So, we have
#intcos(lnx)dx#
We'll make the following substitution:
#u=lnx#
#du=dx/x#
#xdu=dx#
Let's try to get #x# in terms of #u#:
#e^u=e^lnx#
#e^u=x#
Thus,
#e^udu=dx# and we get
#inte^ucosudu#. We now use Integration by Parts twice, making the following selections:
#w=e^u#
#dw=e^udu#
#dv=cosudu#
#v=sinu#
Applying the formula #uw-intvdw#:
#inte^ucosudu=e^usinu-inte^usinudu#
For #inte^usinudu#:
#w=e^u#
#dw=e^udu#
#dv=sinudu#
#v=-cosu#
Thus,
#inte^usinudu=-e^ucosu+inte^ucosudu#
We see our original integral shows back up:
#inte^ucosudu=e^usinu-(-e^ucosu+inte^ucosudu)#
#inte^ucosudu=e^usinu+e^ucosu-inte^ucosu#
#2inte^ucosudu=e^usinu+e^ucosu#
#inte^ucosudu=1/2(e^usinu+e^ucosu)+C# (Don't forget the constant of integration)
Recalling that #u=lnx, e^u=x#:
#intcos(lnx)dx=1/2(xsin(lnx)+xcos(lnx))+C#
