Question

intergrate e^t cos nt

Answer 1

no integrating d thingy? i dont think it's integrable then?

Answer 2

use by-parts! and i guess that 'n' is a 'd'

Answer 3

wouldnt make sense? no argument for cos?

Answer 4

n is just a constant and by if you do it by parts don't the e^t and cos always exist?

Answer 5

You will want to apply integration by parts twice.

Answer 6

does it look like \[\large \int e^t \cos (nt) dt?\]

Answer 7

In case that is a 'd':

(yeah by-parts won't help :/)

Answer 8

yeah thats it lgasallote

Answer 9

by parts seems logical o.O

Answer 10

but neither cos(nt) or e^t reduce down to 1 yeah? Not sure, haven't done this is a while haha

Answer 11

You don't need them to reduce to one. Once you integrate by parts twice, you will end up with the same integral on left and right of the equality. The one on the right will be negative, you can add it to both sides and then divide both sides by two. This is a common trick with infinitely-differentiable functions like this.

Answer 12

The fastest way to evaluate this is to change cosnx to e^(inx) <--- evaluate the integration of e's and ... get only the real part.

Answer 13

Good trick if you are comfortable with the complex plane, yep.

Answer 14

\[u = \cos (nt) \rightarrow du = -n\sin (nt)dt\] \[dv = e^tdt \rightarrow v = e^t\]

\[e^t \cos (nt) + n\int e^t \sin(nt)dt\]

\[u = \sin (nt) \rightarrow du = \cos (nt)dt\] \[dv = e^tdt \rightarrow v = e^t\]

\[e^t \cos (nt) - [e^t \sin (nt) - \int e^t \cos(nt)dt\]

Answer 15

hmm wait...something's fishy...that should be +int of e^t cos (nt)dt

Answer 16

oh i see...it shouldnt be minus quantity lol

Answer 17

i also missed the n sorry

Answer 18

I had u = e^t and v' = cos(nt) :\

Answer 19

are you familiar with LIATE?

Answer 20

no

Answer 21

lgba, you had it right the first time. The integral is subtracted. You can then add it to both sides of the equality, divide both sides by 2, to obtain your solution.

Answer 22

You did miss a constant the second time, though.

Answer 23

\[e^t \cos (nt) + n[e^t \sin (nt) - n\int e^t \cos (nt) dt]\] is that right now @nbouscal ?

Answer 24

and i dont think it's 2 you divide but (n+1)

Answer 25

Yeah, that looks right. And you are correct, with the constant, it will be a different divisor. Good point.

Answer 26

I guess: if that is 'I'
\[ I= \int e^t cos(nt).dt = cos(nt)\int e^t.dt + n \int sin(nt)e^t.dt\]\[=e^t.cos(nt) + n[e^t\int sin(nt).dt - n\int e^t cos(nt).dt]\]
\[or, I = e^t.cos(nt) + ne^t\int sin(nt).dt - nI\]
\[or, I = \frac{e^tcos(nt) - \frac {cos(nt)}{n} }{n+1} + k\]



there may be an error. actually i think there's definitely an error.

Answer 27

@Jgip

LIATE stands for Logarithm-> Inverse trigonemtric functions -> Algebraic expressions -> trigonometric expressions -> exponential expression

this states the order of priority when substituting u

Answer 28

ohhh ok, I never got taught LIATE.

Answer 29

\[
\frac{e^t \cos (nt) + ne^t \sin (nt)}{n^2+1}
\]

Answer 30

This is a common theme in integrating by parts. With smooth functions, you'll never reach 1, so you must reach a point where you get your original integral back again, so that you can solve for it.

Answer 31

if i continue what i was doing... \[\int e^t \cos (nt) dt = e^t \cos (nt) + ne^t \sin (nt) - n^2 \int e^t \cos (nt) dt\]

therefore

\[\int e^t \cos (nt) dt + n^2 \int e^t \cos (nt) dt = e^t \cos (nt) + ne^t \sin (nt)\]

\[(n^2+1) \int e^t \cos (nt) dt = e^t [\cos (nt) + n\sin (nt)]\]

\[\int e^t \cos (nt) dt = \frac{e^t [\cos (nt) + n\sin (nt)]}{n^2 + 1}\]

got it @Jgip

Answer 32

for more info on LIATE you can check out http://openstudy.com/study#/updates/4f9df983e4b000ae9ed2688c

Answer 33

fair warning... \(\uparrow\) that is a tutorial *I* wrote :)

Answer 34

I am LOST *.*

Answer 35

In this case, whether you set the exponential or trigonometric function as \(u\) is actually not really important, because they are both easily integrable. You can do it either way. The key is to do it twice, then combine like terms for the original integral.

Answer 36

Thanks, I'll keep this is mind about integration by parts. just worked through it like above. very good explanations, I appreciate the help heaps!

Answer 37

whenever i see trig functions and exponential functions together i assume already that it goes by that method...i think

Answer 38

i suppose it wouldnt be safe to generalize

Answer 39

O now I get where I was wrong. Stupid me >.<

Answer 40

lol where?