anonymous
  • anonymous
int e^-theta cos(5theta) d theta
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
I have u= cos(5theta) du= -sin(5theta) dv= e^-theta and v= e^-theta
zepdrix
  • zepdrix
woops! du=-5sin(5theta)
zepdrix
  • zepdrix
v=-e^-theta

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zepdrix
  • zepdrix
Let's make sure we start this one off right cause it's a doozy! :)
zepdrix
  • zepdrix
\[\Large \int\limits e^{-\theta}\cos(5\theta)\;d \theta\]
anonymous
  • anonymous
ill try to show you what i have so far
zepdrix
  • 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
  • anonymous
\[uv- intvdu = (\cos5\theta)(e^-\theta) - \int\limits (e^-\theta) (-sinx)\]
anonymous
  • anonymous
then
anonymous
  • anonymous
\[e^-\theta \cos5\theta - [ e^-\theta -\sin5\theta - \int\limits e^-\theta \cos5\theta]\]
anonymous
  • anonymous
thats it. i'm lost! lol
anonymous
  • anonymous
those e's are e to the neg theta
zepdrix
  • 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
  • zepdrix
Your du and v are not correct :( might wanna clean that up before you get too deep into this problem.
anonymous
  • anonymous
ok. I see what you did with the 5
anonymous
  • anonymous
i forgot about that
zepdrix
  • 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
  • anonymous
will everything after the integral sign become positive now?
anonymous
  • anonymous
i see how this is set up. working through it confuses me though
zepdrix
  • zepdrix
ya it's a bit tricky :c
zepdrix
  • 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
  • anonymous
\[ e ^{-\theta}(\cos \theta) + 5 \int\limits e ^{-\theta} (\sin 5\theta) d\]
anonymous
  • anonymous
ok, so the 5 isn't a pos then.
anonymous
  • anonymous
sorry, the equation helper takes me forever
zepdrix
  • 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
  • anonymous
and i forgot the 5 with the cos theta
anonymous
  • anonymous
should i pull the e to the - theta out front?
zepdrix
  • zepdrix
Do you know how to solve this integral?\[\Large \int\limits e^x \cos x\;dx\]
zepdrix
  • 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
  • anonymous
i think so. yes. i would make u cosx and e to the x dv
anonymous
  • anonymous
I saw a similar problem to the one we are doing now and this seems much more complicated
zepdrix
  • zepdrix
Lemme just ummmm...
zepdrix
  • 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
  • zepdrix
If it's not necessary, that's ok.
anonymous
  • anonymous
\[e ^{-\theta} (\cos 5\theta) - [ e ^{-\theta} (\sin 5\theta) - \int\limits \sin 5\theta(e ^{-\theta})] d \theta \]
anonymous
  • anonymous
no not really. I still don't know where to go from here
anonymous
  • anonymous
really confused still
zepdrix
  • zepdrix
You're still missing a negative on the first term. and the second term.. hmm
zepdrix
  • zepdrix
Did you see my \(\Large v\) above? :\
anonymous
  • anonymous
oh yea, - e
anonymous
  • anonymous
all i see is e theta sin and cos over and over
zepdrix
  • zepdrix
I'm seeing a lot of mistakes D: you gotta slow down!!
zepdrix
  • 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
  • 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
  • anonymous
i keep having typos
zepdrix
  • 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
  • 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
  • zepdrix
I know that equation editor can be rough :3 lol it's not nice like paper
anonymous
  • anonymous
the one in front of the square bracket?
zepdrix
  • zepdrix
ya :o
zepdrix
  • 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
  • anonymous
ok. this is where i made the biggest mistake. I never had a 25
zepdrix
  • zepdrix
Yah integrating again gave us an extra factor of 5 :o
anonymous
  • anonymous
so confusing
anonymous
  • anonymous
now do we add the integrals on one side and the e's on the other?
zepdrix
  • 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
  • zepdrix
So we currently have:\[\Large \color{royalblue}{I}=-e^{-\theta}\cos5\theta+5e^{-\theta}\sin5\theta-25 \color{royalblue}{I}\]
zepdrix
  • zepdrix
Yah we want to solve for I :)
anonymous
  • anonymous
i like the variable idea (a lot)
zepdrix
  • zepdrix
Hmm I think that last term should be +25I.. lemme see if we messed up somewhere...
anonymous
  • anonymous
hahaha
zepdrix
  • zepdrix
Ah yes it's + :O I didn't distribute correctly.
zepdrix
  • zepdrix
\[\Large \color{royalblue}{I}=-e^{-\theta}\cos5\theta+5e^{-\theta}\sin5\theta+25 \color{royalblue}{I}\]
zepdrix
  • zepdrix
Nope I think it was subtraction +_+ grrrr
zepdrix
  • zepdrix
so easy to get lost in this one t.t
anonymous
  • anonymous
I had to reload, my computer froze on the equation
zepdrix
  • zepdrix
You get that weird funky blue screen? :o
zepdrix
  • zepdrix
\[\Large \color{royalblue}{I}=-e^{-\theta}\cos5\theta+5e^{-\theta}\sin5\theta-25 \color{royalblue}{I}\]
zepdrix
  • zepdrix
Ok ok ok it's -25I, i made up my mind +_+
anonymous
  • anonymous
\[I = 1/2 [ -e ^{-\theta}(\cos5\theta) + 5e ^{-\theta} (\sin5\theta) - 251] + C\]
anonymous
  • anonymous
?
anonymous
  • anonymous
pardon my french. but... DAMN! This one is really kickin my butt
zepdrix
  • 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
  • zepdrix
We're trying to solve for \(\Large \color{royalblue}{I}\) So now we just divide by 26 right? :o
anonymous
  • 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
  • zepdrix
:c
zepdrix
  • zepdrix
I wish you woulda let me do the example.. oh well
anonymous
  • 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
  • anonymous
you can do the example if you wish
anonymous
  • anonymous
do we divide the 26 by the whole thing?
zepdrix
  • zepdrix
yes
anonymous
  • anonymous
I=5/26 [e^ θ( -sin5θ) −e^−θ (cos5θ)]
anonymous
  • anonymous
?
anonymous
  • anonymous
i understand if i am irritating you. i am irritating me!
anonymous
  • anonymous
I'm really trying to get this one... haha
zepdrix
  • 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
  • anonymous
is it ok if it looks like this? I=1/26 [e^ θ ( 5-sin5θ) −e^−θ (cos5θ)]
zepdrix
  • zepdrix
I'm so confused .. :( so many typos...
anonymous
  • anonymous
with the thetas both neg
zepdrix
  • zepdrix
why is there still a negative on the sine? D:
anonymous
  • 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
  • zepdrix
( 5-sin5θ) should be (5sin5θ) rest of it looks good though! :) yay team!
anonymous
  • anonymous
I = 1/26 [e^ θ ( 5 sin 5θ) −e^−θ (cos5θ)]
zepdrix
  • zepdrix
close enough :) lol
anonymous
  • anonymous
WOW! you are the beesssttt! Triple medals for you! I pray this problem is NOT on ANY test EVER
zepdrix
  • 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
  • anonymous
I did that except for instead of cosx it was sinx and i actually got that one right ;)
zepdrix
  • zepdrix
Oh good :3

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