 Solution

Where $\mathbf{A}=|\mathbf{v}{|}^2\mathbf{w}$ is twice continuously differentiable, the identity divergence of a curl is zero gives $\nabla \cdot (\nabla \times \mathbf{A})=0$. All regular portions of the parameter interval lie in this region, henceas an improper line integral. Besides the $\mathrm{arctanh}(x^2)$ endpoint problem from (i), the curve touches $y=1$ at $\theta =\pi /2$, where the printed real $\arcsin y$ expression has no smooth open neighbourhood. The identity applies on the intervening open segments, and its zero value has a unique continuous extension along the curve. A literal classical integrand at every parameter is not supplied by the printed field.