1. ## Spot the mistake.

$\int \frac{sin(ln(x))cos(ln(x))}{x} dx$

$u = sin(ln(x))$
$du = \frac{cos(ln(x))}{x} dx$

$\int u du$

$\frac{1}{2} u^2 + C$

$\frac{1}{2}sin^2(ln(x)) + C$

2. ## Re: Spot the mistake.

There is no mistake.

3. ## Re: Spot the mistake.

There is none. I assume you are surprised because the answer book or whatever says $-\dfrac{1}{2}\cos^2 (\ln x)+\mathcal{C}$
Both answers only differ by a constant, which disappears once you make the integral definite:

$\displaystyle\int_a^b\frac{\sin (\ln x)\cos (\ln x)}{x}\;dx$

$=\dfrac{1}{2}(\cos^2 (\ln a)-\cos^2 (\ln b))$

$=\dfrac{1}{2}(\sin^2 (\ln b)-\sin^2 (\ln a))$

4. ## Re: Spot the mistake.

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