TITLE: Identity involving an improper integral (with geometric application)
QUESTION [7 upvotes]: Is it (for some reason) true that
$\lim_{c\to 0^+}\int_c^{\pi/2}\frac{c}{t}\sqrt\frac{1+t^2}{t^2-c^2}dt=\frac{\pi}{2}$?  
Numerical evidence (from Mathematica):

when $c=1/5$, the integral is $\approx 1.578$.
when $c=1/10$, the integral is $\approx 1.575$.
when $c=1/100$, the integral is $\approx 1.571$.

Geometric motivation:
By Clairaut's relation, if for some $T$,
$$\int_c^{T}\frac{c}{t}\sqrt\frac{1+t^2}{t^2-c^2}dt=\frac{\pi}{2},$$
then there are two three distinct geodesics between the points $(T,0, \tfrac{1}{2}T^2)$ and $(-T,0, \tfrac{1}{2}T^2)$ on the paraboloid of revolution $z=\tfrac{1}{2}r^2$:  one through the point $(0,0,0)$, $(0, c, \tfrac{1}{2}c^2)$ and one through the point $(0, -c, \tfrac{1}{2}c^2)$.
Edit: Deleted the comment about injectivity radius which concluded my original question -- it was false.  Note that the answers of Nemo and Christian Remling, below, appear to work for any integration bound $T$, not just for $T=\pi/2$: for any $T>0$,
$$\lim_{c\to 0^+}\int_c^{T}\frac{c}{t}\sqrt\frac{1+t^2}{t^2-c^2}dt=\frac{\pi}{2}.$$

REPLY [4 votes]: You can, if you want, calculate the integral. By performing the change of variables
$$
t\mapsto \sqrt{\frac{1+t^2}{t^2-c^2}}.
$$
you end up with a standard integral. For $0<c<\pi/2$ the result is
$$
\int_c^{\pi/2}\frac{c}{t}\sqrt{\frac{1+t^2}{t^2-c^2}}\,dt=\lim_{a\to c^+}\biggl[c\,\text{artanh}\,\sqrt{\frac{t^2-c^2}{1+t^2}}-\arctan\biggl(c\sqrt{\frac{1+t^2}{t^2-c^2}}\biggr)\biggl]_a^{\pi/2}.
$$
The first term will not contribute in the limit $c\to 0^+$, the second one will give you $\pi/2$.