61

hahaha

Answer 62

Ah yes it's + :O I didn't distribute correctly.

Answer 63

\[\Large \color{royalblue}{I}=-e^{-\theta}\cos5\theta+5e^{-\theta}\sin5\theta+25 \color{royalblue}{I}\]

Answer 64

Nope I think it was subtraction +_+ grrrr

Answer 65

so easy to get lost in this one t.t

Answer 66

I had to reload, my computer froze on the equation

Answer 67

You get that weird funky blue screen? :o

Answer 68

\[\Large \color{royalblue}{I}=-e^{-\theta}\cos5\theta+5e^{-\theta}\sin5\theta-25 \color{royalblue}{I}\]

Answer 69

Ok ok ok it's -25I, i made up my mind +_+

Answer 70

\[I = 1/2 [ -e ^{-\theta}(\cos5\theta) + 5e ^{-\theta} (\sin5\theta) - 251] + C\]

Answer 71

?

Answer 72

pardon my french. but... DAMN! This one is really kickin my butt

Answer 73

?? +_+ Hmm not sure what you're doing there.
From here:\[\Large \color{royalblue}{I}=-e^{-\theta}\cos5\theta+5e^{-\theta}\sin5\theta-25 \color{royalblue}{I}\]
We'll add 25I to each side,\[\Large 26\color{royalblue}{I}=5e^{-\theta}\sin5\theta-e^{-\theta}\cos5\theta\]

Answer 74

We're trying to solve for \(\Large \color{royalblue}{I}\)

So now we just divide by 26 right? :o

Answer 75

lol idk what i'm doing either (i guess) I should just give up on this one. I'm all over the board

Answer 76

:c

Answer 77

I wish you woulda let me do the example.. oh well

Answer 78

I keep thinking i have 2 integrals on one side and that i have to divide one of the I's by the other side and this is where my 1/2 comes in

Answer 79

you can do the example if you wish

Answer 80

do we divide the 26 by the whole thing?

Answer 81

yes

Answer 82

I=5/26 [e^ θ( -sin5θ) −e^−θ (cos5θ)]

Answer 83

?

Answer 84

i understand if i am irritating you. i am irritating me!

Answer 85

I'm really trying to get this one... haha

Answer 86

The sine function should be positive.

Only factor a 1/26 out of each term.
Leaving a 5 in front of the sine.

Answer 87

is it ok if it looks like this?
I=1/26 [e^ θ ( 5-sin5θ) −e^−θ (cos5θ)]

Answer 88

I'm so confused .. :( so many typos...

Answer 89

with the thetas both neg

Answer 90

why is there still a negative on the sine? D:

Answer 91

I cant get this equation thing down. it makes me more confused and then i screw up the problem more trying to use it

Answer 92

( 5-sin5θ) should be (5sin5θ)

rest of it looks good though! :) yay team!

Answer 93

I = 1/26 [e^ θ ( 5 sin 5θ) −e^−θ (cos5θ)]

Answer 94

close enough :) lol

Answer 95

WOW! you are the beesssttt! Triple medals for you! I pray this problem is NOT on ANY test EVER

Answer 96

I really recommend you go solve this integral:\[\Large \int\limits e^x \cos x\;dx\]
It's much much easier, but with the same major steps.
Make sure you understand the process :O
Ending up with the intial integral and then doing algebra to solve

Answer 97

I did that except for instead of cosx it was sinx and i actually got that one right ;)

Answer 98

Oh good :3