Given the definite integral:
$$
I(a,\,b) := \int_0^{\frac{\pi}{2}} \frac{\cos\left(a\,\arcsin\left(b\,x\right)\right)}{\sqrt{1 - \left(b\,\sin x\right)^2}}\,\text{d}x
$$
through a substitution of the type $t = \frac{4}{\pi}\,x - 1$ we go back to this other integral:
$$
I(a,\,b) = \int_{-1}^1 \underbrace{\frac{\cos\left(a\,\arcsin\left(b\,\frac{\pi}{4}\,(t + 1)\right)\right)}{\sqrt{1 - \left(b\,\sin \left(\frac{\pi}{4}\,(t + 1)\right)\right)^2}}\,\frac{\pi}{4}}_{:= f(a,\,b,\,t)}\text{d}t
$$
to which it's possible to apply the Legendre-Gauss quadrature.
In particular, opting for the two-point formula:
$$
I(a,\,b) \approx k_1\,f(a,\,b,\,t_1) + k_2\,f(a,\,b,\,t_2)
$$
where is it:
$$
\begin{cases}
k_1\,t_1^0 + k_2\,t_2^0 = \frac{1 + (-1)^0}{1 + 0} \\
k_1\,t_1^1 + k_2\,t_2^1 = \frac{1 + (-1)^1}{1 + 1} \\
k_1\,t_1^2 + k_2\,t_2^2 = \frac{1 + (-1)^2}{1 + 2} \\
k_1\,t_1^3 + k_2\,t_2^3 = \frac{1 + (-1)^3}{1 + 3}
\end{cases}
\; \; \; \Leftrightarrow \; \; \;
\begin{cases}
k_{1,2} = 1 \\
t_{1,2} = \pm \frac{1}{\sqrt{3}}
\end{cases}
$$
it follows that:
$$
I(a,\,b) \approx f\left(a,\,b,\,-\frac{1}{\sqrt{3}}\right) + f\left(a,\,b,\,\frac{1}{\sqrt{3}}\right),
$$
approximation that involves at most an error equal to:
$$
\epsilon(a,\,b) = \frac{1}{135}\underset{-1 \le t \le 1}{\max} \left|\frac{\partial^4 f(a,\,b,\,t)}{\partial t^4}\right|.
$$
As an example:
$$
I\left(3,\,\frac{1}{10}\right) \approx f\left(3,\,\frac{1}{10},\,-\frac{1}{\sqrt{3}}\right) + f\left(3,\,\frac{1}{10},\,\frac{1}{\sqrt{3}}\right) = 1.51672
$$
with a maximum error equal to:
$$
\epsilon\left(3,\,\frac{1}{10}\right) = \frac{1}{135}\underset{-1 \le t \le 1}{\max} \left|\frac{\partial^4 f\left(3,\,\frac{1}{10},\,t\right)}{\partial t^4}\right| = 1.08341\cdot 10^{-4}\,.
$$
Best Answer
I'll be more careful and explicitly write the limits every time.
$$I(a,b,c)=\int_b^c \arccos(a-\cos(y))dy$$
$$y=\arccos z$$
$$dy = -\frac{dz}{\sqrt{1-z^2}}$$
$$I=\int_{\cos b}^{\cos c} \arccos(a-z)\frac{dz}{\sqrt{1-z^2}}$$
$$z=a-u$$
$$u=a-z$$
$$I=\int_{a-\cos c}^{a-\cos b} \arccos(u)\frac{du}{\sqrt{1-(a-u)^2}}$$
$$u=\cos v$$
$$v=\arccos u$$
$$du=- \sin v$$
$$I=\int_{\arccos(a-\cos b)}^{\arccos(a-\cos c)} \frac{v \sin v dv}{\sqrt{1-(a-\cos v)^2}}$$
Yes, this is not fun.
Let's get a little back:
$$I=\int_{a-\cos c}^{a-\cos b} \arccos(u)\frac{du}{\sqrt{1-(a-u)^2}}$$
$$I=\frac{\pi}{2}\int_{a-\cos c}^{a-\cos b} \frac{du}{\sqrt{1-(a-u)^2}}-\int_0^1 \int_{a-\cos c}^{a-\cos b} \frac{u du dt}{\sqrt{1-u^2 t^2}\sqrt{1-(a-u)^2}}$$
First integral is elementary, second integral w.r.t. u is most likely incomplete elliptic integral. Some transformations are required, but this is what I would suggest. Integrating w.r.t. $u$ first, using special functions, then taking the $t$ integral.