Question

(0－1)x(arcsinx)²的定积分怎么算？

Answer 1

令 arcsinx = u, 则 x = sinu， dx = cosudu
I = ∫<0 ,1> x(arcsinx)² dx = (1/2)∫<0 , π/2> u^2sin2u du
= (-1/4)∫<0 , π/2> u^2dcos2u
= (-1/4)[u^2cos2u]<0 , π/2> - ∫<0 , π/2> 2ucos2udu]
= (-1/4)[-π^2/2 - ∫<0 , π/2> udsin2u]
= (-1/4)[-π^2/2 - [usin2u]<0 , π/2> + ∫<0 , π/2> sin2udu]
= (-1/4)[-π^2/2 - 0 - (1/2)[cos2u]<0 , π/2>]
= (-1/4)[-π^2/2 +1] = (1/8)(π^2-2)

Answer 2

方法如下，请作参考：

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