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
8、d)2(22)sin2(1)cos.xxxxxC2(2)sin2(1)cos2 cosxxxxxxdx 22,cosuxx dvdxsin or cospower sinecosine,the integrand is the product of a function()and aorfunctionsin(orcos).For this case,we always regardas the function .kkkkxbxdxxbxdxkNxuxbxbx udvuvvdu22)(2)cosxxxdx Integration by Parts113)lnxxdx Solution21ln
9、ln2xxdxxdx 221lnln2xxx dx 221ln2xxxdxx 2211ln.22xxxCln,;ux dvxdxlnpower loga ,the integrand is the product of afunction()and aof.For this case,we always regarrithmln,not das the function.,xxdxxxxxRu 21ln2xxxdx 211,2dudx vxxudvuvvduIntegration by Parts 12By substitutionSolution Letarcsin,ux.dxdv arcs
10、in xdx arcsin(arcsin)xxxdx 21arcsin1xxxdxx 4)arcsin xdx 2211arcsin(1)21xxdxx 12221arcsin(1)(1)2xxxdx 2arcsin1.xxxCThen,21(arcsin),1dudxx.vx udvuvvduIntegration by Parts13arccos,arctan,arccot,arcsin,arccos,arctan,arccot,xdxxdxxdxxxdxxxdxxxdxxxdxSimilarly,one can solve the following problems by partsr
11、csin.power inverse trigonemetric function (),the integrand is the product of a function()and anof.For this case,we choose()notas the functi arcsin.,on.kkkx aaxdxetckNxax etcxxu 0arcsin,ux ddvxxxd udvuvvdu4)arcsin xdx Integration by Parts 14Solution(I)5)sinxexdx sinxexdx sinxxde sin(sin)xxexe dx sinc
12、osxxexexdx sincosxxexxde sincos(cos)xxxexexe dx(sincos)sinxxexxexdx sinxexdx(sincos).2xexxCLetsin,ux.xxe dxdedvcos,xux dve dx2sin(sincos).xxexdxexxC Note Sometimes,after we use the parts formula,the primal integral will occur again.Then,(sin)cos,dudxx.xve udvuvvduIntegration by Parts 155)sinxexdx So
13、lution(II)Let,xue sin(cos).xdxdxdvsinxexdx(cos)xe dx coscos()xxexxd e coscosxxexexdx cos(sin)xxexe dx )in,(sxudvdexcossinsinxxxexexexdx sinxexdx(sincos).2xexxCor,the integrand is the product of an function and afunction.For this cassin cosexponentiae,we can choose a l trigonemetriny of the two funct
14、ions as the.c axaxebxdxebxdxuPlease evaluatecos.xexdx udvuvvduIntegration by Parts16(ln)(ln)(ln)nnnnIxxxxddxx Evaluate(ln)().nnIxdx nN Solution11(ln)(ln)nnxxx nxdxx 1(ln)(ln)nnxxnxdx 1(ln)nnxxnI 11lnlnln,IxdxxxxdxxxxCxParticularly,2221(ln)2(ln)2(ln);IxxIxxxxxC33232(ln)3(ln)3(ln)6(ln);IxxIxxxxxxxCas
15、so on.Finish.Exercise Evaluate,22(0).()nndxInN axa 1(ln)nnnIxxnI udvuvvduIntegration by Parts 17Suppose that one of the anti-derivatives for()f xis2,xe evaluate().xfx dx Solution()xfx dx ()xdf x ()(),xf xf x dx 2and(),xf x dxeC ()()f x dxf x QBy the method of integration by parts,we haveTake differentiation to both sides of the last equation,we obtain that2()2,xf xxe ()xfx dx ()()xf xf x dx 222xx e 2.xeC
