Here,
#I=intx^3sqrt(16-x^2)dx=intx^2sqrt(16-x^2)*xdx#
Let, #x=4sinu=>x^2=16sin^2u=>2xdx=32sinucosudu#
i.e. #xdx=16sinucosudu#
#=>I=int(16sin^2u)sqrt(16-16sin^2u)(16sinucosu)du#
#I=16xx4xx16intsin^2u(cosu)sinucosudu#
#I=1024intsin^2ucos^2usinu du#
#=-1024int(1-cos^2u)cos^2u(-sinu)du#
#=-1024[int(cosu)^2(-sinu)du-int(cosu)^4(-sinu)du#
#=-1024[(cosu)^3/3-(cosu)^5/5]+Ctowhere,sin^2u=x^2/16#
#=-1024[(sqrt(1-sin^2u))^3/3-(sqrt(1-sin^2u))^5/5]+C#
#=-1024[(sqrt(1-x^2/16))^3/3-(sqrt(1-x^2/16))^5/5]+C#
#=-1024[(sqrt(16-x^2))^3/(3xx64)-(sqrt(16-
x^2))^5/(5xx1024)]+C#
#=-[(16(sqrt(16-x^2))^3)/3-(sqrt(16-x^2))^5/5]+C#
#I=(sqrt(16-x^2))^5/5-(16(sqrt(16-x^2))^3)/3+C#