$\omega = 3dx_1+(2x_1+x_2^2)dx_2$
$\gamma(t)=(|\cos(2t)|,|1-2\sin(t)|+1), t \in [0,\pi/3]$
Calculate $\int\limits_{\gamma}\omega$
So far I showed that $\omega$ is not exact with partial derivatives. And I know that $\int\limits_{\gamma}\omega=\int\limits_{0}^{\pi/3}(3\gamma_1'+(2|\cos(2t)|+(|1-2\sin(t)|+1)^2)\gamma_2')dt$
In class this week we went over line integrals and that kind of stuff, so I feel like I should use $\Gamma(\gamma(t))$ somehow. But we didn't see any examples of that in class, so I'm very confused about everything. Where should I start? Any help is appreciated!
This is just a straight-forward exercise in applying a definition.
Note that $\cos(2t) \ge 0$ on $[0, \pi/4]$, and $\le 0$ on $[\pi/4, \pi/3]$, while $1 - 2\sin(t) \ge 0$ on $[0,\pi/6]$ and $\le 0$ on $[\pi/6,\pi/3]$. So $$\gamma(t) = \begin{cases}(\cos 2t, 2 - 2\sin t),&t\in\left[0,\frac\pi 6\right)\\(\cos 2t, 2\sin t),&t\in\left[\frac\pi 6, \frac \pi 4\right)\\(-\cos 2t, 2\sin t),&t\in \left[\frac\pi 4, \frac \pi 3\right]\end{cases}$$
Break your integral into three parts:
$$\int_\gamma \omega = \int_0^{\pi/6}\omega(\gamma(t)) + \int_{\pi/6}^{\pi/4}\omega(\gamma(t))+ \int_{\pi/4}^{\pi/3}\omega(\gamma(t))$$
In the first part $\gamma'(t) = (-2\sin 2t, - 2\cos t)$, so the integral is $$\int_0^{\pi/6}\omega(\gamma(t)) = \int_0^{\pi/6} 3(-2\sin 2t)\,dt + \int_0^{\pi/6}[2\cos 2t + (2-2\sin t)^2](-2\cos t)\,dt$$ Using a trig identity to write $\cos 2t$ in terms of $\sin t$, these become fairly simple to calculate.
I'll leave the other two parts to you.