evaluate the integral from 0 to infinity for e^(-x) cos 3 x dx the sol'n I came up with is 3/10 is it correct?

Answer 1

The answer is 1/10.

Use integration by parts. I do it slightly different than taught in standard books, i use a tabular method which is awkward to show in this forum. At any rate:

∫℮^-xcos(3x)dx
Differentiate e^-x and integrate cos(3x) until you get back to the function where you started:

Line1: e^-x cos(3x)
Line2: -e^-x (1/3)sin(3x)
Line3: e^-x –(1/9)cos(3x) {stop this is the original function weighted by a constant}

you sort of “cross multiply” changing every other sign:

1st term in the answer: e^-x(1/3)sin(3x)
2nd term in the answer: (-e^-x)(–(1/9)cos(3x))--> e^-x(1/9)cos(3x) BUT you have to change the sign
2nd term in the answer: -e^-x(1/9)cos(3x)

3rd term in the answer: –1/9 ∫ e^-xcos(3x)

So in total we have:

∫e^-xcos(3x)dx = e^-x(1/3)sin(3x) – e^-x(1/9)cos(3x) – (1/9 ∫ e^-x)cos(3x)

And

(10/9) ∫e^-xcos(3x)dx = e^-x(1/3)sin(3x) – e^-x(1/9)cos(3x)

So

∫e^-xcos(3x)dx = (9/10)[e^-x(1/3)sin(3x) – e^-x(1/9)cos(3x)]

Applying the limits you’ll get 1/10.

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Puggy’s slight error is where he adds, + (1/9) Integral ( e^(-x)cos(3x) dx)

Let u = e^(-x). dv = sin(3x) dx.
du = (-1)e^(-x) dx. v = (-1/3)cos(3x).

*************************
vdu =(1/3) ( e^(-x)cos(3x) dx) BUT this should be minus the integral (like he did in the first part)
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i.e. -(1/9) Integral ( e^(-x)cos(3x) dx)

Answer 2

you were close.
from a table,

int(0:inf) e^(cx)cosbxdx = e^(cx)/(c^2+b^2)*(ccosbx +bsinbx)

in your problem, c = -1 and b = 3.

thus the integral equals (from 0 to inf)

e^(-x)/(1+9)*(-cos3x + 3sin3x)
= 1/(10e^x)*(3sin3x - cos3x)

since the trig function in parentheses has an absolute value less than 4, as x approaches inf, the whole expression approaches 0.

so the integral evaluates as

0 - (1/(10e^0)*(3sin0 - cos0)
= -1/10*(-1) = 1/10.

I know I used a formula and but I'm just checking your work, it's not cheating for me.

the link to the tables is in the source box.

I think Puggy got a sign switched somewhere in his work, possibly when he subtracted the 1/9 integral. If he'd factored out 10/9 instead of 8/9, he'd have had it right. Since you had the right denominator you were probably on the right track as well. My suspicion is that you got the trig functions reversed to end up with 3 instead of one. A very small error, but that's all it takes when it's complicated like this.

Answer 3

Let's check.

Integral (e^(-x) cos(3x) dx )

Let u = e^(-x). dv = cos(3x) dx
du = (-1)e^(-x)dx. v = (1/3)sin(3x)

(1/3)e^(-x)sin(3x) + (1/3) Integral (e^(-x)sin(3x) dx)

Let u = e^(-x). dv = sin(3x) dx.
du = (-1)e^(-x) dx. v = (-1/3)cos(3x).

(1/3)e^(-x)sin(3x) + (1/3) [(-1/3)e^(-x)cos(3x)
+ (1/3) Integral (e^(-x)cos(3x) dx)

(1/3)e^(-x)sin(3x) - (1/9)e^(-x)cos(3x) +
(1/9) Integral ( e^(-x)cos(3x) dx)

Now, we use that funky method of subtracting (1/9) of the integral both sides. This gives us

(8/9) Integral (e^(-x) cos(3x) dx ) = (1/3)e^(-x)sin(3x) - (1/9)e^(-x)cos(3x)

Multiply both sides by 9/8,

Integral (e^(-x) cos(3x) dx ) = (9/8)(1/3)e^(-x)sin(3x) - (9/8)(1/9)e^(-x)cos(3x)

Integral (e^(-x) cos(3x) dx ) = (3/8)e^(-x)sin(3x) - (1/8)e^(-x)cos(3x) + C

Remember that improper integrals must be expressed as a limit; therefore, the question is

lim Integral (0 to t, (e^(-x) cos(3x) dx ) )
t -> infinity

Which we now have the answer for.

lim (3/8)e^(-x)sin(3x) - (1/8)e^(-x)cos(3x) {evaluated from 0 to t) }
t -> infinity

lim [(3/8)e^(-t)sin(3t) - (1/8)e^(-t)cos(3t)] - [0 - (1/8)]
t -> infinity

lim [(3/8)sin(3t)/e^t - (1/8)cos(3t)/e^t - 0 + 1/8]
t -> infinity

As t approaches infinity, the first expression is a bounded function over infinity, so it goes to 0. The second expression is also a bounded function over infinity, so that goes to 0. Therefore, evaluating the limit, we get

0 - 0 - 0 + 1/8, or

1/8