So, we want #intsin(lnx)dx#
Let #u=lnx#
#du=dx/x#
#xdu=dx#
Now we want to get #x# in terms of #u.# Recalling that #u=lnx:#
#e^lnx=e^u#
#x=e^u#
So,
#e^udu=dx#
Apply the substitution, yielding
#inte^usin(u)du#
This will require two applications of Integration by Parts.
Let
#w=e^u#
#dw=e^udu#
#dv=sin(u)du#
#v=-cos(u)#
Then, integrating by parts with #wv-intvdw#:
#inte^usin(u)du=-e^ucosu+inte^ucosu#
We will now have to determine #inte^ucosudu#
#w=e^u#
#dw=e^udu#
#dv=cosudu#
#v=sinu#
#inte^ucosudu=e^usinu-inte^usinudu#
So, we see our original integral showed up here.
Recalling that #inte^usin(u)du=-e^ucosu+inte^ucosudu#, substitute in #e^usinu-inte^usinudu#:
#inte^usin(u)du=-e^ucosu+e^usinu-inte^usinudu#
Solve for #inte^usinudu#:
#2inte^usinudu=e^usinu-e^ucosu#
#inte^usinu=1/2(e^usinu-e^ucosu)+C#
Don't forget the constant of integration.
Rewrite in terms of #x:#
#intsin(lnx)dx=1/2(e^(lnx)sin(lnx)-e^(lnx)cos(lnx))+C#
Since #e^lnx=x:#
#intsin(lnx)dx=1/2(xsin(lnx)-xcos(lnx))+C#
