How do you find the integral of #int (cotx)^5(sinx)^4dx#?

2 Answers
Jun 2, 2018

#t^4/4+t^2/2-1/2*ln|t+1|-1/2ln|t-1|+C# where #t=cos(x)#

Explanation:

Note that
#cot^4(x)*sin^4(x)=cos^5(x)/sin^5(x)*sin^4(x)=cos^5(x)/sin(x)#
Substituting

#t=cos(x)#
then we get

#dt=-sin(x)dx#
and our integral will be
#-int t^5/(1-t^2)dt#
and this is
#-int( -t^3-t-1/(2*(t+1))-1/(2*(t-1)))dt#
this is
#-(t^4/4-t^2/2-1/2*ln|t+1|-1/2*ln|t-1|)+C#

Jun 2, 2018

#I=ln|sinx|-sin^2x+1/4sin^4x+c#

Explanation:

Here,

#I=int(cotx)^5(sinx)^4dx#

#=intcos^5x/sin^5x xxsin^4xdx#

#=intcos^5x/sinxdx#

#=int((cos^2x)^2cosx)/sinxdx#

#=int((1-sin^2x)^2cosx)/sinxdx#

Subst. #sinx=u=>cosxdx=du#

So,

#I=int(1-u^2)^2/udu#

#=int(1-2u^2+u^4)/udu#

#=int[1/u-2u+u^3]du#

#=ln|u|-2u^2/2+u^4/4+c#

Subst. back .#u=sinx#

#I=ln|sinx|-sin^2x+1/4sin^4x+c#