How do you integrate #int sin(lnx) dx#?

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Answer 1 · 2018-08-07T13:53:31

#intsin(lnx)dx=x/2[sin(lnx)-cos(lnx)]+C#

Here,

#I=intsin(lnx)dx.................(A)#

Subst. #color(red)(lnx=u=>x=e^u=>dx=e^udu#

#:.I=intsinu*e^udu#

#I=sinuinte^udu-int(d/(du)(sinu)inte^udu)du#

#:.I=sinuxxe^u-intcosue^udu#

#I=e^usinu-{cosuxxe^u-int(-sinue^u)du}#

#:.I=e^usinu-e^ucosu-inte^usinudu+c#

#I=e^usinu-e^ucosu-I+c.to#from#(A)#

#:.I+I=e^usinu-e^ucosu+c#

#:.2I=e^usinu-e^ucosu+c#

#:.I=1/2[e^usinu-e^ucosu]+c/2#

Subst.back #color(red)(lnx=u=>x=e^u#

#intsin(lnx)dx=1/2[x*sin(lnx)-xcos(lnx)]+C#

Hence ,

#intsin(lnx)dx=x/2[sin(lnx)-cos(lnx)]+C#