If we make this substitution:
#x=4sintrArrdx=4cost*dt#,
then:
#int(64sin^3t)/sqrt(16-16sin^2t)*4cost*dt=#
#=int(64sin^3t)/(4sqrt(1-sin^2t))*4cost*dt=#
#=int(64sin^3t)/(4cost)*4cost*dt=64intsin^3t*dt=#
#64intsintsin^2t*dt=64intsint(1-cos^2t)dt=#
#64int(sint-sintcos^2t)dt=64(-cost+cos^3t/3)+c=(1)#.
Now, remembering our substitution:
#sint=x/4rArrcost=sqrt(1-sin^2)=sqrt(1-x^2/16)=1/4sqrt(16-x^2)#.
So:
#(1)=64(-1/4sqrt(16-x^2)+1/64*1/3*sqrt(16-x^2)*(16-x^2))+c=#
#=-16sqrt(16-x^2)+1/3sqrt(16-x^2)(16-x^2)+c=#
#=1/3sqrt(16-x^2)(-48+16-x^2)+c=#
#=-1/3sqrt(16-x^2)*(x^2+32)+c#.