What is #int (ln(sinx))*cosxdx#?

1 Answer
Nov 8, 2015

Answer:

#sin(x)ln(sin(x))-sin(x)+C#
or
#sin(x)(ln(sin(x))-1)+C#

Explanation:

#int(ln(sin(x))*cosxdx#

let # s = sin(x)# (normally I would use "u", but we will use it later)
then #ds = cos(x)dx#

now we have
#intln(s)*ds#

To integrate this we use integration by parts
#int udv = uv - int vdu#

let #u = ln(s)# let #dv = ds#
so #du = 1/sds# and #v=s#

the integral can now be rewritten as
#ln(s)s- int s*1/sds#
#sln(s)- int 1ds#
#sln(s)- s +C#

Now substitute #sin(x)# for # s#

#sin(x)ln(sin(x))-sin(x)+C#
or
#sin(x)(ln(sin(x))-1)+C#