Evaluate integral with substitution and then by parts - wondering what I did wrong Evaluate the integral by using substitution prior to integration by parts
Integral is:
$\int sin(lnx) dx$
$w = lnx$ ....  $dw = \frac 1x$ .... $dx = e^u dw$
Integrating by parts I get
$\int sin(w) dw = sin(w)e^w - \int cos(w)e^w dw$
and I don't know how to go from there. I tried doing integration by parts again but I'm not getting anywhere. Any help is appreciated.
Edit: Integrating by parts again I get:
$e^wsin(w) - e^wcos(w) + \int e^wsin(w) dw$
 A: $$\int \sin(\ln x) dx=\int \sin(u)e^u du=\sin(u)e^u-\int \cos(u)e^u du$$ now by doing same method we have, $$\int \cos(u)e^u du=-\sin(u)e^u+\int \sin(u)e^u du$$ 
combine this two.
A: Your integral is
$$\int {\sin (\ln (x))dx} $$
using the substitution $u=ln(x)$ and considering $dx = {e^u}du$ it becomes
$$\int {\sin (u){e^u}du} $$
Now we use integration by parts two times to get
$$\eqalign{
  & \int {\sin (u){e^u}du}  = \sin (u){e^u} - \int {\cos (u){e^u}du}  + C  \cr 
  & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \sin (u){e^u} - \left( {\cos (u){e^u} - \int { - \sin (u){e^u}du} } \right) + C  \cr 
  & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \sin (u){e^u} - \cos (u){e^u} - \int {\sin (u){e^u}du}  + C \cr} $$
Finally, solve the above equation for $\int {\sin (u){e^u}du} $ which leads to
$$\int {\sin (u){e^u}du}  = {1 \over 2}{e^u}\left( {\sin (u) - \cos (u)} \right) + {1 \over 2}C$$
if you want your final answer in $x$ just substitute $u=ln(x)$ to get
$$\int {\sin (\ln (x))dx}  = {1 \over 2}x\left( {\sin (\ln (x)) - \cos (\ln (x))} \right)   + {1 \over 2}C$$
