Calculate $\int \frac{1}{\sqrt{4-x^2}}dx$ Calculate $$\int \dfrac{1}{\sqrt{4-x^2}}dx$$
Suppose that I only know regular substitution, not trig.
I tried to get help from an integral calculator, and what they did was:
$$\text{Let u = $\frac{x}{2}$} \to\dfrac{\mathrm{d}u}{\mathrm{d}x}=\dfrac{1}{2}$$
Then the integral became:
$$={\displaystyle\int}\dfrac{1}{\sqrt{1-u^2}}\,\mathrm{d}u = \arcsin(u) = \arcsin(\frac{x}{2})$$
And I'm not sure how they accomplished this, where did the 4 go? I understand the arcsin part but not sure how they got rid of the 4? Also how did they know to substitute $\frac{x}{2}$? It doesn't seem very obvious to me.

$\int \dfrac{1}{\sqrt{4-x^2}}dx=\int \dfrac{\frac{1}{2}}{\frac{\sqrt{4-x^2}}{2}}dx=\int \dfrac{\frac{1}{2}}{\sqrt{{\frac{4}{4}-(\frac{x}{2})^2}}}dx=\int \dfrac{\frac{1}{2}}{\sqrt{{1-(\frac{x}{2})^2}}}dx=arcsin (\frac{x}{2}) + C$.
