Question

solve ∫e^-x(cosx)dx

Answer 1

\( \displaystyle \int e^{-x} \cos x \ dx \) ?

Answer 2

How do you get that?

Answer 3

is that what was written or am i incorrect?

Answer 4

oh yeah that's what was written. Sorry.

Answer 5

no problem!

what have you tried so far? this looks like a bit of a tricky problem. :)

Answer 6

seems like a straightforward integration by parts problem

Answer 7

Well I know I have to use the double antidifferentiation by parts. and I know I have to use this equation: ∫udv = uv - ∫vdu. But I don't know what should be what.

Answer 8

try it either way

Answer 9

with a choice of two functions here and looking at how their derivatives / antiderivatives act, the behavior tends to be about the same either way you choose.

Answer 10

so the first one is -e^-x sinx dx?

Answer 11

what were your choices of u and dv?
i feel like there might be a sign mixup somewhere

Answer 12

u=e^-x and dv=cosxdx

Answer 13

when you integrated dv = cos x dx to get v, you got -sin x? this would be an integral and not derivative

Answer 14

ambiguous comment actually: your process was the derivative, d/dx (cos x) = -sin x, but we wanted integral.

Answer 15

no I got sinx.

Answer 16

oh, i see.
didnt see an integral and assumed it was first part! sorry!
that looks good then

Answer 17

Oh sorry! so I think I understand it. But I don't know the derivative of -e^-x

Answer 18

-e^(-x) = -1 * e^(-x) that -1 is just a constant right?

Answer 19

yes so it goes out in front of the integral?

Answer 20

yup, you could do that too

i assume you have the full uv - integral v du <-- to watch for that negative sign out front too

Answer 21

so the answer is e^-xsinx - [-e^-xsinx+1∫-cose^-x] + c?

Answer 22

did you choose u = e^(-x) and dv = sin x dx for this one? or u = sin x and dv = e^(-x) ?

Answer 23

the first one

Answer 24

Please take note that the following mathematical "trick" is quite a common thing and I hope you do understand this method of solving this exercise.

The most common issue with such integrals is finding which part to substitute in that nice formula you were given.

USUALLY we consider u=*whatever function is next to e^x* and v'=*whatever function looks something like e^x* because in this method we will have to apply that formula twice - and chances are that, by using that notation, the second time you apply it you will end up with your original integral or some simplified integral (constant*e^x) which you can easily determine.

(Note that v' is my notation for dv/dx, u' is du/dx and so forth)

Let I=∫cos(x)*e^(-x) dx (what we are trying to find out).

Let u=cos(x) and v'=e^(-x).
We get that u'=-sin(x) and v=-e^(-x).

Applying the *whatever you people call it* formula ∫u*v'=u*v-∫u'v we have :

∫cos(x) * e^(-x) dx = cos(x)*( -e^(-x) ) - ∫ ( -sin(x) ) *( -e^(-x) ) dx.

Let J=∫ ( -sin(x) ) *( -e^(-x) ) dx = ∫ ( sin(x)*e^(-x) ) dx.

Let u=sin(x) and v'=e^(-x).
We get that u'=cos(x) and v=-e^(-x).

Again, applying the *whatever you people call it* formula ∫u*v'=u*v-∫u'v we have :

∫ ( sin(x)*e^(-x) ) dx = sin(x)*( -e^(-x) ) - ∫( cos(x) )*( -e^(-x) ) dx=

sin(x)*( -e^(-x) ) + ∫( cos(x) )*( e^(-x) ) dx.

Voila, we now have something that goes like this:

I=cos(x)*( -e^(-x) ) - J
J=sin(x)*( -e^(-x) ) + I.

Bring both these equations "up" we have that:

I=cos(x)*( -e^(-x) ) - ( sin(x)*( -e^(-x) ) + I )
=cos(x)*( -e^(-x) ) - sin(x)*( -e^(-x) ) - I

So that 2*I=cos(x)*( -e^(-x) ) - sin(x)*( -e^(-x) )

Which means I= ( cos(x)*( -e^(-x) ) - sin(x)*( -e^(-x) ) /2

or in a more academic way: I=(sin(x)-cos(x)) * e^(-x) * 1/2.
.

Answer 25

just had to rewrite this to see why things weren't adding up when i looked through:
Integral of e^(-x) cos x dx
u = e^(-x) du = - e^(-x) dx
dv = cos x dx v = sin x
= e^(-x) (sin x) - integral of -e^(-x) sin x dx <-- at this point, i think you forgot to change the sign of "- integral of ..." when pulling out the negative: - e^(-x) --> - integral e^(-x)
= e^(-x) (sin x) + integral of e^(-x) sin x dx
u = e^(-x) du = - e^(-x) dx
dv = sin x dx v = - cos x
= e^(-x) sin x + (- e^(-x) cos x - integral (-cos x) (- e^(-x)) dx ) <-- i think you wrote sin x here, might've just been typo. also a negative sign again because the formula is: uv - int v du

Answer 26

"so the answer is e^-xsinx \(\color{red}{-}\) [-e^-x\(\color{red}{sinx}\)\(\color{red}{+}\)1∫-cose^-x] + c?"

Answer 27

* I used the exact opposite notation: u=cos(x) and dv=e^(-x) and solved v=-e^(-x) in my head. That's why du=-sin(x).

* There was a double negative inside the first integral: I= ...- ∫ ( -sin(x) ) *( -e^(-x) ) which I simplified in its further notation with J=∫ ( -sin(x) ) *( -e^(-x) ) =∫ ( sin(x)*e^(-x) )

* I intentionally left the "-" before the integral out when calculating J but I brought it back in

I=cos(x)*( -e^(-x) ) - J
J=sin(x)*( -e^(-x) ) + I

Answer 28

Okay thank you!