《高数双语》课件section 4-2.pptx
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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
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