12-267/Tuesday September 11 Notes
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Solving the complicated integral in the Brachistochrone integral
[math]\displaystyle{ \int \sqrt{\frac{d-y}{y}} dy }[/math]
[math]\displaystyle{ = \int \sqrt{\frac{d y-y^2}{y}} dy }[/math]
Complete the square in the integrand:
[math]\displaystyle{ = \int \sqrt{\frac{\frac{d^2}{4} - (y - \frac{d}{2})^2}{y}} }[/math]
Substitute [math]\displaystyle{ u = y-\frac{d}{2} }[/math] and [math]\displaystyle{ du = dy }[/math]:
[math]\displaystyle{ = \int 2 \sqrt{\frac{\frac{d^2}{4} - u^2}{d+2 u}} du }[/math]
Assuming all variables are positive, substitute [math]\displaystyle{ u = \frac{1}{2} d \sin{s} }[/math] and [math]\displaystyle{ du = \frac{1}{2} d \cos{s} ds }[/math]. Then [math]\displaystyle{ \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} }[/math] and [math]\displaystyle{ s = \sin^{-1}{\frac{2u}{d}} }[/math]:
[math]\displaystyle{ = \frac{d^2}{2} \int \frac{\cos^2{s}}{d \sin{s} + d} ds }[/math]
For the integrand substitute [math]\displaystyle{ p=\tan{\frac{s}{2}} }[/math] and [math]\displaystyle{ dp = \frac{1}{2} \sec^2{\frac{s}{2}} ds }[/math]. Then transform the integrand using the substitutions [math]\displaystyle{ \sin{s} = \frac{2p}{p^2 + 1} }[/math], [math]\displaystyle{ \cos{s} = \frac{1-p^2}{p^2 + 1} }[/math] and [math]\displaystyle{ ds = \frac{2 dp}{p^2 + 1} }[/math]:
[math]\displaystyle{ = \frac{d^2}{2} \int 2 \frac{(1-p^2)^2}{(p^2 + 1)^3 (\frac{2 d p}{p^2 + 1} + d)} dp }[/math]
Simplify the integrand [math]\displaystyle{ \frac{2(1-p^2)^2}{(p^2 + 1)^3 (\frac{2 d p}{p^2 + 1} + d)} }[/math] to get [math]\displaystyle{ \frac{2 (p-1)^2}{d p^4 + 2 d p^2 + d} }[/math]:
[math]\displaystyle{ = \frac{d^2}{2} \int \frac{2(p-1)^2}{d p^4+2 d p^2+d} dp }[/math]
[math]\displaystyle{ = d^2 \int \frac{(p-1)^2}{d p^4+2 d p^2+d} dp }[/math]
[math]\displaystyle{ = d^2 \int \frac{(p-1)^2}{d (p^2+1)^2} dp }[/math]
[math]\displaystyle{ = d \int \frac{(p-1)^2}{(p^2+1)^2} dp }[/math]
For the integrand [math]\displaystyle{ \frac{(p-1)^2}{(p^2+1)^2} }[/math] use partial fractions:
[math]\displaystyle{ = d \int (\frac{1}{p^2+1}-\frac{2 p}{(p^2+1)^2}) dp }[/math]
[math]\displaystyle{ = d \int \frac{1}{(p^2+1)} dp - 2 d \int \frac{p}{(p^2+1)^2} dp }[/math]
For the integrand [math]\displaystyle{ \frac{p}{(p^2+1)^2} }[/math], substitute [math]\displaystyle{ w = p^2+1 }[/math] and [math]\displaystyle{ dw = 2 p dp }[/math]:
[math]\displaystyle{ = d \int \frac{1}{p^2+1} dp - d \int \frac{1}{w^2} dw }[/math]
The integral of [math]\displaystyle{ \frac{1}{p^2+1} }[/math] is [math]\displaystyle{ \tan^{-1}{p} }[/math]:
[math]\displaystyle{ = d \tan^{-1}{p}-d \int \frac{1}{w^2} dw }[/math]
[math]\displaystyle{ = d \tan^{-1}{p}+ \frac{d}{w}+constant }[/math]
Substitute back for [math]\displaystyle{ w = p^2+1 }[/math]:
[math]\displaystyle{ = \frac{d ((p^2+1) tan^{-1}{p}+1)}{p^2+1}+C }[/math]
Substitute back for [math]\displaystyle{ p = \tan{\frac{s}{2}} }[/math]:
[math]\displaystyle{ = \frac{1}{2} d (\cos{s}+2 \tan^{-1}{\tan{\frac{s}{2}}}+1)+C }[/math]
Substitute back for [math]\displaystyle{ s = \sin^{-1}{\frac{2 u}{d}} }[/math]:
[math]\displaystyle{ = 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 }[/math]
Substitute back for [math]\displaystyle{ u = y-\frac{d}{2} }[/math]:
[math]\displaystyle{ = 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 }[/math]
Factor the answer a different way:
[math]\displaystyle{ = \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 }[/math]
Which is equivalent for restricted y and d values to:
[math]\displaystyle{ = 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 }[/math]
Syjytg 23:00, 11 September 2012 (EDT)
