How is the integral (x*(cosx)^2) / (senx)^4 ?

1 Answer
May 17, 2018

The answer is #=-(xcot^3x)/3-1/6csc^2x+1/3ln(|cscx|)+C#

Explanation:

The integral is

#I=int(x(cos^2xdx))/sin^4x=intx(cos^2x/sin^2x)(dx/sin^2x)#

#=intxcot^2xcsc^2xdx#

Apply integration by parts

#intuv'=uv-intu'v#

#u=x#, #=>#, #u'=1#

#v'=cot^2xcsc^2x#, #=>#, #v=-cot^3x/3#

Therefore,

#I=-(xcot^3x)/3-int-(cot^3xdx)/3#

#=-(xcot^3x)/3+int(cot^3xdx)/3#

#I_1=int(cot^3xdx)/3=1/3intcotx(csc^2x-1)dx)#

Let #u=cscx#, #du=-cotxcscxdx#

Therefore,

#I_1=-(1/3int((u^2-1)du)/u)=-1/3(intudu-1/3int(du)/u)#

#=-1/3*u^2/2+1/3ln(u)#

#=-1/6csc^2x+1/3ln(cscx)#

Finally,

#I=-(xcot^3x)/3-1/6csc^2x+1/3ln(|cscx|)+C#