If we define $\arcsin(x)=\int_0^x 1/\sqrt{1-t^2}dt$, how to show $\arcsin(x)$ is continuous and increasing Specifically, want to show that $\arcsin(x)$ is an increasing continuous function from $[-1,1]$ onto $[-\pi/2, \pi/2]$. My thought is that there is not much work to do since the integrand is positive, the function must be increasing? How does one show continuity?
From the fundamental theorem, $F(x)=\int_0^xf(t)\,\mathrm dt$ with $f$ continuous, implies $F'(x)=f(x)$. So $\arcsin(x)$ is not just continuous, it is differentiable!

Note that we have some problems at $x=\pm1 $ because the integrand grows to infinity thaer. But that implies that we have an improper integral for these $x$, so that by definition $\arcsin(\pm1)=\lim_{x\to\pm1}\arcsin(x)$.
The following result may be helpful for you:


  $f^{-1}(x)$ is increasing and continuous iff $f(x)$ is increasing and continuous.