设 x=acosθ,y=bsinθ ,则 x'=-asinθ,y'=bcosθ ,
x'^2+y'^2=a^2sin^2θ+b^2cos^2θ)
椭圆周长=∫（θ从0到2π）根号[a^2sin^2θ+b^2cos^2θ]dθ
=4∫（θ从0到π/2）根号[a^2sin^2θ+b^2cos^2θ]dθ
=4∫（θ从0到π/2）bcosθ根号[a^2tan^2θ/b^2+1]dθ
【 设（atanθ）/b=tanψ ,asec^2θdθ=bsec^2ψdψ]
原式=4∫（ψ从0到π/2）b^2cos^3θ/acos^3ψdψ