12-267/Tuesday September 11 Notes

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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)