1、Integration by Substitution 换元积分法1Integration by the First Substitution2The first type of substitutionSince we obtain()()()()uxfxx dxf u du Suppose(1)()()F uf u ;(2)()ux is differentiable.()()()()(),uxuxf u duF uCFxC Running the Chain Rule BackwardsBy the rule of differentiation of composite functio
2、n,we have ()()()Fxfxx so that ()()()fxx dxFxC Integration by the First Substitution3Theorem Suppose that f is continuous,is continuously differentiable,and the range of is included in the domain of definition of f,then ()()()()uxfxx dxf u du Integration by the First Substitutions4Note that Solutions
3、incosdxxdx sinsincossinxxexdxedx,thensinxu,uue dueC sinsincossinxxexdxedx Let.Since we obtainsinuuxe du sin()uuxeC sin.xeC(1)Evaluatesincosxexdx Integration by the First Substitutions ()(where,)maxbdxa bR mN(21)2dxdx331(21)(21)(21).2xdxxdx21xu341,4u duuC 331(21)(21)(21)2xdxxdx Note that,then,we have
4、Let.Since we haveSolution32112uxu du 41 12 4uC41(21).8xC(2)Evaluate3(21).xdx 111()()()()1mmmaxbaxbdxaxbd axbCaam Integration by the First Substitutions2221111dxdxaxaaxa 2111xdaaxa 1arctan.xCaa22211dxxdaaxxa arcsin.xCa221112dxdxaxaaxax111()()2d axd axaaxax 1ln|ln|2axaxCa1ln.2axCaax Solution:(3)(4)(5)
5、Integration by the First Substitutions(6)Evaluate223xdxx Solution22222233133333xxdxdxdxdxxxxx 221arctandxxCaaax 3arcsin.3xxC 221arctandxxCaaax 22arcsindxxCaax 221ln2dxaxCaaxax 7Integration by the First SubstitutionsSolutionsincosxdxx (7)Evaluatetan.xdx tan xdx(cos)cosdxx coslnuxuC ln cos.xC tan xdx
6、ln cos.xC cot xdx ln sin.xCIntegration by the First Substitutions9Solution(I)1sindxx (8)Evaluatecsc.xdx csc xdx 12sincos22dxxx 212tancos22xdxx 1tan2tan2xdx ln tan2xCln csccot.xxCIntegration by the First Substitutions10Solution(II)csc xdx 1sindxx 2sinsinxdxx 21(cos)1cosdxx cosux 211duu 111211duuu 11l
7、n21uCu 11cosln.21cosxCx csc xdx ln csccot.xxCsec xdx ln sectan.xxCcsc xdx sec xdx 11cosln.21cosxCx 11sinln.21sinxCx (8)Evaluatecsc.xdx Integration by the First Substitutions11Solution(9)Evaluate25sincos.xxdx 25sincosxxdx 24sincos(sin)xxdx 222sin(1sin)(sin)xxdx 246(sin2sinsin)(sin)xxx dx 357121sinsin
8、sin.357xxxC()()()()uxfxx dxf u du Integration by the First Substitutions12(10)Evaluate1.(12ln)dxxx Solution1(12ln)dxxx 1(ln)12lndxx 11(12ln)212lndxx 12lnux112duu 1ln2uC1ln(12ln).2xC1(ln)xx Integration by the First Substitutions323uxue du 1331xedxx 1()2xx (11)Evaluate31.xedxx Solution3)2(xedx 32(3)3x
9、edx 23ueC32.3xeCIntegration by the First Substitutions21(arctan)1duduu 2arctan(1)xdxxx11()()22xdxdxxx (12)Evaluate 2arctan.(1)xdxxx Solution 21(arctan)()21xdxx 21(arctan)21uxuduu 2(arctan)(arctan)u du 32(arctan)3uC32(arctan).3xCIntegration by the First Substitutions212uuCfind Suppose that().f xSolut
10、ion Let 2sinux 2cos1,xu ()1f uu ()1f uu du 21().2f xxxC22(sin)cos,fxx Integration by the Second Substitutions16Theorem:Suppose that f is continuous,is continuously derivable and does not change sign.Then 1()()()()txf x dxftt dt where 1 is the inverse function of .If the integral()f x dx is difficult
11、 to evaluate directly,we may change the function x into()xt ,then the indefinite integral becomes to ()()ftt dt .The second type of substitutionIntegration by the Second Substitutions171.()dtdxt 1()()()()txf x dxftt dt ()()df x dxf xdx ()()()()dddtftt dtftt dtdxdtdxProofTake differentiation to the b
12、oth sides of we haveBy the definition of indefinite integral,we know the conclusion is true.1 does not change sign,we have that function exists and Since 1()()()fttt ()ft ()f x Integration by the Second Substitutions1853.106ttC21(1),0,2xtt2112(1)2xxdxtt tdt Evaluate12.xxdx SolutionLet12,txthen21(1)2
13、dxdttdt53121()2 53ttCC421()2t dtt dtIntegration by the Second Substitutions19Solution Evaluate3222(0).()dxaax Letsinxat cosdxatdt32221()dxax 322221cossinatdtaat 22cosdtat 21tantCatax22ax 2221.xCaax,2 2t sinxta 22tanxtax Integration by the Second Substitutions20Solution Let2sinxt,2 2t 324.xx dx 2cosd
14、xtdt 324xx dx 322sin44sin2costttdt 3232 sincosttdt 2232 sin(1cos)costttdt 2432(coscos)costt dt 351132(coscos)35ttC t2x24x 35224144.35xxC EvaluateIntegration by the Second Substitutions21Solution221(0).dxaxa Lettanxat 2secdxatdt221dxxa 21secsecatdtat sectdt 1ln|sectan|ttCtax22xa 221lnxxaCaa,2 2t 22ln
15、 xxaC EvaluateIntegration by the Second Substitutions22Solution,letsecxat 0,2t 221(0).dxaxa sec tandxattdt 221dxxa sectantanattdtat sectdt 1ln|sectan|ttCtax22xa 221lnxxaCaa Ifxa 22ln.xxaC EvaluateIntegration by the Second Substitutions23Solution(continued)Ifxa ,then,xu ,letua 22122221lndudxuuaCxaua
16、221lnxxaC 1221lnCxxa2212lnxxaCa22ln xxaC Evaluate221(0).dxaxa Integration by the Second Substitutions24By previous examples,we have seen that we can use following formulas:(1)22sinaxxat;(2)22tanaxxat;(3)22secxaxat.Also,we can use hyperbolic functions:(4)22sinhaxxat Integration by the Second Substitu
17、tions25sinhxat cosh.dxatdt 221coshcoshatdxdtatxa ,then ThusLetSolution1dttC 221lnxxaCaa 1arcsinxhCa 22ln.xxaC Evaluate221.dxxa Integration by the Second Substitutions2671(2)dxx x Let1xt 21,dxdtt 71(2)dxx x 72112tdttt 6712tdtt 71ln|12|14tC 711ln|2|ln|.142xxC SolutionIf the denominator of a fraction has high order with respect to the independent variable,we may use substitution as 1.xt EvaluateSolutionIntegration by the Second Substitutions27121uduu 421.1dxxx 4211dxxx Let1xt 21,dxdtt 24211111dxttt 321tdtt 222121tdtt 2ut 11121uduu 111(1)21u duu 31113uuC 322111.3xxCxx Evaluate
