The integral
$$
\int d\theta=\int\frac{r_0}{R}\sqrt{\frac{R^2-r^2}{r^2-r_0^2}}\frac{dr}{r} \tag{1}
$$
appears in the solution of Venezian to the terrestrial brachistochrone problem. (Venezian, Giulio. "Terrestrial brachistochrone." American Journal of Physics 34.8 (1966): 701-701.) Here, $R$ is the radius of the Earth, and $r_0$ is the deepest point in the tunnel:
To solve, one must set $$ u= \sqrt{\frac{r^2-r_0^2}{R^2-r^2}} \tag{2} $$ with the result that (1) becomes $$ \int \frac{(R^2-r_0^2)}{(1+u^2)(r_0^2+R^2u^2)}du $$ and one can finish the job with the usual methods.
Is the substitution of (2) a "one-off" or is there a general rule to attack the type of integral given in (1)?
I would say that my first reaction is always to try to get rid of the radicals.
So, if you use $(2)$ $$u= \sqrt{\frac{r^2-r_0^2}{R^2-r^2}} \implies r=\frac{\sqrt{R^2 u^2+{r_0}^2}}{\sqrt{u^2+1}}\implies dr=\frac{u (R-{r_0}) (R+{r_0})}{\left(u^2+1\right)^{3/2} \sqrt{R^2 u^2+{r_0}^2}}$$ making $$I=\int\frac{r_0}{R}\sqrt{\frac{R^2-r^2}{r^2-r_0^2}}\frac{dr}{r}=\int \frac{{r_0} \left(R^2-{r_0}^2\right)}{R \left(u^2+1\right) \left(R^2 u^2+{r_0}^2\right)}\,du$$ and using partial fraction decomposition leads to $$I=\int{r_0} \left(\frac{R}{R^2 u^2+{r_0}^2}-\frac{1}{R (u^2+1)}\right)\,du$$ which is effectively simple.
In the same spirit, we could also do $$v= \sqrt{\frac{R^2-r^2}{r^2-r_0^2}}$$ to arrive to $$I=\int {r_0} \left(\frac{1}{R (v^2+1)}-\frac{R}{{r_0}^2 v^2+R^2}\right)\,dv$$