I dropped out of university for 3 years or so and now I am want to come back so I started solving some calculus assignments in order to wash off my rusty knowledge of it. Anyway, as you can probably guess I started with the most simple indefinite integrals in order to recall and practice the main integral table formulas, where I met this one $\int \frac{\arccos(\frac{x}{2})}{\sqrt{4-x^2}}dx$ as you can see it looks pretty simple when the place where I've taken this integral tries to convince me that $\sqrt{4-x^2} = \sqrt{1-(\frac{x}{2})^2}$ sadly I cannot see it (maybe because of my long gone mathematics skills and intuition).
Can you please guys try to elaborate on this? I tried wolfram but one of it steps is
and I fail to understand how $du= - \frac{1}{\sqrt{4-x^2}}dx$
No, it is not true that $\sqrt{4-x^2}=\sqrt{1-\left(\frac x2\right)^2}$. What you have is$$\sqrt{4-x^2}=\sqrt{4-4\left(\frac x2\right)^2}=2\sqrt{1-\left(\frac x2\right)^2}.$$On the other hand, since $\arccos'(x)=-\frac1{\sqrt{1-x^2}}$, then, if $u=\arccos\left(\frac x2\right)$, then$$\mathrm du=-\frac12\cdot\frac{\mathrm dx}{\sqrt{1-\left(\frac x2\right)^2}}=-\frac{\mathrm dx}{\sqrt{4-x^2}}.$$