Question

integrate by parts e^-x cos 2xdx.

Answer 1

Did you mean
\[\int\limits_{}^{}e^{-x}*Cos(2x)dx\]

Answer 2

yes

Answer 3

oaky working on it

Answer 4

thank you!!

Answer 5

So in this case you use integration by parts.
Let:
\[u = e^{-x}\], then
\[du = -e^{-x}*dx\]

Answer 6

\[dv = Cos(2x)dx\]
\[v = (Sin(2x))/2\]

Answer 7

Integration by parts: u*v - Integral(v*du)
\[e^{-x}*(1/2)*(Sin(2x)) - \int\limits_{}^{}(1/2)(Sin(2x))*(-e^{-x})dx\]

Answer 8

Simplify (take 1/2 out):
\[(1/2) (e^{-x}Sin(2x) + \int\limits_{}^{}Sin(2x)e^{-x}dx)\]
Now for now ignore everything except:
\[I =\int\limits_{}^{}Sin(2x)*e^{-x}*dx\]

Answer 9

I will just call it I, solve it and then plug it back into the equation above. Do you get it?

Answer 10

\[I =\int\limits_{}^{}Sin(2x)*e^{-x}*dx\]
use integration by parts again:
u = e^(-x); du = -e^(-x)*dx

Answer 11

hey, sorry I hate typing out integral problems, especially long ones, can I write it out, take a picture and email it to you?

Answer 12

my page just crashed and I lost half the stuff I entered

Answer 13

yeah no problem

Answer 14

okay workin on it

Answer 15

i understand how to get to:

=e^-x cos 2x + 2e^-x sin 2x -4 integral e^-x cos 2xdx,

Answer 16

Oh then its all good. Now move -4 integral e^-x cos 2xdx, to the left

Answer 17

So you'll get:
5*integral e^-x cos 2xdx = e^-x cos 2x + 2e^-x sin 2x

Answer 18

So for the answer:
integral e^-x cos 2xdx = [ (e^-x cos 2x + 2e^-x sin 2x) / 5 ] + C

Answer 19

do you get it? I mean as long as your solution is correct, then that's the answer. Wait, actually I just checked it and you're off by a minus sign somewhere

Answer 20

[ (-e^-x cos 2x + 2e^-x sin 2x) / 5 ] + C
See, you were right, but just check your arithmetic. Remember this:
When you have two functions that cannot be differentiated down to 0, you keep integrating, one side until you see something that looks like the original integral.

Answer 21

i think i get it! :)

Answer 22

So in this case you keep integrating by parts until one side has the original integral in it. Once you get that original integral, just move it over and solve for it.

Answer 23

awesome thank you very much!!!

Answer 24

I remember having trouble with integration by parts back in HS. Especially if you have two functions that can't be differentiated down to 0, and kept wonderin which one to pick for u and dv, the answer is, it doesn't matter, your still goin around in circles. Even if u pick Cos(2x) for u and e^(-x) dx for dv, u will get the same answer. Click on Become a fan if I helped, lol I wanna see who I become after superhero! Thanks.

Answer 25

i did! yeah i never know what to pick for u or dv

Answer 26

lol 40 fans... and still Superhero, well maybe at 50 =D!