如题所述
看第五题的答案
就是用个性质：∫(0→π) xƒ(sinx) dx = (π/2)∫(0→π/2) ƒ(sinx) dx
温馨提示：答案为网友推荐，仅供参考
第1个回答 2016-06-12
consider
∫ xsinxcosx dx
=-(1/4)∫ xdcos2x
=-(1/4)xcos2x +(1/4)∫ cos2x dx
=-(1/4)xcos2x +(1/8)sin2x + C
∫(0->π) { x√[(sinx)^2-(sinx)^4] } dx
=∫(0->π/2) xsinxcosx dx - ∫(π/2->π) xsinxcosx dx
=(1/8)[ -2xcos2x + sin2x ]|(0->π/2) - (1/8)[ -2xcos2x + sin2x ]|(π/2->π)
=(1/8)(4π-1 )本回答被网友采纳
∫ xsinxcosx dx
=-(1/4)∫ xdcos2x
=-(1/4)xcos2x +(1/4)∫ cos2x dx
=-(1/4)xcos2x +(1/8)sin2x + C
∫(0->π) { x√[(sinx)^2-(sinx)^4] } dx
=∫(0->π/2) xsinxcosx dx - ∫(π/2->π) xsinxcosx dx
=(1/8)[ -2xcos2x + sin2x ]|(0->π/2) - (1/8)[ -2xcos2x + sin2x ]|(π/2->π)
=(1/8)(4π-1 )本回答被网友采纳
