Question

定积分 （a^2 - x^2）^(1/2) 范围是0到2/π （π是圆周率）

Answer 1

let

x = asinb

dx = acosb db

x=0, b=0

x=π/2, arcsin(π/(2a))

∫(0,π/2)（a^2 - x^2）^(1/2) dx

= ∫(0,arcsin(π/2a)) a^2(cosb)^2 db

= (1/2)∫(0,arcsin(π/2a)) a^2(1+cos2b) db

= (a^2/2)[b+sin2b/2] (0,arcsin(π/(2a)))

= (a^2/2) [ arcsin(π/(2a)) + (π(4a^2-π^2)^(1/2)/(4a^2)) ]