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

∫x*Sin[u] du = -x*Cos[u] + ∫x*Cos[u] du

Wrong.

∫x*Sin[u] du = -x*Cos[u] + ∫Cos[u]dx

Oh and dx = xdu.  Fuck.

Your work and answer are correct, but for the love of god do not leave an "x" in the integrand after doing substitution. It's easy to make a mistake (especially when doing two integration by parts afterwards), and it's much harder for whoever is grading it to interpret it.

>>1
Wrong.

The correct method is: mathematica.

(%i1) integrate(sin(log(x)),x);
(%o1) (x*(sin(log(x))-cos(log(x))))/2

USE THE WAAAAAAAAAAAAAAAAGH!

OP here.  I realized I didn't need the substitution at all since adding an x just confused me.  Without using the substition, I used integration by parts in a slightly different manner:

u = Sin[Log[x]], du/dx = Cos[Log[x]]/x, v = x, dv/dx = 1
∫u dv/dx = uv - ∫v du/dx

∫Sin[Log[x]] dx = x Sin[Log[x]] - ∫Cos[Log[x]] dx
∫Sin[Log[x]] dx = x Sin[Log[x]] - (x Cos[Log[x]] + ∫Sin[Log[x]] dx)
2∫Sin[Log[x]] dx = x(Sin[Log[x]] - Cos[Log[x]])

∫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.

>>9
You can treat x as a function of u and arrive at a proper answer - as the OP did. It's pointless and bad notation, but possible.

http://integrals.wolfram.com/

I feel the need, the need for weed!

Marijuana MUST be legalized.