Question

求解这个定积分

Answer 1

解：设cosx=tant，则dcosx=(sect)^2dt，t∈[-π/4,π/4]，
∫(0,π)√[1+(cosx)^2]dcosx=-∫(-π/4,π/4)sectdtant
而∫sectdtant=secttant-∫sect[(sect)^2-1]dt=secttant-∫sectdtant+∫sectdt=secttant-∫sectdtant+ln丨sect+tant丨，∴∫sectdtant=(1/2)(secttant+ln丨sect+tant丨)+C，
∴原式=-π(secttant+ln丨sect+tant丨)(t=-π/4,π/4)=2π[ln(√2-1)-√2]。供参考。

Answer 2

let
y=2sinu
dy = 2cosu du
y=0, u=0
y=√2, u=π/4
∫(-√2->√2) √(8-2y^2) dy
=2∫(0->√2) √(8-2y^2) dy
=2∫(0->π/4) (2√2cosu) (2cosu du)
=8√2 ∫(0->π/4) (cosu)^2 du
=4√2 ∫(0->π/4) (1+cos2u) du
=4√2 [ u+(1/2)sin2u]|(0->π/4)
=4√2 ( π/4+1/2 )
=√2. π + 2√2