Evaluating $\oint_c x\,dx-y\,dy$ (and others) over unit circle $c$. I'm not sure I understand how those separate derivatives are supposed to work. The question says:

Solve the following line integrals, where $c$ is the circle with radius $1$ centered at the origin, that can be parametrized by $ c : [0,2\pi]\rightarrow \mathbb{R}^2 : t \mapsto c(t) = (\cos(t),\sin(t))$.
(i) $\;\oint_c x\,dx-y\,dy$
(ii) $\;\oint_c x\,dx+x\,dy$
(iii) $\;\oint_c y\,dx$
(iv) $\;\oint_c dy$

I think I'm supposed to use Green's theorem and solve these integrals as a double integral on the region of a circle, and then use polar coordinate substitution? But I'm not sure I understand how those separate derivatives are supposed to work like in "$x\,\mathrm{d}x-y\,\mathrm{d}y$" for example.
 A: I think the best explanation for this $\mathrm{d}x$ notation is in terms of differential forms, but that may be too advanced, so let's use simpler notions.
Suppose $\gamma:[a,b]\longrightarrow\mathbb{R}^2$ is a continuously differentiable curve and suppose further that $f:U\longrightarrow\mathbb{C}$ is a differentiable function defined on some open neighborhood of the image of $\gamma$ (so, in the case of the circle, it would be defined at least on some thick annular region around the circle). Then if $g$ is any continuous function defined on the image of $\gamma$, we set
$$
\int_\gamma g\,\mathrm{d}f = \int_a^b g(\gamma(t)) \nabla\!f(\gamma(t))\cdot\frac{\mathrm{d}\gamma(t)}{\mathrm{d}t}\,\mathrm{d}t
$$
What happens if we reparametrize $\gamma$? If $[c,d]$ is another interval and $\phi:[c,d]\longrightarrow [a,b]$ is continuously differentiable with $\phi(c)=a$, $\phi(d)=b$, we have that
\begin{align*}
\int_c^d \!\!g(\gamma(\phi(s))) \nabla\!f(\gamma(\phi(s)))&\cdot\frac{\mathrm{d}\gamma(\phi(s))}{\mathrm{d}s}\mathrm{d}s=\!\!\int_c^d\!\!g(\gamma(\phi(s))) \nabla\!f(\gamma(\phi(s)))\cdot\frac{\mathrm{d}\gamma}{\mathrm{d}t}\bigg\vert_{t=\phi(s)}\!\frac{\mathrm{d}\phi(s)}{\mathrm{d}s}\mathrm{d}s=\\
&= \int_a^b g(\gamma(t)) \nabla\!f(\gamma(t))\cdot\frac{\mathrm{d}\gamma(t)}{\mathrm{d}t}\,\mathrm{d}t
\end{align*}
by the change of variables formula, so the integral is unchanged.
Morally speaking,
$$
\mathrm{d}f=\nabla\!f(\gamma(t))\cdot\frac{\mathrm{d}\gamma(t)}{\mathrm{d}t}\,\mathrm{d}t
$$
The dot product of the gradient with the tangent vector of the curve tells us how much the function $f$ changes when we move infinitesimally along the curve. The integral then just sums all these infinitesimal changes. Now if we have several pairs of functions $g_1, f_1$, $g_2, f_2$, ..., we can just set
$$
\int_\gamma g_1\mathrm{d}f_1+g_2\mathrm{d}f_2+ ...=\int_\gamma g_1\mathrm{d}f_1 + \int_\gamma g_2\mathrm{d}f_2+...
$$
as a sum of integrals.
Let's examine your example with $x\,\mathrm{d}x-y\,\mathrm{d}y$: we have $x(c(t))=\cos(t)$, $y(c(t))=\sin(t)$, $\nabla x(c(t))=(1,0)$, $\nabla y(c(t))=(0,1)$, $\frac{\mathrm{d}c}{\mathrm{d}t}=(-\sin(t),\cos(t))$. Hence
$$
\int_{\mathbb{S}^1}x\,\mathrm{d}x-y\,\mathrm{d}y = \int_0^{2\pi} (-\cos(t)\sin(t)-\sin(t)\cos(t))\,\mathrm{d}t=-\int_0^{2\pi}\sin(2t)\,\mathrm{d}t=0
$$
As for Green's formula, if $\gamma(a)=\gamma(b)$ and $\gamma$ is injective, then your curve bounds a region $D$. The formula is then that for any continuously differentiable $g_1, g_2$ defined on a neighborhood of $D$:
$$
\int_\gamma g_1\mathrm{d}x + g_2\mathrm{d}y = \iint_D \bigg(\frac{\partial g_2}{\partial x}-\frac{\partial g_1}{\partial y}\bigg)\,\mathrm{d}x\,\mathrm{d}y
$$
Importantly, the curve must be oriented counterclockwise, otherwise there would be a factor of $-1$ in this equality. In the case we just looked at, $g_1=x$ and $g_2=y$, so the partial derivatives entering into the double integral are $0$ and the answer is again $0$, as it should be.
A: 
Since there is no indication to apply Green's theorem
it rather seems we are expected to calculate the line integral by using the given closed path parametrisation of the unit circle
\begin{align*}
c:[0,2\pi]\rightarrow \mathbb{R}^2 : t \mapsto c(t) = (\cos(t),\sin(t))
\end{align*}
as it is stated.
So we want to calculate the four line integrals
\begin{align*}
\oint_c x\,dx-y\,dy&\qquad\qquad\oint_c x\,dx+x\,dy\tag{1.1,$\quad$1.2}\\
\oint_c y\,dx&\qquad\qquad\oint_c dy\tag{1.3,$\quad$1.4}
\end{align*}

Note, the line notation used here is just a space-saving notation for the vector-based notation which is often used when parametrisation comes into play. Since then we can conveniently write, given the parameter $t$ varies in the interval $[t_1 ,t_2]$ and $c=(x(t),y(t))$
\begin{align*}
\color{blue}{\oint_c a(x,y)\,dx+b(x,y)\,dy=\int_{t_1}^{t_2} \left(a(x(t),y(t))\dot{x}(t)+b(x(t),y(t))\dot{y}(t)\right)dt}\tag{2}
\end{align*}
In fact we have a shortcut for some of the four line-integrals in (1.1) to (1.4). We recall a fundamental theorem of line integrals which states that if we integrate within a simply connected region the integral
\begin{align*}
\color{blue}{\oint_c a(x,y)\,dx+b(x,y)\,dy=0}\qquad\mathrm{iff}\qquad
\color{blue}{\frac{\partial a(x,y)}{\partial y}=\frac{\partial b(x,y)}{\partial x}}\tag{3}
\end{align*}

*

*Example 1.1: We have according to (2)
\begin{align*}
\oint_c a(x,y)\,dx+b(x,y)\,dy=\oint_c x\,dx-y\,dy
\end{align*}
It follows
\begin{align*}
\frac{\partial a(x,y)}{\partial y}=\frac{\partial x}{\partial y}=0
\qquad\mathrm{and}\qquad
\frac{\partial b(x,y)}{\partial x}=\frac{\partial (-y)}{\partial x}=0\\
\end{align*}
and we conclude according to (3)
\begin{align*}
\color{blue}{\oint_c x\,dx-y\,dy=0}
\end{align*}


*Example 1.2: We have according to (2)
\begin{align*}
\color{blue}{\oint_cx\,dx+x\,dy}
&=\int_0^{2\pi} \left(\cos (t)\left(-\sin(t)\right)+\cos^2(t)\right)\,dt\\
&=-\int_{0}^{2\pi}\cos(t)\sin(t)\,dt+\int_{0}^{2\pi}\cos^2(t)\,dt\\
&=-\frac{1}{2}\int_0^{2\pi}\sin(2t)\,dt+\frac{1}{2}\int_{0}^{2\pi}\left(\cos(2t)+1\right)\,dt\\
&=-\frac{1}{4}\cos(2t)\Big|_{0}^{2\pi}+\frac{1}{4}\sin(2t)\Big|_{0}^{2\pi}+\frac{1}{2}t\Big|_{0}^{2\pi}\\
&\,\,\color{blue}{=\pi}
\end{align*}


*Example 1.3: We have according to (2) and example 1.2
\begin{align*}
\color{blue}{\oint_c y\,dx}
&=\int_0^{2\pi} \sin (t)\left(-\sin(t)\right)\,dt\\
&=-\int_{0}^{2\pi}\left(1-\cos^2(t)\right)\,dt\\
&=-t\Big|_{0}^{2\pi}+\pi\\
&\,\,\color{blue}{=-\pi}
\end{align*}


*Example 1.4: We have according to (2)
\begin{align*}
\oint_c a(x,y)\,dx+b(x,y)\,dy=\oint_c dy
\end{align*}
It follows
\begin{align*}
\frac{\partial a(x,y)}{\partial y}=0
\qquad\mathrm{and}\qquad
\frac{\partial b(x,y)}{\partial x}=\frac{\partial y}{\partial x}=0\\
\end{align*}
and we conclude according to (3)
\begin{align*}
\color{blue}{\oint_c dy=0}
\end{align*}
