First, try integration-by-parts with #u=cos(ln(x))#, #du=-sin(ln(x))*1/x#, #dv=dx#, and #v=x# to get:
#\int cos(ln(x))\ dx=uv-\int v\ du=x cos(ln(x))+\int sin(ln(x))\ dx#.
Now use integration-by-parts again on this second integral with #u=sin(ln(x))#, #du=cos(ln(x))*1/x#, #dv=dx#, and #v=x# to get
#\int cos(ln(x))\ dx=x cos(ln(x))+(xsin(ln(x))-\int cos(ln(x))\ dx)#.
Adding #\int cos(ln(x))\ dx# to both sides of this last equation, dividing both sides by 2, and including the #+C# at the very end gives the answer:
#\int cos(ln(x))\ dx=1/2xcos(ln(x))+1/2xsin(ln(x))+C#.
