Question

∫x^2arcsinx/(√1-x^2)dx上限为1下限为0

计算这个反常积分

Answer 1

∫(0->1) x^2arcsinx/(√1-x^2) dx

let
x = siny
dx = cosy dy
x=0, y=0
x=1 , y =π/2
∫(0->1) x^2arcsinx/(√1-x^2) dx
=∫(0->π/2) y(siny)^2 dy
=(1/2)∫(0->π/2) y( 1-cos2y) dy
= (1/2) [y^2/2](0->π/2) -(1/2) ∫(0->π/2) ycos2y dy
= π^2/16 - (1/4)∫(0->π/2) y dsin2y
=π^2/16 - (1/4)[ ysin2y](0->π/2) + (1/4)∫(0->π/2) sin2y dy
=π^2/16 - (1/8)[cos2y](0->π/2)
=π^2/16 +1/4

Answer 2

反常积分是通过查表得出的，积分算不出来的、、、