Rewrite with a root:
#intx^2/sqrt(16-x^2)dx#
For integrals involving the root #sqrt(a^2-x^2),# we use the substitution #x=asintheta.#
Here, #a^2=16, a=4, x=4sintheta, dx=4costhetad theta# and we get
#int(16sin^2theta4costhetad theta)/sqrt(16(1-sin^2theta))#
Recalling that #1-sin^2theta=cos^2theta,#we get
#=16int(sin^2thetacosthetad theta)/sqrt(cos^2theta)#
#=16int(sin^2thetacancelcostheta)/(cancelcostheta)d theta#
#=16intsin^2thetad theta#
Recalling that #sin^2theta=1/2(1-cos2theta),# we get
#16/2int(1-cos2thetad theta)=8(theta-1/2sin2theta)+C#
We need to get things in terms of #x#.
Recalling that #x=4sintheta, sintheta=x/4, theta=arcsin(x/4)#
To determine #1/2sin2theta,# recall the identity #1/2sin2theta=sinthetacostheta.# To use this, we need to determine the cosine using the below Pythagorean identity:
#sin^2theta+cos^2theta=1#
#cos^2theta=1-sin^2theta#
#cos^2theta=1-x^2/16#
#costheta=sqrt(16-x^2)/4#
Thus, #1/2sin2theta=(x/4)sqrt(16-x^2)/4=(xsqrt(16-x^2))/16#
And
#intx^2/(16-x^2)^(1/2)dx=8arcsin(x/4)-(xsqrt(16-x^2))/2+C#
