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

$\displaystyle u = sin(ln(x))$

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

$\displaystyle \int u du$

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

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

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- Jan 1st 2013, 01:18 PMNervousSpot the mistake.
$\displaystyle \int \frac{sin(ln(x))cos(ln(x))}{x} dx$

$\displaystyle u = sin(ln(x))$

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

$\displaystyle \int u du$

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

$\displaystyle \frac{1}{2}sin^2(ln(x)) + C $ - Jan 1st 2013, 01:28 PMemakarovRe: Spot the mistake.
There is no mistake.

- Jan 1st 2013, 01:31 PMLordoftheFliesRe: Spot the mistake.
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:

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

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

$\displaystyle =\dfrac{1}{2}(\sin^2 (\ln b)-\sin^2 (\ln a))$ - Jan 1st 2013, 07:25 PMibduttRe: 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