∫xarcsinx/√1-x2,上限为1/2,下限为-1/2(利用奇偶性计算)

let
f(x)= xarcsinx/√(1-x^2)
f(-x)= f(x)
∫(-1/2 -> 1/2) xarcsinx/√(1-x^2) dx
=2∫(0 -> 1/2) xarcsinx/√(1-x^2) dx
=-∫(0 -> 1/2) arcsinx d√(1-x^2)
=- [ √(1-x^2) .arcsinx ] |(0 -> 1/2) +∫(0 -> 1/2) dx
=- (√3/2) ( π/6) + [x]|(0->1/2)
=-(√3/12)π + 1/2本回答被网友采纳