《高数双语》课件section 4-3.pptx
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1、Integration by Parts 分部积分法 1Integration by Parts 2When u and v are continuously differentiable functions of x,the Product Rule for differentiation tell us that(),or().uvuvvud uvudvvduIntegrating both sides with respect to x,we have()uv dxuv dxvu dx.uvvu dx uv dxuvvu dx().uvuvvuTransposing terms,()uv
2、 dxvCu ;v dxdv u dxduIntegration by Parts 3Note:This formula express one integral,udv,in terms of a second integral,vdu.With a proper choice of u and v,the second integral may be easier to evaluate than the first.When this equation is written in the simpler differential notation,we obtain the follow
3、ing formula.Formula for Integration by Parts udvuvvduIntegration by Parts 4 Evaluate.xxe dx Solutionxxxe dxxde xxxee dx xxxeeC(1).xexCudvuvvdu,xxde dxuxvedCan we choose as the for the?xxxeudxe 22211()222xxxxxe dxde xx e dxxeQuestion,xdudx veThe problem becomes worse.Integration by Parts 5cos.xxdx So
4、lution(I)Letcos,ux 212xdxdxdvcosxxdx 22cossin22xxxxdx Solution(II)Let,ux cossinxdxdxdvcosxxdx sinxdx sinsinxxxdx sincos.xxxCthe problem may becomes worse.Indeed,uIt is obviously,if we do not make a proper choice of dv and,udvuvvdu21sin,2duxdx vx ,sindudx vx Evaluate Integration by PartsThe method of
5、 the integration by parts is suitable to apply to the following integrals:6;sin;cos;ln;ln;arcsin;arccos;arctan;sin;cos;kaxkkaamkkkaxaxx e dxxbxdxxbxdxxxdxxxdxxaxdxxaxdxxbxdxebxdxebxdx udvuvvduIntegration by Parts7Compute the following integrals:221);2)(2)cos;3)ln;4)arcsin;5)sin.xxx e dxxxxdxxxdxxdxe
6、xdx udvuvvduIntegration by Parts821)xx e dx Solution22xxx e dxx de 22()xxx ee d x 22xxx exe dx 2,xuxdvde22xxxx exee dx,xux dvde222.xxxx exeeCpower exponential,the integrand is the product of afunction()and anfunction.For this case,we always regard as the function.kaxkaxkkx e dxuxxeN udvuvvduIntegrat
7、ion by Parts922)(2)cosxxxdx Solution Let22,uxxcossin.xdxdxdv22(2)cos(2)sinxxxdxxx dx22(2)sinsin(2)xxxxd xx 2(2)sin2(1)sinxxxxxdx(1)sin(1)(cos)xxdxxdx(1)coscos(1)xxxd x co(s1),()vuxdxd Then,2(2)22,dud xxxsin.vx udvuvvduIntegration by Parts1022(2)cos(2)sin2(1)cos2 cos(1)xxxdxxxxxxxd xSolution(continue
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