Recognise arcsine and arctangent forms

$\int\frac1{\sqrt{4-x^2}}\,dx=\int\frac1{\sqrt{2^2-x^2}}\,dx=\sin^{-1}(x/2)+C$, for $|x|<2$.

$\int\frac1{9+x^2}\,dx=\int\frac1{3^2+x^2}\,dx=\frac13\tan^{-1}(x/3)+C$ for every real $x$.

Factor the radical: $\sqrt{1-4x^2}=\sqrt{4(1/4-x^2)}=2\sqrt{(1/2)^2-x^2}$. Hence $\int\frac1{\sqrt{1-4x^2}}\,dx=\frac12\int\frac1{\sqrt{(1/2)^2-x^2}}\,dx=\frac12\sin^{-1}\bigl(\frac{x}{1/2}\bigr)+C=\frac12\sin^{-1}(2x)+C$.

Take $f(x)=2x$, $f^{\prime}(x)=2$ and $a=1$. Then $\int\frac1{\sqrt{1-(2x)^2}}\,dx=\frac12\int\frac2{\sqrt{1-(2x)^2}}\,dx=\frac12\sin^{-1}(2x)+C$. Both methods require $|x|<1/2$.