= Solution
{c}
Set $u=Br/A$, retaining $r_0$ in $B$. Replacing only the lower integration limit by zero gives
$$
\tau\simeq\frac{A}{B^{3/2}}\int_0^1\sqrt{\frac u{1-u}}\,du
=\frac{\pi A}{2B^{3/2}}.
$$
Substitution of the constants yields
$$
\boxed{\tau\simeq\frac{\pi GM}{(2G\delta M/r_0)^{3/2}}
=\frac{\pi M r_0^{3/2}}{2\sqrt{2G}\,(\delta M)^{3/2}}.}
$$
To quantify the approximation, write $\epsilon=\delta M/M\ll1$. The true lower limit is $u_0=Br_0/A=\epsilon$, and an elementary <integration by substitution>, $u=\sin^2\chi$, gives
$$
\tau_{\rm exact}=\frac{A}{B^{3/2}}
\left[\frac\pi2-\arcsin\sqrt\epsilon+\sqrt{\epsilon(1-\epsilon)}\right].
$$
Since $\arcsin\sqrt\epsilon-\sqrt{\epsilon(1-\epsilon)}=\tfrac23\epsilon^{3/2}+O(\epsilon^{5/2})$, the approximate time is too large by a relative fraction $\tfrac4{3\pi}\epsilon^{3/2}+O(\epsilon^{5/2})$. This explains why changing the lower limit introduces little error for a small excess mass. The <cosmological turnaround> time diverges as $\delta M\to0^+$, consistently with the absence of finite <cosmological turnaround> in an exactly critical <Einstein-de Sitter universe>.
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