I'm asked to calculate $$\int_{S^1} i^* (xdy-ydx)$$
Consider the unit disk $D:= \{v \in \mathbb{R}^2 \mid \Vert v \Vert \leq 1\}$. Then $S^1 = \partial M$ and thus by Stokes' theorem we have:
$$\int_{S^1} i^*(xdy -ydx) = \int_D d(xdy-ydx) $$ $$= \int_D dx \wedge dy - dy \wedge dx = 2 \int _Ddx \wedge dy$$
I'm having trouble calculating the integral $\int_D dx \wedge dy$. Do I do this with the definition of integral? I.e. with the pullback of a chart?
Thanksk in advance for your help.
The final integral is just the integral over the unit disk of the volume form $dV = dx \wedge dy$ and so your answer evaluates to $2\times \int_{D}dV = 2\times\textrm{vol}(D) = 2\pi$.
This can be verified by parameterising $S^{1}$ by $x = \cos\theta$, $y = \sin\theta$ from which the line integral becomes $$\int_{0}^{2\pi}d\theta (\cos^{2}\theta + \sin^{2}\theta) = \int_{0}^{2\pi}d\theta = 2\pi.$$