= Solution
A fluid particle follows a <Lagrangian trajectory>, so its position satisfies $\dot x=u(x,t)$ rather than the <velocity> evaluated permanently at its initial position. Assume $\omega\ne0$, and write
$$
x(t)=X_0+u_0a(t)+u_0^2b(t)+O(u_0^3),\qquad a(0)=b(0)=0.
$$
Set $\alpha=kX_0$ and $\theta=\alpha-\omega t$. A <Taylor expansion> of the <velocity> about $X_0$ gives
$$
u_0\dot a+u_0^2\dot b=u_0\sin\theta+ku_0^2a\cos\theta+O(u_0^3).
$$
Thus $\dot a=\sin\theta$, and
$$
a(t)=\frac{\cos\theta-\cos\alpha}{\omega}.
$$
At the next order,
$$
\dot b=\frac k\omega(\cos^2\theta-\cos\alpha\cos\theta),
$$
so the required <second-order trajectory in a sinusoidal velocity field> is
$$
\boxed{\begin{aligned}
x(t)={}&X_0+\frac{u_0}{\omega}(\cos\theta-\cos\alpha)\\
&+\frac{ku_0^2}{\omega}\left[\frac t2+
\frac{\sin2\alpha-\sin2\theta}{4\omega}
-\frac{\cos\alpha(\sin\alpha-\sin\theta)}{\omega}\right]+O(u_0^3).
\end{aligned}}
$$
Both corrections vanish at $t=0$. All retained terms have units of length. The secular contribution gives the <Stokes drift>
$$
\boxed{\overline{\dot x}=\frac{ku_0^2}{2\omega}+O(u_0^3),}
$$
although the <Eulerian time average> of the given <velocity> is zero. The particle samples the <wave> at its displaced position, producing this second-order drift. The expansion is in the dimensionless parameter $|ku_0/\omega|\ll1$, for fixed times; it ceases to be uniform when the accumulated drift changes the <wave> phase by order unity.
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