How do I calculate the area $\int_0^2 \frac{1}{\sqrt{2x-x^2}} dx$? I have the integral

$$\int_0^2 \frac{1}{\sqrt{2x-x^2}}dx$$

I know the answer is $\pi$ but I have problems with limits $0$ and $2$. How can I use $\varepsilon$ in this problem?
$$2x-x^2=-(x-1)^2+1\implies \sqrt{2x-x^2}=\sqrt{1-(x-1)^2}\implies$$

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

Be careful here: with limits, if one of them is zero or two this is then an improper integral, though it looks nice
$$\int_0^2 \frac{dx}{\sqrt{2x - x^2}}
=
\int_{- 1}^1 \frac{dz}{\sqrt{1 - z^2}}
=
\int_{- \frac{\pi}{2}}^ {\frac{\pi}{2}} \frac{\cos t dt}{\sqrt{1 - \sin^2 t}}
=
\int_{- \frac{\pi}{2}}^ {\frac{\pi}{2}} dt = \pi.$$