$\displaystyle \int \frac{sin(ln(x))cos(ln(x))}{x} dx$
$\displaystyle u = sin(ln(x))$
$\displaystyle du = \frac{cos(ln(x))}{x} dx$
$\displaystyle \int u du$
$\displaystyle \frac{1}{2} u^2 + C $
$\displaystyle \frac{1}{2}sin^2(ln(x)) + C $
$\displaystyle \int \frac{sin(ln(x))cos(ln(x))}{x} dx$
$\displaystyle u = sin(ln(x))$
$\displaystyle du = \frac{cos(ln(x))}{x} dx$
$\displaystyle \int u du$
$\displaystyle \frac{1}{2} u^2 + C $
$\displaystyle \frac{1}{2}sin^2(ln(x)) + C $
There is none. I assume you are surprised because the answer book or whatever says $\displaystyle -\dfrac{1}{2}\cos^2 (\ln x)+\mathcal{C}$
Both answers only differ by a constant, which disappears once you make the integral definite:
$\displaystyle \displaystyle\int_a^b\frac{\sin (\ln x)\cos (\ln x)}{x}\;dx$
$\displaystyle =\dfrac{1}{2}(\cos^2 (\ln a)-\cos^2 (\ln b))$
$\displaystyle =\dfrac{1}{2}(\sin^2 (\ln b)-\sin^2 (\ln a))$
Both the answers are right, You could have made the substitution as
Just change the substitution and you would get the desired answer.
u = cos(ln(x))
Now du = (-sin(ln(x)))/x dx
Thus the integral becomes
-∫▒udu = -u^2/2+C = - 〖cos〗^2 (ln(x) )+C