Solving the Euler-Lagrange equation for the brachistochrone problem with friction This Wolfram Alpha Page  contains a derivation of the parametric form of the brachistochrone curve that result from either assuming friction or its absence. 
I am asking for help understanding how the solution to the differential equation obtained from applying the Euler-Lagrange equation to the integrand of the the integral representing the total time of descent is obtained. This differential equation can be found on step (30) of the page. I am asking for help in understanding the next step, how setting $\frac{dy}{dx} = \cot(1/2\cdot\theta)$ allows for the equation to be solved, obtaining the parametric equations for $x$ and $y$, shown in steps (32) and (33). 
 A: You want to know how $x,\,y$ are obtained in terms of $\theta$ from $\frac{1+y^{\prime 2}}{(1+\mu y^\prime)^2}=\frac{C}{y-\mu x},\,y^\prime=\cot\frac{\theta}{2}$. The former equation reduces to $$y-\mu x=C\frac{(1+\mu y^\prime)^2}{1+y^{\prime 2}}=C\left(\sin\frac{\theta}{2}+\mu\cos\frac{\theta}{2}\right)^2,$$because $1+y^{\prime 2}=\csc^2\frac{\theta}{2}$. Because of the forms of rotation matrices and compound-angle formulae, the effect of $\mu$ in this equation is clearly that of a rotation, thereby mixing $x$ and $y$ and adding a constant to $\theta$. So the frictionless result $$x=\frac{k^2}{2}[\theta-\sin\theta],\,y=\frac{k^2}{2}[1-\cos\theta]$$must generalise to$$x=\frac{k^2}{2}[\theta-\sin\theta+\mu (1-\cos\theta)],\,y=\frac{k^2}{2}[1-\cos\theta+\mu (\theta-\sin\theta)].$$
A: The friction effect should be included as a dissipation force $-\mu y'$
$$
\frac{\partial f}{\partial y}-\frac{d}{dx}\frac{\partial f}{\partial y'} = -\mu y'
$$
then from here
$$
f-y'\frac{\partial f}{\partial y'}=-\mu y + C_0
$$
or
$$
\frac{1}{\sqrt{2gy}\sqrt{1+y'^2}}=-\mu y + C_0
$$
or
$$
y'=\pm \sqrt{\frac{1}{\sqrt{2gy}(\mu y - C_0)^2}-1}
$$
etc.
