integral question?

Question

I would like to know why this integral is = .5[(e^x)(-cos(x) + sin(x)]
as opposed to 0, which is what I keep getting using tabular/integration by parts method.

lim
b --> infinity from left
integral from 0 to b of [ (e^-x)cos(x) ] dx

any help is greatly appreciated. this one is really baffling me

thanks, I definitely found the integral of cosx to be -sinx instead of sinx for v in the beginning....

commencing banging head against brick wall

Answer 1

First, let us find the antiderivative of e^(-x) cos x dx:

∫e^(-x) cos x dx

First, integrate by parts. u=e^(-x), du=-e^(-x) dx, v=sin x, dv=cos x dx. So:

∫e^(-x) cos x dx = e^(-x) sin x + ∫e^(-x) sin x dx + C

Now integrate by parts again: u=e^(-x), du=-e^(-x) dx, v=-cos x, dv=sin x dx

∫e^(-x) cos x dx = e^(-x) sin x - e^(-x) cos x - ∫e^(-x) cos x dx + C

Now we employ simple algebra. Add ∫e^(-x) cos x dx to both sides:


2 ∫e^(-x) cos x dx = e^(-x) sin x - e^(-x) cos x + C

Dividing by 2:

∫e^(-x) cos x dx = (e^(-x) sin x - e^(-x) cos x)/2 + C

Or factoring:

∫e^(-x) cos x dx = e^(-x)(sin x - cos x)/2 + C

Now, as for evaluating the integral:

[b→∞]lim [0, b]∫e^(-x) cos x dx
[b→∞]lim e^(-x)(sin x - cos x)/2 + C |[0, b]
[b→∞]lim e^(-b)(sin b - cos b)/2 - e^(-0)(sin 0 - cos 0)/2
[b→∞]lim e^(-b)(sin b - cos b)/2 + 1/2

And taking the limit as b→∞

0 + 1/2
1/2