Let x=½sinθ, dx=½cosθdθ; √(1-4x2)=√(1-sin2θ)=cosθ; 4x2=sin2θ.
Integrand becomes sin2θcosθ(½cosθdθ)=½sin2θcos2θdθ.
Since sin(2θ)≡2sinθcosθ, ⅛sin2(2θ)≡½sin2θcos2θ.
cos(2A)≡1-2sin2A where a is any angle, sin2A≡½(1-cos(2A)).
If A=2θ, sin2(2θ)≡½(1-cos(4θ)).
⅛∫sin2(2θ)dθ=(1/16)∫(1-cos(4θ))dθ=(1/16)[θ-¼sin(4θ)]+C, where C is integration constant.
sin(4θ)=2sin(2θ)cos(2θ)=2(2sinθcosθ)(1-2sin2θ)=
4sinθcosθ(1-2sin2θ)=8x(1-8x2)√(1-4x2).
θ=arcsin(2x), so ∫4x2√(1-4x2)dx=(1/16)[arcsin(2x)-2x(1-8x2)√(1-4x2)]+C.
(arcsin is the same as sin-1.)