= Solution
Model the uniformly wound air-core <solenoid> as coaxial circular turns with <solenoid turn density> $n=N/L$. Put its ends at $x'=0,L$ and an internal axial point at $x\in[0,L]$. By the <superposition principle>, integrating the <on-axis magnetic field of a circular current loop> gives
$$
B(x)=\frac{\mu_0nIa^2}{2}\int_0^L\frac{dx'}{[a^2+(x-x')^2]^{3/2}}.
$$
Since $d[u/\sqrt{a^2+u^2}]/du=a^2/(a^2+u^2)^{3/2}$, this evaluates to the <on-axis magnetic field of a finite solenoid>:
$$
B(x)=\frac{\mu_0NI}{2L}\left[
\frac{x}{\sqrt{a^2+x^2}}+
\frac{L-x}{\sqrt{a^2+(L-x)^2}}\right].
$$
The source's acute endpoint <angles> satisfy $\cos\alpha=x/\sqrt{a^2+x^2}$ and $\cos\beta=(L-x)/\sqrt{a^2+(L-x)^2}$. Hence
$$
\boxed{B=\frac{\mu_0NI}{2L}(\cos\alpha+\cos\beta).}
$$
Although the paper calls this a “flux”, $B$ is <magnetic flux density>, measured in teslas. <Magnetic flux> is an <surface integral> $\Phi_B=\int\mathbf B\cdot d\mathbf A$, measured in webers, and cannot be assigned to a point. For example, two differently sized small disks at the same point have the same limiting $B$ but different <magnetic fluxes>, approximately $B$ times their normal <areas>. The displayed plus sign uses the internal-point angle convention shown in the source. Outside the coil, the <integral> expression still works, but one must retain signed axial projections rather than adding two acute-angle cosines.
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