Integrate $\sqrt{4 - x^2}\cdot\operatorname{sgn}(x-1)$

I'm trying to learn calculus through self study and I happened upon the following exercise:

$$ \int_{0}^{2} \sqrt{4 - x^2}\cdot\operatorname{sgn}(x-1) \,dx $$

Seeing the sgn I thought: well this is easy and concluded that since $ \int_{0}^{2} \operatorname{sgn}(x-1) \,dx = 0$, the answer should be zero. But of course then I remembered that $ \int f(x)g(x) \,dx \neq \int f(x) \,dx \int g(x) \,dx$ and got lost.

I know how to solve $ \int_{0}^{2} \sqrt{4 - x^2}dx$, since that geometrically corresponds to a quarter circle of radius $2$ it's just $\pi$, but how do I approach the product of these things?

That integral is equal to$$-\int_0^1\sqrt{4-x^2}\,\mathrm dx+\int_1^2\sqrt{4-x^2}\,\mathrm dx.$$So, all you have to do is to compute those two integrals. In order to do that, it is usful to know that$$\int\sqrt{4-x^2}\,\mathrm dx=\frac12x\sqrt{4-x^2}+2\arcsin\left(\frac{x}{2}\right).$$