怎样用洛必达法则求函数的极值？

Question

如题所述

Answer 1

let
u=π-x
du=-dx
x=π/2, u=π/2
x=π , u=0
∫(0->π) √[(sinx)^2-(sinx)^4 ] dx
=∫(0->π/2) √[(sinx)^2-(sinx)^4 ] dx +∫(π/2->π) √[(sinx)^2-(sinx)^4 ] dx
=∫(0->π/2) √[(sinx)^2-(sinx)^4 ] dx +∫(π/2->0) √[(sinu)^2-(sinu)^4 ] (-du)
=∫(0->π/2) √[(sinx)^2-(sinx)^4 ] dx +∫(0->π/2) √[(sinu)^2-(sinu)^4 ] du
=∫(0->π/2) √[(sinx)^2-(sinx)^4 ] dx +∫(0->π/2) √[(sinx)^2-(sinx)^4 ] dx
=2∫(0->π/2) √[(sinx)^2-(sinx)^4 ] dx