How do you integrate #int x^3sqrt(16-x^2)# by trigonometric substitution?

Let #x=4sint#, then #sqrt(16-x^2)=4cost# and #dx=4costdt#

and #intx^3sqrt(16-x^2)dx#

= #int64sin^3t*4cost*4costdt#

= #1024intsin^3tcos^2tdt#

= #1024int(1-cos^2t)cos^2tsintdt#

= #1024int(cos^2t-cos^4t)sintdt#

Let #cost=u# then #du=-sintdt# and our integral becomes

#-1024int(u^2-u^4)du#

= #-1024(u^3/3-u^5/5)#

i.e. #1024(-1/3cos^3t+1/5cos^5t)# (substituting #u=cost#)

and as #cost=1/4sqrt(16-x^2)#, our integral becomes

#1024(-1/3*1/64(16-x^2)^(3/2)+1/5*1/1024(16-x^2)^(5/2))+c#

= #-16/3(16-x^2)^(3/2)+1/5(16-x^2)^(5/2)+c#

Here,

#I=intx^3sqrt(16-x^2)dx#

Let, #x=4sinu=>dx=4cosudu#

#and sinu=x/4=>cosu=sqrt(1-sin^2u)=sqrt(1-x^2/16#

#=>cosu=sqrt(16-x^2)/4...to(A)#

#I=int64sin^3usqrt(16-16sin^2u)xx4cosudu#

#=256intsin^3uxx4cosuxxcosudu#

#I=1024intsin^3ucos^2udu#

#I=4^5int(1-cos^2u)cos^2uxxsinudu#

#=4^5[intcos^2usinudu-intcos^4usinudu]#

#=4^5[-int(cosu)^2(-sinu)du+int(cosu)^4(-sinu)du]#

#=4^5[-(cosu)^3/3+(cosu)^5/5]+c#

#=4^5[(cosu)^5/5-(cosu)^3/3]+c...to#from #(A)#

#=4^5[(sqrt(16-x^2))^5/(5xx4^5)-(sqrt(16-x^2))^3/(3xx4^3)]+c#

#=(sqrt(16-x^2))^5/5-4^2xx(sqrt(16-x^2))^3/3+c#

#I=(sqrt(16-x^2))^5/5-(16(sqrt(16-x^2))^3)/3+c#