# 12-267/Tuesday September 11 Notes

### Solving the complicated integral in the Brachistochrone integral

$\int \sqrt{\frac{d-y}{y}} dy$

$= \int \sqrt{\frac{d y-y^2}{y}} dy$

Complete the square in the integrand:

$= \int \sqrt{\frac{\frac{d^2}{4} - (y - \frac{d}{2})^2}{y}}$

Substitute $u = y-\frac{d}{2}$ and $du = dy$:

$= \int 2 \sqrt{\frac{\frac{d^2}{4} - u^2}{d+2 u}} du$

Assuming all variables are positive, substitute $u = \frac{1}{2} d \sin{s}$ and $du = \frac{1}{2} d \cos{s} ds$. Then $\sqrt{\frac{d^2}{4} - u^2} = \sqrt{\frac{d^2}{4} - \frac{1}{4} d^2 \sin^2{s}} = \frac{1}{2} d \cos{s}$ and $s = \sin^{-1}{\frac{2u}{d}}$:

$= \frac{d^2}{2} \int \frac{\cos^2{s}}{d \sin{s} + d} ds$

For the integrand substitute $p=\tan{\frac{s}{2}}$ and $dp = \frac{1}{2} \sec^2{\frac{s}{2}} ds$. Then transform the integrand using the substitutions $\sin{s} = \frac{2p}{p^2 + 1}$, $\cos{s} = \frac{1-p^2}{p^2 + 1}$ and $ds = \frac{2 dp}{p^2 + 1}$:

$= \frac{d^2}{2} \int 2 \frac{(1-p^2)^2}{(p^2 + 1)^3 (\frac{2 d p}{p^2 + 1} + d)} dp$

Simplify the integrand $\frac{2(1-p^2)^2}{(p^2 + 1)^3 (\frac{2 d p}{p^2 + 1} + d)}$ to get $\frac{2 (p-1)^2}{d p^4 + 2 d p^2 + d}$:

$= \frac{d^2}{2} \int \frac{2(p-1)^2}{d p^4+2 d p^2+d} dp$

$= d^2 \int \frac{(p-1)^2}{d p^4+2 d p^2+d} dp$

$= d^2 \int \frac{(p-1)^2}{d (p^2+1)^2} dp$

$= d \int \frac{(p-1)^2}{(p^2+1)^2} dp$

For the integrand $\frac{(p-1)^2}{(p^2+1)^2}$ use partial fractions:

$= d \int (\frac{1}{p^2+1}-\frac{2 p}{(p^2+1)^2}) dp$

$= d \int \frac{1}{(p^2+1)} dp - 2 d \int \frac{p}{(p^2+1)^2} dp$

For the integrand $\frac{p}{(p^2+1)^2}$, substitute $w = p^2+1$ and $dw = 2 p dp$:

$= d \int \frac{1}{p^2+1} dp - d \int \frac{1}{w^2} dw$

The integral of $\frac{1}{p^2+1}$ is $\tan^{-1}{p}$:

$= d \tan^{-1}{p}-d \int \frac{1}{w^2} dw$

$= d \tan^{-1}{p}+ \frac{d}{w}+constant$

Substitute back for $w = p^2+1$:

$= \frac{d ((p^2+1) tan^{-1}{p}+1)}{p^2+1}+C$

Substitute back for $p = \tan{\frac{s}{2}}$:

$= \frac{1}{2} d (\cos{s}+2 \tan^{-1}{\tan{\frac{s}{2}}}+1)+C$

Substitute back for $s = \sin^{-1}{\frac{2 u}{d}}$:

$= 1/2 (\sqrt{d^2-4 u^2}+2 d \tan^{-1}(\frac{2 u}{d \sqrt{1-\frac{4 u^2}{d^2}}+1)})+d)+C$

Substitute back for $u = y-\frac{d}{2}$:

$= d (-\tan^{-1}{\frac{d-2 y}{2 d \sqrt{\frac{y (d-y)}{d^2}}+d}})+\sqrt{y (d-y)}+\frac{d}{2}+C$

Factor the answer a different way:

$= \frac{1}{2} (-2 d \tan^{-1}{\frac{d-2 y}{2 d \sqrt{\frac{y (d-y)}{d^2}}+d}}+2 \sqrt{y (d-y)}+d)+C$

Which is equivalent for restricted y and d values to:

$= y \sqrt{\frac{d}{y}-1}-\frac{1}{2} d \tan^{-1}{\frac{\sqrt{\frac{d}{y}-1} (d-2 y)}{2 (d-y)}}+C$

Syjytg 23:00, 11 September 2012 (EDT)