I read here that we may define function,
$$ \arcsin a := \int_{0}^a \frac{1}{\sqrt{1-x^2}} \, dx$$ This is what I understood: I believe the domain is for $ a \in [0,1)$, since integrand is continuous on $[0,a]$ so the Riemann Integral makes sense. When $a=1$, we define $$\arcsin 1= \frac{\pi}{2} := \int_{0}^1 \frac{1}{\sqrt{1-x^2}} \,dx$$
Why can we do this? Firstly, we don't know the integral exists?
I do see by MCT of Lebesgue Integral, $\arcsin a \rightarrow \arcsin 1^+$. But is this finite?
The function $$\frac{1}{\sqrt{1-x^2}}= \frac{1}{\sqrt{1-x}\sqrt{1+x}}\le \frac{1}{\sqrt{1-x}}$$ is summable near the point $x=1$ (check this by integrating the last expression).