ppcq.net
当前位置:首页 >> ∫E的 x次方sin2xDx >>

∫E的 x次方sin2xDx

∫e^-x *sin2xdx=-∫e^-x *sin2xd(-x)=-∫sin2xde^-x=-e^-xsin2x+∫e^-x*cos2x*2dx=-e^-xsin2x-2∫e^-x*cos2xd(-x)=-e^-xsin2x-2∫cos2xde^-x=-e^-xsin2x-2e^-xcos2x+2∫e^-x*(-sin2x)*2dx=-e^-xsin2x-2e^-xcos2x-4∫e^-x*sin2xdx ∴5∫e^-x *sin2xdx=-e^-xsin2x-2e^-xcos2x ∫e^-x *sin2xdx=-e^-x *(sin2x+2cos2x )/5

∫e^xsin2xdx=∫sin2xde^x=e^xsin2x-∫e^xdsin2x=e^xsin2x-2∫e^xcos2xdx=e^xsin2x-2∫cos2xde^x=e^xsin2x-2e^xcos2x+2∫e^xdcos2x=e^xsin2x-2e^xcos2x-4∫e^xsin2xdx所以∫e^xsin2xdx=(e^xsin2x-2e^xcos2x)/5+C

原式=-∫sin2xd[e^(-x)]=-e^(-x)sin2x+∫e^(-x)dsin2x=-e^(-x)sin2x+2∫e^(-x)cos2xdx=-e^(-x)sin2x-2∫cos2xd[e^(-x)]=-e^(-x)sin2x-2e^(-x)cos2x+2∫e^(-x)dcos2x=-e^(-x)sin2x-2e^(-x)cos2x-4∫e^(-x)sin2xdx 令∫(e^-x)sin2xdx=T,则有 T=-e^(-x)sin2x-2e^(-x)cos2x-4T 所以T=[-e^(-x)sin2x-2e^(-x)cos2x]/5

∫e^x*sinxdx=e^sinx-∫e^cosxdx=e^xsinx-(e^xcosx+∫e^xsindx)=e^x (sinx-cosx)-∫e^xsinxdx所以2∫e^xsinxdx=e^x(sinx-cosx)+C1 ∫e^xsinxdx=e^x(sinx-cosx)/2+C希望可以帮到你祝学习快乐!O(∩_∩)O~

令a=∫ e^(-x)sin2xdxa=∫ e^(-x)sin2xdx = - ∫ sin2xd(e^(-x))=-e^(-x)sin2x+∫ e^(-x)d(sin2x) =-e^(-x)sin2x+2∫ e^(-x)cos2xdx =-e^(-x)sin2x-2∫ cos2xd(e^(-x)) =-e^(-x)sin2x-2cos2x*e^(-x)+2∫ e^(-x)d(cos2x) =-e^(-x)sin2x-2cos2x*e^(-x)-4∫ e^(-x)sin2xdx =-e^(-x)

∫(e^x)sin(x/2) dx=∫sin(x/2) d(e^x)=(e^x)sin(x/2) - (1/2)∫(e^x)cos(x/2) dx=(e^x)sin(x/2) - (1/2)∫cos(x/2) d(e^x)=(e^x)sin(x/2) - (1/2)(e^x)cos(x/2) - (1/4)∫(e^x)sin(x/2) dx 将 -(1/4)∫(e^x)sin(x/2) dx移到等式左边与左边合并,除去系数,得:∫(e^x)sin

解: ∫e^xsin2xdx=e^xsin2x-2∫e^xcos2xdx=e^xsin2x-2[e^xcos2x+2∫e^xsin2x]dx=e^xsin2x-2e^xcos2x-4∫e^xsin2x dx得:5∫e^xsin2xdx=e^xsin2x-2e^xcos2x+C1故:∫e^xsin2xdx=1/5e^x(sin2x-2cos2x)+C 满意请采纳,谢谢~

原式=-∫sin2xd[e^(-x)]=-e^(-x)sin2x+∫e^(-x)dsin2x=-e^(-x)sin2x+2∫e^(-x)cos2xdx=-e^(-x)sin2x-2∫cos2xd[e^(-x)]=-e^(-x)sin2x-2e^(-x)cos2x+2∫e^(-x)dcos2x=-e^(-x)sin2x-2e^(-x)cos2x-4∫e^(-x)sin2xdx令∫(e^-x)sin2xdx=t,则有t=-e^(-x)sin2x-2e^(-x)cos2x-4t所以t=[-e^(-x)sin2x-2e^(-x)cos2x]/5

令√x=t, 则x=t 原式=∫sintdt=∫2tsintdt=-2∫tdcost=-2(tcost-∫costdt)=-2(tcost-sint)+c=2sint-2tcost+c =2sin√x-2√x*cos√x+c

∫e^(sinx)sin2xdx=2∫e^(sinx)sinxcosxdx=2∫e^(sinx)sinxd(sinx)令t=sinx,=2∫te^td(t)=2【te^t-∫e^td(t)】 分部积分法=2(t-1)e^t代入得,=2(sinx-1)e^(sinx)+c

网站首页 | 网站地图
All rights reserved Powered by www.ppcq.net
copyright ©right 2010-2021。
内容来自网络,如有侵犯请联系客服。zhit325@qq.com