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## Answers

*r=a(1+cos x)*

*r^2=a^2(1+cos x)^2 = a^2 + 2[a^2][cos(x)] + [a^2][cos^2(x)]*

*dr/dx = r' = -a sin(x)*

*(r')^2 = [a^2][sin^2 (x)]*

*Therefore perimeter (s) of curve r=a(1+cos x) in polar coordinate with x vary from 0 to Pi (due to curve is symmetry on axis x=0) is*

*s = 2 Int { sqrt[r^2 + (r')^2] } dx where x vary from 0 to Pi.*

*Thus*

*sqrt[r^2 + (r')^2]*

*= sqrt { a^2 + 2[a^2][cos(x)] + [a^2][cos^2 (x)] + [a^2][sin^2 (x)] }*

*= sqrt { (2a^2)[1+cos(x)] }*

*= [sqrt(2)]a {sqrt [1+cos(x)]}*

*Then*

*s = 2[sqrt(2)]a . Int {sqrt [1+cos(x)]} dx*

*Let 1+cos(x) = 1+2cos^2 (x/2) - 1 = 2cos^2 (x/2)*

*s = 2[sqrt(2)]a . Int {sqrt [2cos^2 (x/2)]} dx*

*s = 4a . Int [cos(x/2)] dx where x vary from 0 to Pi*

*s = 4a [sin(x/2)]/(1/2)*

*s = 8a [sin(Pi/2) - sin(0)]*

*s = 8a*