How do you integrate #int ln(sint)cost# by integration by parts method?
1 Answer
Jan 15, 2017
Explanation:
First let
#I=intln(sint)costdt=intln(w)dw#
Now we should apply integration by parts. Let:
#{(u=ln(w)" "=>" "du=1/wdw),(dv=dw" "=>" "v=w):}#
Thus:
#I=wln(w)-intw1/wdw=wln(w)-intdw#
Integrating and factoring:
#I=wln(w)-w=w(ln(w)-1)#
From
#I=sint(ln(sint)-1)+C#