Indefinite Integral of Sin[Log[x]]

I solved this problem, but I'm wondering if the method I used was correct, or if I just got lucky and ended up with the correct answer by chance.  I used a substitution for the Log[x] and left it there through solving the entire problem, which didn't feel right as I did it, but ended up with the right answer... is that correct?
My work:
∫Sin[Log[x]] dx = ∫x*Sin[u] du where u = Log[x]
Using integration by parts:
∫x*Sin[u] du = -x*Cos[u] + ∫x*Cos[u] du
Integration by parts again:
∫x*Sin[u] du = -x*Cos[u] + x*Sin[u] - ∫x*Sin[u] du
2∫x*Sin[u] du = -x*Cos[u] + x*Sin[u]
∫x*Sin[u] du = x*(Sin[u]-Cos[u])/2
therefore, ∫Sin[Log[x]]dx = x*(Sin[Log[x]-Cos[Log[x]])/2

>>4
it's not just about grading and interpretation. it's just plain wrong to leave x in the integrand when it is du, because then x becomes a constant, and you're left with a totally different integral.