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.
 A: Since $\text{d}x=2\text{d}u$, we have $2$ both in the numerator and in the denominator and it reduces.
A: $$\int \frac{\text{d}x}{\sqrt{4-x^2}}=\int \frac{2 \ \text{d}u}{\sqrt{4-(2u)^2}}=\int \frac{2 \ \text{d}u}{\sqrt{4(1-u^2)}}=\int \frac{2 \ \text{d}u}{2\sqrt{1-u^2}}=\int \frac{ \text{d}u}{\sqrt{1-u^2}} $$
Why especially this substitution: Notice that 
$$\int \frac{\text{d}x}{\sqrt{4-x^2}}=\int \frac{\text{d}x}{\sqrt{4\left(1-\frac14x^2\right)}}=\int \frac{\text{d}x}{2\sqrt{1-\left(\frac{x}{2} \right)^2}}$$
so you can see that it is quite nice to substitute $u=\frac{x}{2}$; we get a function $\frac{1}{\sqrt{1-u^2}}$ and we already know the integral to this one.
A: $\int \dfrac{1}{\sqrt{4-x^2}}dx=\int \dfrac{\frac{1}{2}}{\frac{\sqrt{4-x^2}}{2}}dx=\int \dfrac{\frac{1}{2}}{\sqrt{{\frac{4}{4}-(\frac{x}{2})^2}}}dx=\int \dfrac{\frac{1}{2}}{\sqrt{{1-(\frac{x}{2})^2}}}dx=arcsin (\frac{x}{2}) + C$.
A: 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}
$
