Using power series with integration to find the length of x^2/a^2+y^2/b^2=1. assume b>0.  Solution hint as below: consider there two case(a>b, or a

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Using power series with integration to find the length of x^2/a^2+y^2/b^2=1. assume b>0. 

Solution hint as below: consider there two case(a>b, or a<b) with different solution

Set x=acosu,y=bsintu,0<=u<=2pi, 

L=integral(2pi,0()sqrt((dx/du)^2+(dy/dt)^2)dt

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