= Solution
In the small-slope <Monge representation>, $s=x+O(y_x^2)$, $\mathbf t=(1,y_x)+O(y_x^2)$, and the leading transverse velocity is $y_t$. With no imposed axial prestress, the induced <filament tension> is second order in transverse amplitude, so its transverse contribution is higher order. The transverse component of force balance reduces to the <small-slope elastohydrodynamic filament equation>
$$
\boxed{c_\perp y_t=-By_{xxxx}.}
$$
An externally imposed zeroth-order tension would instead add $Ty_{xx}$.
For the transverse motion, the off-diagonal part of the <resistive-force theory> tensor gives axial propulsive force density $(c_\perp-c_\parallel)y_xy_t$ to second order. Its integral is
$$
\boxed{F=(c_\perp-c_\parallel)\int_0^L y_xy_t\,dx.}
$$
Substitute the bending equation and integrate by parts:
$$
\int_0^L y_xy_{xxxx}\,dx=\left[y_xy_{xxx}\right]_0^L-\int_0^L y_{xx}y_{xxx}\,dx=\left[y_xy_{xxx}-\frac12y_{xx}^2\right]_0^L.
$$
The <boundary expression for transverse filament thrust> is therefore
$$
\boxed{F=B\left(1-\frac{c_\parallel}{c_\perp}\right)\left[\frac12y_{xx}^2-y_xy_{xxx}\right]_0^L.}
$$
Only endpoint slope, bending moment, and shear enter this expression. Exact <inextensibility> also generates second-order longitudinal material motion; if that motion is retained in instantaneous total axial drag, it adds $-c_\parallel\int v_x\,dx$. For a periodic deformation with fixed axial base position, that additional term has zero cycle average. Thus the displayed formula is the transverse propulsive contribution and also gives the cycle-averaged propulsive force.
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