Use integration by parts to evaluate the improper integral.

∞
∫e^(-x)cos(x) dx
0
When I used integration by parts, I somehow end up with a non-existent integral.

u = e^(-x)
du = -e^(-x) * dx
dv = cos(x) * dx
v = sin(x)

int(e^(-x) * cos(x) * dx) = e^(-x) * sin(x) - int(-e^(-x) * sin(x) * dx)
int(e^(-x) * cos(x) * dx) = e^(-x) * sin(x) + int(e^(-x) * sin(x) * dx)

u = e^(-x)
du = -e^(-x) * dx
dv = sin(x) * dx
v = -cos(x)

int(e^(-x) * cos(x) * dx) = e^(-x) * sin(x) + (-e^(-x) * cos(x) - int(e^(-x) * cos(x) * dx))
int(e^(-x) * cos(x) * dx) = e^(-x) * sin(x) - e^(-x) * cos(x) - int(e^(-x) * cos(x) * dx)
2 * int(e^(-x) * cos(x) * dx) = e^(-x) * (sin(x) - cos(x)) + C
int(e^(-x) * cos(x) * dx) = (1/2) * e^(-x) * (sin(x) - cos(x)) + C

From 0 to infinity

(1/2) * e^(-inf) * (sin(inf) - cos(inf)) - (1/2) * e^(-0)* (sin(0) - cos(0))

Now, sin(inf) and cos(inf) will be some value between -1 and 1, and e^(-inf) = 1/e^inf = 1/inf = 0, so we have:

(1/2) * 0 * (Constant value) - (1/2) * 1 * (0 - 1)
0 + (1/2) * 1
1/2

Let I=integral(e^(-x)cos(x))dx
Let u=e^(-x) => du=-e^(-x)dx
dv=cos(x)dx => v=-sin(x)
I=e^(-x)sin(x)-(integral(e^(-x)sin(x)dx… [the negatives in "du" and "v" cancel out when i multiply them.
I must apply integration by parts again.
u=e^(-x) => du=-e^(-x)dx
dv=sin(x)dx => v=cos(x)
I=e^(-x)sin(x)-e^(-x)cos(x)-integral(e^…
At first it appears integration by parts has failed. However, notice my original integral is present. If treat this integral as an unknown and solve for it, I get.
2I=e^(-x)sin(x)-e^(-x)cos(x)
I=e^(-x)sin(x)-e^(-x)cos(x)
At our limits we get:
1/2[0-0]1/2-[e^(-0)sin(0)-e^(-0)cos(x)]… its obvious e^(-x)cos(x)-->0 and e^(-x)cos(x)-->0, as x-->infinity
1/2*(0-[-1])
1/2