Perform the substitution
#x=4sintheta#, #=>#, #dx=4costhetad theta#
#sqrt(16-x^2)=sqrt(16-16sin^2theta)=4costheta#
Therefore,
The integral is
#I=intx^2sqrt(16-x^2)dx=int16sin^2theta*4costheta*4costheta d theta#
#=256intsin^2thetacos^2theta d theta#
#2sinthetacostheta=sin(2theta)#
#I=64intsin^2(2theta)d theta#
#cos(4theta)=1-2sin^2(2theta)#
#sin^2(theta)=(1-cos(4theta))/2#
Therefore,
#I=32int(1-cos(4theta))d theta#
#=32(theta-sin(4theta)/4)#
#=32arcsin(x/4)-8sin(4theta)#
#sin(4theta)=costheta(4sintheta-8sin^3theta)#
#=sqrt(16-x^2)/4(x-1/8x^3)#
And finally,
#intx^2sqrt(16-x^2)dx=32arcsin(x/4)-2sqrt(16-x^2)(x-1/8x^3)+C#
