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

- anonymous

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

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- anonymous

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

- zepdrix

woops!
du=-5sin(5theta)

- zepdrix

v=-e^-theta

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## More answers

- zepdrix

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

- zepdrix

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

- anonymous

ill try to show you what i have so far

- zepdrix

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! :(

- anonymous

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

- anonymous

then

- anonymous

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

- anonymous

thats it. i'm lost! lol

- anonymous

those e's are e to the neg theta

- zepdrix

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

- zepdrix

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

- anonymous

ok. I see what you did with the 5

- anonymous

i forgot about that

- zepdrix

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

- anonymous

will everything after the integral sign become positive now?

- anonymous

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

- zepdrix

ya it's a bit tricky :c

- zepdrix

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

- anonymous

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

- anonymous

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

- anonymous

sorry, the equation helper takes me forever

- zepdrix

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.

- anonymous

and i forgot the 5 with the cos theta

- anonymous

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

- zepdrix

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

- zepdrix

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

- anonymous

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

- anonymous

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

- zepdrix

Lemme just ummmm...

- zepdrix

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?

- zepdrix

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

- anonymous

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

- anonymous

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

- anonymous

really confused still

- zepdrix

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

- zepdrix

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

- anonymous

oh yea, - e

- anonymous

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

- zepdrix

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

- zepdrix

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:

- anonymous

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

- anonymous

i keep having typos

- zepdrix

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

- zepdrix

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

- zepdrix

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

- anonymous

the one in front of the square bracket?

- zepdrix

ya :o

- zepdrix

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

- anonymous

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

- zepdrix

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

- anonymous

so confusing

- anonymous

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

- zepdrix

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

- zepdrix

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

- zepdrix

Yah we want to solve for I :)

- anonymous

i like the variable idea (a lot)

- zepdrix

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

- anonymous

hahaha

- zepdrix

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

- zepdrix

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

- zepdrix

Nope I think it was subtraction +_+ grrrr

- zepdrix

so easy to get lost in this one t.t

- anonymous

I had to reload, my computer froze on the equation

- zepdrix

You get that weird funky blue screen? :o

- zepdrix

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

- zepdrix

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

- anonymous

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

- anonymous

?

- anonymous

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

- zepdrix

?? +_+ 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,\[\Large 26\color{royalblue}{I}=5e^{-\theta}\sin5\theta-e^{-\theta}\cos5\theta\]

- zepdrix

We're trying to solve for \(\Large \color{royalblue}{I}\)
So now we just divide by 26 right? :o

- anonymous

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

- zepdrix

:c

- zepdrix

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

- anonymous

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

- anonymous

you can do the example if you wish

- anonymous

do we divide the 26 by the whole thing?

- zepdrix

yes

- anonymous

I=5/26 [e^ θ( -sin5θ) −e^−θ (cos5θ)]

- anonymous

?

- anonymous

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

- anonymous

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

- zepdrix

The sine function should be positive.
Only factor a 1/26 out of each term.
Leaving a 5 in front of the sine.

- anonymous

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

- zepdrix

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

- anonymous

with the thetas both neg

- zepdrix

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

- anonymous

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

- zepdrix

( 5-sin5θ) should be (5sin5θ)
rest of it looks good though! :) yay team!

- anonymous

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

- zepdrix

close enough :) lol

- anonymous

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

- zepdrix

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

- anonymous

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

- zepdrix

Oh good :3

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