= Solution
Take $x$ tailward, so forward swimming has laboratory <velocity> $-U\mathbf i$. For an inextensible travelling waveform, material motion in the swimmer frame is $V\mathbf i-c\mathbf t$: a material point passes through the wave along its tangent while the wave pattern translates at $V$. Thus the relative <velocity> of otherwise stationary fluid is
$$
\mathbf w=c\mathbf t-(V-U)\mathbf i.
$$
Put $q=V-U$ and choose $\mathbf t=(X_s,Y_s)$, $\mathbf n=(-Y_s,X_s)$, with $X_s^2+Y_s^2=1$. Then
$$
w_t=c-qX_s,\qquad w_n=qY_s.
$$
On the forward-swimming branch $0<U<V<c$ and $X_s>0$, the tangential slip is positive. A law for reversed tangential slip would also need its sign, not just its absolute magnitude.
The quadratic normal law represents inertial cross-flow <drag>: dynamic <pressure> scales as fluid density times $w_n^2$, while $|w_n|w_n$ chooses the <force> direction. Its coefficient contains the cross-sectional size and an empirical <drag> coefficient. The $|w_t|^{3/2}$ exponent is characteristic of an attached laminar <boundary layer>: its thickness scales as $\delta\sim(\nu s/|w_t|)^{1/2}$ and its shear as $\mu|w_t|/\delta$. However, this gives local shear proportional to $|w_t|^{3/2}s^{-1/2}$, not $s^{1/2}$. The positive power of $s$ is genuinely printed in this paper and repeated in its target formulas. Below it is treated as the prescribed phenomenological weight, rather than misidentified as standard local <laminar skin-friction scaling along a slender swimmer>.
The normal axial <force> on an element is
$$
dF_{N,x}=K_N|qY_s|qY_s(-Y_s)ds=-K_Nq^2|Y_s|^3ds.
$$
The prescribed tangential <force> gives $dF_{T,x}=K_T(c-qX_s)^{3/2}X_s s^{1/2}ds$. Zero mean axial <force> therefore gives the <nonlinear resistive-force balance for a travelling filament>
$$
\boxed{K_N(V-U)^2\int_0^L|\widehat Y_s|^3ds
=K_T\int_0^L|c-(V-U)X_s|^{3/2}X_s s^{1/2}ds.}
$$
The absolute values are present in the original PDF. In a periodically beating, constant-mean-speed model, phase-average the integrals where necessary; the leading approximation below is phase independent. Local <force> laws neglect nonlocal hydrodynamic interactions and assume nearly steady local response.
For the sinusoidal displacement, let $\varepsilon=\beta k\ll1$. Then $Y_s=\varepsilon\cos ks$ and $X_s=(1-Y_s^2)^{1/2}=1+O(\varepsilon^2)$. Over one wavelength,
$$
\int_0^L|Y_s|^3ds=\frac{\beta^3k^3}{k}\int_0^{2\pi}|\cos z|^3dz=\frac83\beta^3k^2,
\qquad \int_0^Ls^{1/2}ds=\frac23L^{3/2}.
$$
Replace $X_s$ by one in the leading tangential term and balance the <forces>. After multiplication by $3/2$,
$$
\boxed{4\beta^3k^2K_N(V-U)^2\simeq K_TL^{3/2}(c-V+U)^{3/2}.}
$$
This replacement also requires $c\varepsilon^2\ll c-V+U$ for a uniformly small relative error inside the fractional power. If the tangential slip is itself $O(c\varepsilon^2)$, its variation along the body must be retained. Periodicity gives $V/c=\langle X_s\rangle=1-\varepsilon^2/4+O(\varepsilon^4)$, so small geometric slope alone does not guarantee that additional slip condition.
For <Lighthill elongated-body theory>, put $h(x,t)\simeq\beta\sin[k(x-Vt)]$ and $w_\perp=h_t+Uh_x$. The transverse fluid <kinetic energy> per unit length is $M_Aw_\perp^2/2$. In the usual tail-flux approximation, with negligible leading-end contribution, the lateral work flux is $M_AU\langle h_tw_\perp\rangle$, whereas <kinetic energy> carried into the wake is $M_AU\langle w_\perp^2\rangle/2$. Their difference is the useful propulsive power $TU$. Consequently
$$
T=M_A\left\langle h_tw_\perp-\frac12w_\perp^2\right\rangle
=\frac{M_A}{2}\langle h_t^2-U^2h_x^2\rangle
=\frac{M_A}{4}\beta^2k^2(V^2-U^2).
$$
Balancing this against the <drag> of the printed tangential model, $D=(2/3)K_TL^{3/2}(c-V+U)^{3/2}$, gives
$$
\boxed{\frac38M_A\beta^2k^2(V^2-U^2)=K_TL^{3/2}(c-V+U)^{3/2}.}
$$
For a genuine local laminar skin-friction estimate, write $dD=C_Tu_t^{3/2}s^{-1/2}ds$, where $C_T$ has the appropriate different dimensions. Its integrated <drag> is $2C_Tu_t^{3/2}\sqrt L$, and the corresponding elongated-body balance is instead $M_A\beta^2k^2(V^2-U^2)/8=C_T\sqrt L\,(c-V+U)^{3/2}$. The printed $L^{3/2}$ expression is therefore a result of its stated <force> model, not of that standard boundary-layer law.
The two thrust models describe different mechanisms: cross-flow <resistive-force theory> uses dissipative local <drag>, while <elongated-body theory> uses reactive acceleration of <added mass>. Both suppress details of the wake, body-body interactions, finite slenderness, skin-friction unsteadiness, separation and possible turbulent transition. The tail-only reactive result also requires a negligible leading-end flux, usually supplied by a nose taper or small leading-end motion; a blunt constant-cross-section body waving equally at both ends cannot automatically omit its leading-end contribution. Speed fluctuations and recoil may matter, and a planar constant-amplitude waveform is an idealization of a real sea snake.
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