Expanding a differential equation in powers of $\epsilon$ How does one expand the right hand side of

$$\frac{da}{d\theta}=\frac{\epsilon h\sin(\theta)}{1+\epsilon a^{-1}h\cos(\theta)}$$

in powers of $\epsilon$ and obtain 

$$\frac{da}{d\theta}=\epsilon h\sin(\theta)+O(\epsilon^{2})?$$

Note: For brevity, $h$ stands for $h(a\cos(\theta), a\sin(\theta))$ and $\epsilon$ is a small parameter, i.e., $|\epsilon|\ll 1$.

Context: I am trying to work along with the textbook Nonlinear Ordinary Differential Equations by Jordan and Smith and came across this step in section 4.2 Amplitude and frequency estimates: polar coordinates. We are looking at a differential equation in the form 

$$\ddot{x} + \epsilon h(x,\dot{x})+x = 0, $$

which has an equivalent system

$$\dot{x}=y, \dot{y}=-\epsilon h(x,y)-x. $$

The system itself can be represented parametrically by time-dependent polar coordinates $a(t), \theta(t)$ which allows the polar form of the system

$$\dot{a}=-\epsilon h\sin(\theta)$$ 

$$\dot{\theta}=-1-\epsilon a^{-1}hcos(\theta).$$

The goal is to eventually derive some estimates for the amplitude and frequency of the periodic time solutions of $\ddot{x} + \epsilon h(x,\dot{x})+x = 0$, if any, which are represented by closed paths on its corresponding phase diagram.
$\dfrac{da}{d\theta}=\dfrac{\epsilon h\sin\theta}{1+\epsilon a^{-1}h\cos\theta}$

$(a\epsilon h\sin\theta)\dfrac{d\theta}{da}=a+\epsilon h\cos\theta$

$-a\epsilon h\dfrac{d(\cos\theta)}{da}=a+\epsilon h\cos\theta$

$-a\epsilon h\dfrac{du}{da}=a+\epsilon hu$ by letting $u=\cos\theta$

$\dfrac{du}{da}+\dfrac{u}{a}=-\dfrac{1}{\epsilon h}$

$\dfrac{d(au)}{da}=-\dfrac{a}{\epsilon h}$

$au=C-\dfrac{a^2}{2\epsilon h}$

$\cos\theta=\dfrac{C}{a}-\dfrac{a}{2\epsilon h}$
$\dfrac{da}{d\theta}=\dfrac{\epsilon h\sin\theta}{1+\epsilon a^{-1}h\cos\theta}$

$(a\epsilon h\sin\theta)\dfrac{d\theta}{da}=a+\epsilon h\cos\theta$

$-a\epsilon h\dfrac{d(\cos\theta)}{da}=a+\epsilon h\cos\theta$

$-a\epsilon h\dfrac{du}{da}=a+\epsilon hu$ by letting $u=\cos\theta$

$\dfrac{du}{da}+\dfrac{u}{a}=-\dfrac{1}{\epsilon h}$

$\dfrac{d(au)}{da}=-\dfrac{a}{\epsilon h}$

$au=C-\dfrac{a^2}{2\epsilon h}$

$\cos\theta=\dfrac{C}{a}-\dfrac{a}{2\epsilon h}$