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\title{Solving Brachistochrone Problem}
\author{Adrian D'Costa}
\begin{document}
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$\displaystyle \int \sqrt{\frac{y}{c-y}}\,\,dy.......\text{(i)}$
\newline
Let:
\begin{align*}&u^{2} = \frac{y}{c-y}......\text{(ii)}\\& u^{2}c - u^{2}y = y\\& y + u^{2}y = u^{2}c\\& y(1 + u^{2}) = u^{2}c\\& \therefore y = \frac{u^{2}c}{1 + u^{2}} \end{align*}
Now:
\begin{align*}\frac{dy}{du} &= \frac{(1 + u^{2})2uc - u^{2}c(2u)}{(1+u^{2})^{2}}\\&= \frac{2uc + 2u^{3}c - 2u^{3}c}{(1+u^{2})^{2}}\\&= \frac{2uc}{(1+u^{2})^{2}}\end{align*}
$\therefore dy = \frac{2uc}{(1+u^{2})^{2}}\,\,du......\text{(iii)}$
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Plugging in $(ii)$ and $(iii)$ into $(i)$ we get:
\newline
$\displaystyle 2 \int \frac{u^{2}c}{(1+u^{2})^{2}}......\text{(iv)}$
\includegraphics[width=\textwidth]{FigureA.jpg}
Again let:
\begin{align*}&u = \cot{(\theta)}\\& du = -\csc^{2}(\theta)\,\,d\theta\\& \text{And } 1 +u^{2} = 1 + \cot^{2}(\theta) \end{align*}
By plugging in all three of these values into $(iv)$ we get:
\newline
\begin{align*}&-2c\int \frac{\csc^{2}(\theta)\cot^{2}(\theta)}{\csc^{4}(\theta)}\,\,d\theta\\&= -2c \int \frac{\cot^{2}(\theta)}{\csc^{2}(\theta)}\,\,d\theta\\&= -c \int 2\cos^{2}(\theta)\,\,d\theta\\&= -c \int (\cos(2\theta) + 1)\,\,d\theta\\&= -c\left(\frac{\sin(2\theta)}{2} + \theta\right) + C\\&= -c\left(\sin(\theta)\cos(\theta) + \theta \right) + C\\&= -c\left(\frac{u}{1 + u^{2}} - \cot^{-1}\left(\sqrt{\frac{y}{c-y}}\right)\right) + C\\&= \left(\sqrt{\frac{y}{c-y}}(y-c) + c \cot^{-1}\left(\sqrt{\frac{y}{c-y}}\right)\right) + C...\text{[Assuming both c and y} > 0 \text{]}\\&\text{[Answer]}\end{align*}
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