Question #f9641

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Answer 1 · 2018-02-14T06:11:01

#int\ cos(x)/(sin^2(x)+sin(x))\ "d"x=ln|sin(x)/(sin(x)+1)|+C#

#\ \ \ \ \ \ int\ cos(x)/(sin^2(x)+sin(x))\ "d"x#

Substitute #u=sin(x)# and #"d"u=cos(x)\ "d"x#. This gives
#=int\ ("d"u)/(u^2+u)#
#=int\ ("d"u)/(u(u+1))#

Separate to partial fractions since #1/(u(u+1))=1/u-1/(u+1)#:
#=int\ (1/u-1/(u+1))\ "d"u#

#=ln|u|-ln|u+1|+C#

#=ln|u/(u+1)|+C#

Substitute back #u=sin(x)#:

#=ln|sin(x)/(sin(x)+1)|+C#