12-267/Tuesday September 11 Notes: Difference between revisions

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===Solving the complicated integral in the Brachistochroe integral===
===Solving the complicated integral in the Brachistochroe integral===


integral sqrt((d-y)/y) dy
<math>\int \sqrt{\frac{d-y}{y}} dy </math>

= integral sqrt(d y-y^2)/y dy
For the integrand sqrt(d y-y^2)/y, complete the square:
<math> = \int \sqrt{\frac{d y-y^2}{y}} dy </math>

= integral sqrt(d^2/4-(y-d/2)^2)/y dy
Complete the square in the integrand:
⚫
For the integrand sqrt(d^2/4-(y-d/2)^2)/y, substitute u = y-d/2 and du = dy:

= integral (2 sqrt(d^2/4-u^2))/(d+2 u) du
<math> = \int \sqrt{\frac{\frac{d^2}{4} - (y - \frac{d}{2})^2}{y}} </math>
= 2 integral sqrt(d^2/4-u^2)/(d+2 u) du

⚫
For the integrand sqrt(d^2/4-u^2)/(d+2 u), (assuming all variables are positive) substitute u = 1/2 d sin(s) and du = 1/2 d cos(s) ds. Then sqrt(d^2/4-u^2) = sqrt(d^2/4-1/4 d^2 sin^2(s)) = 1/2 d cos(s) and s = sin^(-1)((2 u)/d):
Substitute <math>u = y-\frac{d}{2}</math> and <math> du = dy </math>:
= d^2/2 integral (cos^2(s))/(d sin(s)+d) ds

⚫
For the integrand (cos^2(s))/(d sin(s)+d), substitute p = tan(s/2) and dp = 1/2 sec^2(s/2) ds. Then transform the integrand using the substitutions sin(s) = (2 p)/(p^2+1), cos(s) = (1-p^2)/(p^2+1) and ds = (2 dp)/(p^2+1):
<math> = \int 2 \sqrt{\frac{\frac{d^2}{4} - u^2}{d+2 u}} du </math>
= d^2/2 integral (2 (1-p^2)^2)/((p^2+1)^3 ((2 d p)/(p^2+1)+d)) dp

⚫
Simplify the integrand (2 (1-p^2)^2)/((p^2+1)^3 ((2 d p)/(p^2+1)+d)) to get (2 (p-1)^2)/(d p^4+2 d p^2+d):
⚫
Assuming all variables are positive, substitute <math>u = \frac{1}{2} d \sin{s}</math> and <math>du = \frac{1}{2} d \cos{s} ds</math>. Then <math>\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>s = \sin^{-1}{\frac{2u}{d}}</math>:
⚫
= d^2/2 integral (2 (p-1)^2)/(d p^4+2 d p^2+d) dp

⚫
= d^2 integral (p-1)^2/(d p^4+2 d p^2+d) dp
<math> = \frac{d^2}{2} \int \frac{\cos^2{s}}{d \sin{s} + d} ds</math>
⚫
= d^2 integral (p-1)^2/(d (p^2+1)^2) dp

⚫
= d integral (p-1)^2/(p^2+1)^2 dp
⚫
For the integrand substitute <math> p=\tan{\frac{s}{2}}</math> and <math>dp = \frac{1}{2} \sec^2{\frac{s}{2}} ds</math>. Then transform the integrand using the substitutions <math>\sin{s} = \frac{2p}{p^2 + 1}</math>, <math>\cos{s} = \frac{1-p^2}{p^2 + 1}</math> and <math>ds = \frac{2 dp}{p^2 + 1}</math>:
⚫
For the integrand (p-1)^2/(p^2+1)^2, use partial fractions:

⚫
= d integral (1/(p^2+1)-(2 p)/(p^2+1)^2) dp
= d integral 1/(p^2+1) dp-2 d integral p/(p^2+1)^2 dp
<math> = \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>

For the integrand p/(p^2+1)^2, substitute w = p^2+1 and dw = 2 p dp:
⚫
Simplify the integrand <math>\frac{2(1-p^2)^2}{(p^2 + 1)^3 (\frac{2 d p}{p^2 + 1} + d)}</math> to get <math>\frac{2 (p-1)^2}{d p^4 + 2 d p^2 + d}</math>:
= d integral 1/(p^2+1) dp-d integral 1/w^2 dw

⚫
The integral of 1/(p^2+1) is tan^(-1)(p):
⚫
<math> = \frac{d^2}{2} \int \frac{2(p-1)^2}{d p^4+2 d p^2+d} dp </math>
= d tan^(-1)(p)-d integral 1/w^2 dw

⚫
= d tan^(-1)(p)+d/w+constant
⚫
<math> = d^2 \int \frac{(p-1)^2}{d p^4+2 d p^2+d} dp </math>
⚫
Substitute back for w = p^2+1:

⚫
= (d ((p^2+1) tan^(-1)(p)+1))/(p^2+1)+C
⚫
<math> = d^2 \int \frac{(p-1)^2}{d (p^2+1)^2} dp </math>
⚫
Substitute back for p = tan(s/2):

⚫
= 1/2 d (cos(s)+2 tan^(-1)(tan(s/2))+1)+C
⚫
<math> = d \int \frac{(p-1)^2}{(p^2+1)^2} dp </math>
⚫
Substitute back for s = sin^(-1)((2 u)/d):

⚫
= 1/2 (sqrt(d^2-4 u^2)+2 d tan^(-1)((2 u)/(d (sqrt(1-(4 u^2)/d^2)+1)))+d)+C
⚫
For the integrand <math>\frac{(p-1)^2}{(p^2+1)^2}</math> use partial fractions:
⚫
Substitute back for u = y-d/2:

⚫
= d (-tan^(-1)((d-2 y)/(2 d sqrt((y (d-y))/d^2)+d)))+sqrt(y (d-y))+d/2+C
⚫
<math> = d \int (\frac{1}{p^2+1}-\frac{2 p}{(p^2+1)^2}) dp</math>

<math> = d \int \frac{1}{(p^2+1)} dp - 2 d \int \frac{p}{(p^2+1)^2} dp</math>

⚫
For the integrand <math>\frac{p}{(p^2+1)^2}</math>, substitute <math>w = p^2+1</math> and <math>dw = 2 p dp</math>:

<math> = d \int \frac{1}{p^2+1} dp - d \int \frac{1}{w^2} dw </math>

⚫
The integral of <math>\frac{1}{p^2+1}</math> is <math>\tan^{-1}{p}</math>:

<math> = d \tan^{-1}{p}-d \int \frac{1}{w^2} dw</math>

⚫
<math> = d \tan^{-1}{p}+ \frac{d}{w}+constant</math>

⚫
Substitute back for <math>w = p^2+1</math>:

⚫
<math> = \frac{d ((p^2+1) tan^{-1}{p}+1)}{p^2+1}+C </math>

⚫
Substitute back for <math>p = \tan{\frac{s}{2}}</math>:

⚫
<math> = \frac{1}{2} d (\cos{s}+2 \tan^{-1}{\tan{\frac{s}{2}}}+1)+C </math>

⚫
Substitute back for <math>s = \sin^{-1}{\frac{2 u}{d}}</math>:

⚫
<math> = 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>u = y-\frac{d}{2}</math>:

⚫
<math> = 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:
Factor the answer a different way:

= 1/2 (-2 d tan^(-1)((d-2 y)/(2 d sqrt((y (d-y))/d^2)+d))+2 sqrt(y (d-y))+d)+C
<math> = \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:
Which is equivalent for restricted y and d values to:

= y sqrt(d/y-1)-1/2 d tan^(-1)((sqrt(d/y-1) (d-2 y))/(2 (d-y)))+C
<math> = 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>

[[User:Syjytg|Syjytg]] 23:00, 11 September 2012 (EDT)
[[User:Syjytg|Syjytg]] 23:00, 11 September 2012 (EDT)

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Solving the complicated integral in the Brachistochroe 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)

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