The integral ∫cos(logex)dx is equal to :(where C is a constant of integration)

I=∫cos(lnx)dx
Let u = lnx, then x = eu and dx = eudu
I = ∫cos(u)eudu
Using integration by parts:
∫fg' = fg - ∫f'g
Let f = cos(u) and g' = eu, then f' = -sin(u) and g = eu
I = cos(u)eu + ∫sin(u)eudu
Now, let's integrate ∫sin(u)eudu using integration by parts again:
Let f = sin(u) and g' = eu, then f' = cos(u) and g = eu
∫sin(u)eudu = sin(u)eu - ∫cos(u)eudu
Substitute this back into the expression for I:
I = cos(u)eu + sin(u)eu - ∫cos(u)eudu
I = cos(u)eu + sin(u)eu - I + C
2I = eu[cos(u) + sin(u)] + C
I = (1/2)eu[cos(u) + sin(u)] + C
Substitute back u = lnx and x = eu:
I = (1/2)x[cos(lnx) + sin(lnx)] + C
Therefore, the integral ∫cos(logex)dx = (1/2)x[cos(lnx) + sin(lnx)] + C