Area enclosed by cardioid using Green's theorem Let $$\gamma(t) = \begin{pmatrix} (1+\cos t)\cos t \\ (1+ \cos t) \sin t \end{pmatrix}, \qquad t \in [0,2\pi].$$ Find the area enclosed by $\gamma$ using Green's theorem.
So the area enclosed by $\gamma$ is a cardioid, let's denote it as $B$. By Green's theorem we have for $f=(f_1, f_2) \in C^1(\mathbb{R}^2, \mathbb{R}^2):$
$$\int_B \text{div} \begin{pmatrix} f_2 \\ -f_1 \end{pmatrix} d(x,y) = \int_{\partial B} f \cdot ds$$
So if we choose $f(x,y) = \begin{pmatrix} -y \\ 0 \end{pmatrix}$ for example, we get
$$\begin{eqnarray}
\text{Area of $B$} &=& \int_{\partial B} f \cdot ds
\\&=& \int_{\gamma} f(\gamma(t)) \cdot \gamma'(t) dt
\\&=& \int_0^{2\pi} \begin{pmatrix} -(1+ \cos t) \sin t \\ 0 \end{pmatrix} \cdot\begin{pmatrix} \sin t ( 1 - 2 \cos t) \\ \cos^2t - \sin^2t + \cos t \end{pmatrix} dt ,\end{eqnarray}$$
which I guess we can evaluate but if I keep going, this will become very nasty and tedious.
There must be a nicer way to do this. Please help me see it.

As a rule of thumb, it's better to use $$\frac12\int_{\partial B}(-y\,dx+x\,dy)$$ whenever trig functions are involved. That will give a simpler expression once you use standard trig identities.
Comment: It's more straight forward in this particular case to do the area integral directly in polar coordinates: $$\int_0^{2\pi}\int_0^{1+\cos\theta} r\,drd\theta.$$
