Question

求定积分！~！！！。！！！

这两个式子相等 但为什么求出来答案不一样

Answer 1

如图所示

Answer 2

(cosx)^2 开方不是 cosx ! 规范解法 :
I = ∫<0, π> x√[(sinx)^2-(sinx)^4]dx = ∫<0, π> x√(sinx)^2(cosx)^2]dx
= ∫<0, π> x|sinxcosx|dx = ∫<0, π/2> xsinxcosxdx + ∫<π/2, π> x(-sinxcosx)dx
= (1/2)∫<0, π/2> xsin2xdx - (1/2)∫<π/2, π> xsin2xdx
= (-1/4)∫<0, π/2> xdcos2x + (1/4)∫<π/2, π> xdcos2x
= (-1/4)[xcos2x-(1/2)sin2x]∫<0, π/2> + (1/4)[xcos2x-(1/2)sin2x]<π/2, π>
= π/8 + π/4 + π/8 = π/2

Answer 3

要讨论cosx