Forming similar differential/integral equations Please help in expressing 
$$ y^{'}(x)-\frac{\sin y}{\int \cos y\, dx} \tag{1}$$  in terms of $ \sin y$ and $ \cos y, $ given that:
$$ y^{'}(x) +\frac{\sin y}{\int \cos y\, dx} = \sin ^{3} y.  \tag{2}$$ 
 A: I am not sure exactly what you want (and too long for a comment) but lets have a go.
Take the derivative of the second equation
$$
y'' +\frac{\cos y}{\int \cos y\, dx}y' -\frac{\sin y}{\int \cos y\, dx}\frac{\cos y}{\int \cos y\, dx} = 3\sin^2 y \cos y\, y'
$$
This leads to
$$
\left[y'-\frac{\sin y}{\int \cos y\, dx}\right]\frac{\cos y}{\int \cos y\, dx}=-y''+3\sin^2 y \cos y \,y'=-y'\dfrac{d}{dy}y' +y'\dfrac{d}{dy}\sin^3 y = -y'\dfrac{d}{dy}\left(y'-\sin^3 y\right)
$$
Then using Eq. 2 again we have
$$
y'-\sin^3 y = -\frac{\sin y}{\int \cos y\, dx}
$$
so we get
$$
y'-\frac{\sin y}{\int \cos y\, dx} =-y'\frac{\frac{d}{dy}\left(-\frac{\sin y}{\int \cos y\, dx}\right)}{\frac{\cos y}{\int \cos y\, dx}}
$$
or
$$
y'-\frac{\sin y}{\int \cos y\, dx} =\left(\sin^3y-\frac{\sin y}{\int \cos y\, dx}\right)\frac{\frac{d}{dy}\left(\frac{\sin y}{\int \cos y\, dx}\right)}{\frac{\cos y}{\int \cos y\, dx}} = \left(\sin^3y-\frac{\sin y}{\int \cos y\, dx}\right)\left(1+\frac{\sin y}{\cos y}\frac{\sin y}{\int \cos y \, dx}\right)
$$
A: If I well understand the question, the result must be a formula containing $\cos y$ and $\sin y$, but not $\int \cos y$

