For $c \colon [0, 2π] \to \mathbb{R}^3$ with the helix $c(t) = (\cos t,\sin t, t)$ and the 1-form on $\mathbb{R}^3$
$$\alpha = 2x_1 x_2 \,dx_1+ (x_1)^2\, dx_2 + x_3\,dx_3$$
How do you find the exterior derivative and show that $dα = 0$?
So far I've got
$$d \alpha = d(2x_1x_2) \wedge dx_1 + d((x_1)^2) \wedge dx_2 + dx_3 \wedge dx_3$$
so then
$$d \alpha = d(2x_1x_2) \wedge dx_1 + d((x_1)^2) \wedge dx_2 + 0$$
but I don't know how to make this equal to $0$