How do you integrate #int ln(sint)cost# by integration by parts method?

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Answer 1 · 2017-01-15T23:49:37

#intln(sint)costdt=sint(ln(sint)-1)+C#

First let #w=sint#, implying that #dw=costdt#. This will help simplify the problem before we attempt integration by parts. Through this substitution:

#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 #w=sint#:

#I=sint(ln(sint)-1)+C#