D∈ (-1/2,1/2) ; first only the integral , \(\displaystyle I=\int \dfrac{\sin^2 2x }{4\sqrt{4-(2x)^2 }}d(2x)=\int \dfrac{\sin^2 t-1+1 }{4\sqrt{4-t^2 }}dt=\int \dfrac{dt }{4\sqrt{4-t^2 }}-\int \dfrac{\cos^2 t }{4\sqrt{4-t^2 }}dt\).

$I=c+ \dfrac{\arcsin\left(\frac{x}{2}\right)}{4} -\displaystyle \int \dfrac{\cos^2 t }{4\sqrt{4-t^2 }}dt $ ; \(\displaystyle \; \) , the integral that remains is \(\displaystyle \int_D \dfrac{\cos^2 t }{4\sqrt{4-t^2 }}dt\). where t=2x.

Or better apply : \(\displaystyle \sin^2 t = 1/2 (1-\cos 2t)\).