Identity involving an improper integral (with geometric application) 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 $T>0$,
$$\lim_{c\to 0^+}\int_c^{T}\frac{c}{t}\sqrt\frac{1+t^2}{t^2-c^2}dt=\frac{\pi}{2}.$$
 A: Since the main contribution to the integral comes from $t<<1$, analytically one has
\begin{align}
\lim_{c\to 0^+}\int_c^{\pi/2}\frac{c}{t}\sqrt\frac{1+t^2}{t^2-c^2}dt&=\lim_{c\to 0^+}\int_c^{\pi/2}\frac{c}{t}\sqrt\frac{1}{t^2-c^2}dt\\
&=\lim_{c\to 0^+}\int_{2/\pi}^{1/c}c\sqrt\frac{1}{1-c^2t^2}dt\\
&=\lim_{c\to 0^+}\int_{2c/\pi}^{1}\sqrt\frac{1}{1-t^2}dt\\
&=\int_{0}^{1}\sqrt\frac{1}{1-t^2}dt\\
&=\frac\pi{2}
\end{align}
A: Your guess is correct. Let's write $t=cs$, so the integral becomes
$$
\int_1^{\pi/(2c)} \sqrt{\frac{1+c^2 s^2}{s^2-1}}\, \frac{ds}{s} .
$$
Fix a small $\epsilon>0$, and consider first the integral from $1$ to $\epsilon/c$. On this interval, if we drop the $c^2s^2$, we'll be off by not more than a multiplicative correction of order $(1+\epsilon)$, and then we're left with
$$
\int_1^{\epsilon/c} \frac{ds}{s\sqrt{s^2-1}} = - \arctan\frac{1}{\sqrt{s^2-1}}\Bigr|_1^{\epsilon/c}\to \frac{\pi}{2}
$$
as $c\to 0+$.
The remaining part of the integral, from $\epsilon/c$ to $\pi/(2c)$ is easily seen to be $O(c\log c^{-1})$, so this will go to zero when $c\to 0+$. Finally, we send $\epsilon\to 0$ to obtain the claim.
A: 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 $\pi/2$.
