A while ago, I saw a post for a solution of the time of the brachistochrone. Sadly, I do not get how the integral shown was substituted. Can anybody help?
The original integral was
$$T=\sqrt{\frac{1}{2g}}\int_0^b \sqrt{\frac{1+(y'(x))^2}{y(x)}}dx$$
with
$$\frac{dx}{d\theta}=A(1-\cos\theta)$$
and
$$\frac{dy}{dθ}=A\sinθ$$
With $$x=A(\theta - \sin\theta)$$ and $$y=A(1-\cos\theta)$$ one came to the solution
$$\begin{align}T&=\sqrt{\frac 1{2g}}\int_{\theta=0}^{\theta_0}\sqrt{\frac{1+(y')^2}y}\frac{dx}{d\theta}\,d\theta\\&=\sqrt{\frac A{2g}}\int_{\theta=0}^{\theta_0}\sqrt{\frac{(1-\cos\theta)^2+\sin^2\theta}{1-\cos\theta}}\,d\theta\\&=\sqrt{\frac Ag}\int_{\theta=0}^{\theta_0}\sqrt{\frac{1-\cos\theta}{1-\cos\theta}}\,d\theta\\&=\sqrt{\frac Ag}\theta_0. \end{align}$$
Solving this is no problem. But I do not get, how the integral was changed.
$$\begin{align}\sqrt{\frac{1+(y')^2}y}\frac{dx}{d\theta}&=\sqrt{\frac{1+\left(\frac{A\sin\theta}{A(1-\cos\theta)}\right)^2}{A(1-\cos\theta)}}A(1-\cos\theta)\\&=\sqrt A\sqrt{\frac{(1-\cos\theta)^2+\sin^2\theta}{1-\cos\theta}}\\&=\sqrt{2A}. \end{align}$$ And $\theta_0$ is determined by $$b=x(\theta_0)=A(\theta_0-\sin\theta_0).$$