Let $$\omega=\frac{x\text{d}y-y\text{d}x}{x^2+y^2}$$ $$c_0,c_1: [0,1]\to\mathbb R^2$$ $$c_0(t)=\Big(\cos(\pi t),\sin(\pi t)\Big), \ \ c_1(t)=\Big(\cos\big((1+t)\pi\big),\sin\big((1+t)\pi\big)\Big)$$
I am trying to prove that $c_0$ and $-c_1$ are not homotopic using the following theorem:
If $c_0,c_1$ are homotopic and $\omega$ is a closed $1$-Form, then $\int_{c_0}\omega = \int_{c_1}\omega$.
So let's assume we already know that $\omega$ is a closed $1$-Form.
Calculation of $\int_{c_0}\omega$:
\begin{align*} c_0^*\omega&=c_0^*(\frac{x\text{d}y-y\text{d}x}{x^2+y^2})) \\ &=c_0^*(x\ \text{d}y-y\ \text{d}x) \\ &=c_0^*(x\ \text{d}y)-c_0^*(y\ \text{d}x) \\ &=\cos(\pi t)\ \text{d}(\sin(\pi t))-\sin(\pi t)\ \text{d}(\cos(\pi t)) \\ &=\pi\cos^2(\pi t)\ \text{d}t+\pi \sin^2(\pi t)\ \text{d}t \\ &=\pi\ \text{d}t \end{align*} $\implies \int_{c_0}\omega =\int_0^1\pi\ \text{d}t=\pi$
Calculation of $\int_{c_1'}\omega$ (with $c_1'=-c_1)$: \begin{align*} c_1'^*\omega&=c_1'^*(\frac{x\text{d}y-y\text{d}x}{x^2+y^2})) \\ &=c_1'^*(x\ \text{d}y-y\ \text{d}x) \\ &=c_1'^*(x\ \text{d}y)-c_0^*(y\ \text{d}x) \\ &=-\cos((1+t)\pi)\ \text{d}(-\sin((1+t)\pi))+\sin((1+t)\pi)\ \text{d}(-\cos((1+t)\pi)) \\ &=\pi\cos^2((1+t)\pi)\ \text{d}t+\pi \sin^2((1+t)\pi)\ \text{d}t \\ &=\pi\ \text{d}t \end{align*} $\implies \int_{c_1'}\omega =\int_0^1\pi\ \text{d}t=\pi$
what makes no sense since I am to prove that $c_0$ and $-c_1=c_1'$ are not homotopic. Does anyone see the mistake?
Note that for any $\theta\in\Bbb R$, $$\cos(\theta +\pi)=-\cos \theta\qquad\text{and}\qquad\sin(\theta +\pi)=-\sin \theta$$ Therefore $c_0=-c_1$.
Note also that it should be $\Bbb R^2\setminus\{0\}$ instead of $\Bbb R^2$, for not only $\omega$ is only defined on $\Bbb R^2\setminus\{0\}$, but also any two paths with common endpoints are homotopic in $\Bbb R^2$.