I have a feeling it could, or with some other theorem.
$F(x,y,z) = (2xyz + \sin x)i + (x^2z)j + (x^2y)k$
$$\int_{c} F.ds$$ where $c(t) = (\cos^5(t),\sin^3(t),t^4)$
I tried it in differential form as follows, but I am getting an integral that seems way to hard. I put it in differential form and got this:
$$\int_{0}^{2\pi} 10\cos^9(t)\sin^4(t)t^4 + 5\cos^4(t)\sin(t)\sin(\cos^5(t)) + \cos^{10}(t)t^4(3\sin^2(t)) + \cdots $$
There is no way there isn't an easier way to solve this problem. Stokes theorem, Greens theorem. Something?
You are in luck, by stokes theorem, the integral is equivalent to:
$\int_R(\nabla\times F)\cdot n dA$, where I believe $R$ is the region enclosed by $c$.
But we see that $\nabla\times F=0$, so the integral is zero.
$\nabla\times F=\left| \begin{array}{ccc} \underline{i} & \underline{j} & \underline{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ 2xyz+\sin(x) & x^2z & x^2y \end{array} \right|=(x^2-x^2,2xy-2xy,2xz-2xz)=(0,0,0)$