= Solution
Introduce $\omega=\dot\theta$. The <pendulum with quadratic damping> is the <autonomous planar system>
$$
\boxed{\dot\theta=\omega,\qquad\dot\omega=-\sin\theta-c\omega|\omega|.}
$$
Its <equilibrium points> are $(k\pi,0)$. Since the derivative of $\omega|\omega|$ is zero at zero, the <linearization of a dynamical system> at each <equilibrium point> is
$$
J_k=\begin{pmatrix}0&1\\-\cos(k\pi)&0\end{pmatrix}.
$$
At an odd multiple of $\pi$ its <eigenvalues> are $\pm1$, so it is a <saddle equilibrium>; the stable and unstable tangents are $\omega=-(\theta-k\pi)$ and $\omega=\theta-k\pi$. At an even multiple the <eigenvalues> are $\pm i$. This linearization is a centre and does not decide the nonlinear stability. In particular it would be incorrect to infer a linear damped focus from this Jacobian.
For the full system the mechanical energy is a <Lyapunov function> near an even <equilibrium point>:
$$
H(\theta,\omega)=\frac12\omega^2+1-\cos\theta,\qquad\boxed{\dot H=-c|\omega|^3\leq0.}
$$
A sufficiently small sublevel set is compact and contains only that <equilibrium point>. The only invariant motion with $\dot H=0$ has $\omega=0$ and $\sin\theta=0$, hence is the <equilibrium point> itself. The <LaSalle invariance principle> proves <asymptotic stability>. Near it the trajectories turn clockwise and spiral inward, as the near-linear angular motion persists while energy decreases. Thus \b[even equilibria are nonhyperbolic asymptotically stable spiral points; odd equilibria remain saddles].
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2015/ia/paper-2-quadratic-damping.png]
{title=Local phase trajectories near a stable pendulum equilibrium and an unstable saddle with quadratic damping}
For explicit trajectories put $W=\omega^2$ on a branch where $\omega\ne0$. The <chain rule> gives $dW/d\theta=2\dot\omega$, so the two signs yield the <first-order linear differential equations> $W'+2cW=-2\sin\theta$ for $\omega>0$, and $W'-2cW=-2\sin\theta$ for $\omega<0$. Their <integrating factors> give
$$
\boxed{\begin{aligned}\omega^2&=K_+e^{-2c\theta}+\frac{2(\cos\theta-2c\sin\theta)}{1+4c^2},&&\omega>0,\\\omega^2&=K_-e^{2c\theta}+\frac{2(\cos\theta+2c\sin\theta)}{1+4c^2},&&\omega<0.\end{aligned}}
$$
Only portions with a nonnegative right-hand side are physical. At a turning point the constants of the two branches are matched using $\omega=0$; neither expression is one global trajectory through both signs. When $c=0$ they reduce to energy contours, with closed orbits around the even <equilibrium points>. For every $c>0$, the <quadratic damping> destroys those closed orbits and makes the centres attracting spirals. Its effect is higher order in amplitude, so decay is weaker near rest than with linear damping. The <saddle equilibria> remain unstable, though their separatrices change shape.
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