= Solution
Write $b_n=4^{-n}\binom{2n}{n}$. The given limiting identity and the <Cauchy coefficient formula> from part (a) give the boundary <contour integral>
$$
J_n:=\int_{\gamma(0,1)}\frac{dz}{z^{n+1}\sqrt{1-z}}=2\pi i b_n.
$$
The <contour> passes through $z=1$, so its value here is an <improper integral> using the one-sided branch values. Parametrize its open arc by $z=e^{i\theta}$, $0<\theta<2\pi$. Substitution of the <analytic branch of a square root> calculated in part (b) gives
$$
J_n=\frac{i}{1-i}\int_0^{2\pi}\frac{e^{-i(n+1/4)\theta}}{\sqrt{\sin(\theta/2)}}\,d\theta.
$$
The factor $e^{-in\theta}$ comes from cancelling $e^{i\theta}$ in $dz$ against $z^{n+1}$; the extra $e^{-i\theta/4}$ comes from the chosen branch. Keeping both phases prevents a change in the half-integer frequency.
Now put $t=\theta/2$. Then
$$
J_n=\frac{2i}{1-i}\int_0^\pi\frac{e^{-i(2n+1/2)t}}{\sqrt{\sin t}}\,dt=(i-1)\int_0^\pi\frac{e^{-i(2n+1/2)t}}{\sqrt{\sin t}}\,dt.
$$
Dividing by $i-1$ gives
$$
\int_0^\pi\frac{e^{-i(2n+1/2)t}}{\sqrt{\sin t}}\,dt=\frac{2\pi i b_n}{i-1}=\pi b_n(1-i).
$$
Taking the <real part> therefore proves
$$
\boxed{\int_0^\pi\frac{\cos\bigl((2n+\tfrac12)t\bigr)}{\sqrt{\sin t}}\,dt=\frac{\pi}{4^n}\binom{2n}{n},\qquad n=0,1,2,\ldots.}
$$
All these <improper integrals> converge absolutely: the numerators are bounded, while $(\sin t)^{-1/2}$ is of order $t^{-1/2}$ at zero and $(\pi-t)^{-1/2}$ at $\pi$. Hence the substitutions and passage to the <real part> are legitimate. The <imaginary part> also shows that the corresponding sine <integral> has the same positive value, which checks the phase calculation.
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