`intsqrt(1-4x^2)dx`
`=intsqrt(1-(2x)^2)dx`
Now apply the integral substitution,
Let u=2x,
`=>du=dx`
`=intsqrt(1-u^2)du`
Now using the standard integral,
`intsqrt(a^2-x^2)dx=(xsqrt(a^2-x^2))/2+a^2/2arcsin(x/a)+C`
`=(usqrt(1-u^2))/2+1/2arcsin(u/1)+C`
Now substitute back u=2x,
`=(2xsqrt(1-(2x)^2))/2+1/2arcsin((2x)/1)+C`
`=xsqrt(1-4x^2)+1/2arcsin(2x)+C`
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