www.5615.net > 定积分∫xsinxDx

定积分∫xsinxDx

∫xsinxdx=-∫xdcosx=-xcosx+∫cosxdx=-xcosx+sinx+C

解: ∫(0,x)xsinxdx =-∫(0,x)xd(cosx) =-xcosx|(0.x)+∫(0,x)cosxdx =-xcosx+sinx|(0,x) =-xcosx+sinx 希望对你有帮助,望采纳,谢谢~

积分上下限为π2和0,算式中没写,用分步积分:∫xsinxdx=∫xd(-cosx)=-xcosx-∫(-cosx)dx=sinx-xcosx=1

原式=∫-xd(cosx) =-xcosx+∫cosxd(-x) =-xcosx-∫cosxdx =-xcosx-sinx

∫u(x)dv(x)=u(x) v(x)-∫v(x)du(x)∫xsin xdx=-∫xdcosxu(x)=x v(x)=-cosx所以∫xsin xdx=-∫xdcosx=-[-xcosx-∫cosxdx]=-[-xcosx-sinx+c]=xcosx+sinx+cc不分正负,最后只需+c

xsinx 是 偶函数, 则∫ xsinxdx = 2∫ xsinxdx= -2∫ xdcosx= -2[xcosx] + 2∫ cosxdx= 2π + 2[sinx] = 2π

原式=-∫xd(cosx) =-xcosx+∫cosxdx (分部积分法) =-xcosx+sinx+C (C是积分常数).

∫ xsin xdx=-∫ xdcosx=-xcosx+∫ cosx*dx=-xcosx+∫ dsinx=-xcosx+sinx +C

∫xsinxdx=-xcosx+sinx+C

∫x xsinxdx/2 =-1/2∫x^2dcosx=-1/2[x^2cosx-∫cosxdx^2]=-1/2x^2cosx+∫xcosxdx=-1/2x^2cosx+∫xdsinx=-1/2x^2cosx+xsinx-∫sinxdx=-1/2x^2cosx+xsinx+cosx+c

网站地图

All rights reserved Powered by www.5615.net

copyright ©right 2010-2021。
www.5615.net内容来自网络,如有侵犯请联系客服。zhit325@qq.com