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.

Recall that $\frac{d}{dx}(arcsin x) =\frac{1}{\sqrt{1-x^2}}$ and hence posing $u=\frac{x}{2}$ and hence $dx=2du$ substituting you have
$
\int \frac{du}{\sqrt{1-u^2}} = \arcsin u = \arcsin \frac{x}{2}
$
