Question

any one knows how to integrate e^-2t cost dt ?

Answer 1

by parts i believe

Answer 2

you end up getting the same integral after a few rounds that you can use to your advantage

Answer 3

\[\int e^{-2t}cos(t)dt= -e^{-2t}sin(t)-\int 2e^{-2t}sin(t) dt\]
\[\int e^{-2t}cos(t)dt= -e^{-2t}sin(t)-2\int e^{-2t}sin(t) dt\]
\[\int e^{-2t}cos(t)dt= -e^{-2t}sin(t)-2(e^{-2t}cos(t)-\int e^{-2t}cos(t) dt)\]

Answer 4

hopefully i kept that straight

Answer 5

nope, i see an error on the last one

Answer 6

hmm sorry to interrupt but which 1 u use as u and which 1 u use as v for intrating by parts? u = exponential and v= trigo?

Answer 7

\[\int e^{-2t}cos(t)dt= -e^{-2t}sin(t)-2\left( e^{-2t}cos(t)-\int -2e^{-2t}cos(t) dt \right)\]
\[\int e^{-2t}cos(t)dt= -e^{-2t}sin(t)-2\left( e^{-2t}cos(t)+2\int e^{-2t}cos(t) dt \right)\]
\[\int e^{-2t}cos(t)dt= -e^{-2t}sin(t)-2e^{-2t}cos(t)-4\int e^{-2t}cos(t) dt\]
\[4\int e^{-2t}cos(t) dt+\int e^{-2t}cos(t)dt= -e^{-2t}sin(t)-2e^{-2t}cos(t)\]
\[5\int e^{-2t}cos(t) dt= -e^{-2t}sin(t)-2e^{-2t}cos(t)\]
\[\int e^{-2t}cos(t) dt= \frac{-e^{-2t}sin(t)-2e^{-2t}cos(t)}{5}\]
maybe??

Answer 8

it shouldnt matter which one you use since they both cycle

Answer 9

I used the e^ to go down, and the cos to go up

Answer 10

......at least thats what I had in mind .... might of got things crossed in my head tho ;)

Answer 11

integrate by parts

Answer 12

thx dude i will read that out million thx man

Answer 13

youre welcome; if anything, redo it and go thru 2 integration by parts; then you can use the process of swapping sides and such

Answer 14

ok thx :)

Answer 15

yeah, I see what I did. I inadvertently differentiated both of them instead so my answer is off on that regards; but the concept is valid lol

Answer 16

u mean u differentiated both instead of differentiating 1 and integrating 1?

Answer 17

yeah, trying to work in in my head i got a few things crossed; let me try it again correctly tho

Answer 18

okok i will do it too hope i can get it right..

Answer 19

\begin{array}l
&&cos(t)\\
+&e^{-2t}&sin(t)\\
-&-2e^{-2t}&-cos(t)\\
+&4e^{-2t}
\end{array}

\[\int e^{-2t}cos(t)dt= e^{-2t}sin(t)-\int-2e^{-2t}sin(t)\ dt\]
\[\int e^{-2t}cos(t)dt= e^{-2t}sin(t)+2\int e^{-2t}sin(t)\ dt\]
\[\int e^{-2t}cos(t)dt= e^{-2t}sin(t)+2\left(-e^{-2t}cos(t)-\int 2e^{-2t}cos(t)\ dt\right)\]
\[\int e^{-2t}cos(t)dt= e^{-2t}sin(t)-2e^{-2t}cos(t)-4\int e^{-2t}cos(t)\ dt\]
now we add 4(int: e^(-2t) cos(t)) to both sides.
\[4\int e^{-2t}cos(t)\ dt+\int e^{-2t}cos(t)dt= e^{-2t}sin(t)-2e^{-2t}cos(t)\]combine like terms
\[5\int e^{-2t}cos(t)\ dt= e^{-2t}sin(t)-2e^{-2t}cos(t)\]and divide of the 5
\[\int e^{-2t}cos(t)\ dt=\frac{ e^{-2t}sin(t)-2e^{-2t}cos(t)}{5}\]
and then we dbl chk with the wolf ;)

Answer 20

hey we got the same answer yess but it has a constant rite? do i need to add a + K?

Answer 21

http://www.wolframalpha.com/input/?i=integrate+%28e%5E%28-2t%29cos%28t%29%29
and its good

Answer 22

yes, the +C at the end is good to put there since this has no definite result

Answer 23

yes okk man thxx !