Find or evaluate the integral

The integral from 1 to e^(pi/2) of the function: cos[ln(x)]dx
I know you need to integrate by parts, but I don't see how to on this one.
If you could show me how you got the answer that would be helpful

∫ cos[ln(x)]dx from 1 to e^(π/2)

Let u = cos[ln(x)]
du = (-1/x)sin[ln(x)] dx
dv = dx
v = x

∫ cos[ln(x)]dx = xcos[ln(x)] + ∫sin[ln(x)] dx

Do it again:

u = sin[ln(x)]
du = (1/x)cos[ln(x)] dx
dv = dx
v = x

∫sin[ln(x)] dx = xsin[ln(x)] - ∫ cos[ln(x)] dx

So:

∫ cos[ln(x)]dx = xcos[ln(x)] + xsin[ln(x)] - ∫ cos[ln(x)] dx

Add the integral to both sides

2 ∫ cos[ln(x)]dx = xcos[ln(x)] + xsin[ln(x)]
∫ cos[ln(x)]dx = (1/2)xcos[ln(x)] + (1/2)xsin[ln(x)]

Now evaluate:

= (1/2)[e^(π/2)]cos[ln(e^(π/2))] + (1/2)[e^(π/2)]sin[ln(e^(π/2))] - (1/2)(1)cos[ln(1)] + (1/2)(1)sin[ln(1)]
= (1/2)[e^(π/2)]cos[π/2] + (1/2)[e^(π/2)]sin[π/2] - (1/2)cos[0] + (1/2)sin[0]
= (1/2)[e^(π/2)] - (1/2)

Done!