设 u＝x ,v'=sinx
则 u'=1 ,v=-cosx
则原积分∫(π/4,0)xsinxdx
＝⦗-xcosx⦘(π/4,0)-∫(π/4,0) -cosx dx
=(-π/4)×(√2/2) + ⦗sinx⦘(π/4,0)
=(4√2-√2π)/8
用分部积分法求∫(π/4,0)xsinxdx
设 u＝x ,v'=sinx
则 u'=1 ,v=-cosx
则原积分∫(π/4,0)xsinxdx
＝⦗-xcosx⦘(π/4,0)-∫(π/4,0) -cosx dx
=(-π/4)×(√2/2) + ⦗sinx⦘(π/4,0)
=(4√2-√2π)/8