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1第三章 一元函數(shù)積分學(xué) (不定積分)一. 求下列不定積分:1. ???dxx1ln2解. ?l2 cxxd????????????21ln4l1ln22. cxdxx ????? 22 art2arctarct1arctn3. ?osi)os(i2解. cxxddxx ?????????????? 22 os1incos1insic1in)c1(i4. )(8d解. 方法一: 令 , tx1? cttdtxd ???????????????? )1ln(81)1( 887828= cx?8ln方法二: ??? ????dxxdxd)1()1()1( 87878= =cx?)ln(|l88 cx????????81ln5． dxdx??????cosi12i2)sin(2cosin1??dxin1cosi12xxx??? 2cos2sin)(22tan1t2|cosin1|l2 xdxx?????c||l|i|l二. 求下列不定積分:1. ???2)1(2xxd解. ?? ??1)()1()( 222 xdxx txan?令 ?tdseco2=???ccttdsinsico22. ?241x解. 令 x = tan t,???? ?????? cttdtdttdxd sin1i3sinisincoetanc1 24434224= cxx?????????23213. ?221)(xd解. 令 tan????? ?????? tdttdttx 222222 sin1cosinsec)1an()(= xct?2rtsinar4. (a > 0)??2xd解. 令 tasin???? ?????? ctatdtatadx 2sin412cos1cosi2223= cxaxa????????222rcsin5. ??dx32)1(解. 令 tsin????? ????dttdttd42cos214)2cos1(cos)(432= ? tttttt in3si83)(8in1= cx?2os41sarc83= tt??)in(cin2i= cxx?518arcsi326. ??dx421解. 令 t???? ??????????dttdttdx224242 11usin?令 ??ud2cosi= cxcu??323)(os7. ???dxx12解. 令 tdtansec,sec???? ??????? ctdtttdxx sin)cos1(1122x??arcos2三. 求下列不定積分:41. ???dxex1243解. ?? ????????? ceedxe xxxx )arctn(1)(12222432. ??)1(xd解. 令 , t2?lntcttdtttddx ???????????????? 2lnarl112ln)1()41( 22= cxx??)artln四. 求下列不定積分:1. ??dx105)2(解. ?? ????? dxxxxd 949595105 )2()2()2(9)(= ??????dx 98398495 4)()2(= 96297398495 )(345)2(5)2(5)( ?????? xxxc??????? 9495 )(64)(67832. ??41xd解. ???? ?????? 24424 )(111/ tdtdtttx令 cxcuduut ??? 2422 ln|setan|lsecan令5五. 求下列不定積分:1. ?xd2cos解. ??????xdxd2sin41)2cos1(2?xsiin42cx8s2. ?xd3sec解. ????? xdxxtansecttansectasec= ???? xddx 32 sec|t|l)1(tncxxd?? |tasec|ltasec1sec33. x2)(ln解. ??? ?????dxxdd23323 )(ln)(l1)(ln)(l?x223l6l)(l ???dx23 6ln)(llcx???ln)(l)(ln234. ?dx)cos(l解. ?? ???? dxxxdx )cos(ln)]sin(l)[co(l)sin(l)co(ln? x?? ][2)cs(l5. ??? ?????? dxxxdxd 2sin81si2sin81cosin81sino 23434cx??????ot4i2i4s2六. 求下列不定積分:61. ???dxx2)1(ln解. ????222 1)ln(1)(l xd= ???xxx 222ln1ta?令 tdt222 sec1an1)(l ??= dtx???22sico)1(ln= ?t22in1)(l= ctx????sil4)1(ln2= x21ln)(l22. ??dx21arctn解. ?? ????? dxxxdx 2222 1arctn1arctnrt= c??? )l(rt1rt1223. ?dxe2arctn解. dxeedexxxxx? ?????? ?2222 1arctn1arctn1rtexxx???2rt1 xxxx )(rt22ceedee xxxxxx ????? ??? artnarctn(1)1(arctn2 227七. 設(shè) , 求 .??????xexf)32(1ln) 0???dxf)(解. ????dxdf )(l)(2?????????12 222)4( 3)]ln([1lncexcxx 0??考慮連續(xù)性, 所以c =－1+ c 1, c1 = 1 + c?dxf)(???????????excxx)4( 3)]1ln([2ln220??八. 設(shè) , (a, b 為不同時為零的常數(shù)), 求 f(x).aefxcossin)('解. 令 , , 所以ttl?， )cos(ln)in(l)(' ttaf???dxxxf ]cssi[)(= cbba?)]s(l)()l2九. 求下列不定積分:1. ??dxx)3(2解. ?????cxxx 3ln)(232222. ??d)1(53(232解. )523()523(???????xdxxxc22)(513. dxx??21)ln(8解. ?? ?????? cxxdxdxx )1(ln21)ln()1ln(1)ln( 22224. )1l()(22x解. cxxdxd ???????? |)1ln(|)1ln()ln()1( 2222十. 求下列不定積分:1. ?dx)(arct2解. ????????? 12222 )(arctn1)()1(arctn)1(rtn xdxd?? dxx 2222 )(tarctrttxtdxt?????? cos1arctn1os1rtna 222令cecttx??? in4ti841rc2xe221arnt2. ??dx1arcsin解. 令 tt2an,ri?則??? ????? ctttdttdtx ana1arcsin 2222xxcx???? 1rcsi)1(arcsinri3. ???dxx221acsin解. ?? ??????? dttdtttx )1(cscosin1sinri 22222令9??? ??????? ctdtdtt 21coco21|sin|lcxxx????2)(arsin|larci4. dx??)1(tn2解. ?? ??? dttdtttx )1(cssecan)(arct 2222令22ott1os ttdtdt ????cxxxctt ?????? 22 )(artn1||lnar1|in|lcocxx??22)(art1lart十一. 求下列不定積分:1. ?d234解. ????? dttdtttxx 233 cosincos2sin8i令dt ????? 532co)cs1(32xx??2523)4(42. ?a2解. ?????? dtadtattx 22 cos1nsecsec令xat ???rtn23. dxex??21)(解. udutdttdtxx cosin1si11)()( 222??????????令令10cecuxx??????21arsinos4. (a > 0)?dx2解. ?xu?令 ??dua24tasin2?令 ?td42si8= ? ?ttdta )cos1(24)cos1(8 22= ctattadtatt ?????? 4sinsi34cosin222= ctttta ?)in1(sinci43 222= a??coosn3= caxxaxa ???222rcsi32= ca???)(3rin2十二. 求下列不定積分:1. ??xdcos1sin解. ??????????? xdxdxd 222 cos1cos1sin)(co1insi?? )()(cos1 22uux令? ?????? cdu||ln1)2(xx?|cos12|lcos12. ??d2in解. ????xdxxcos2)(cos2cos11tx?2tan令 ?? ????? |cos2|ln32|cos|ln12 xtdxtdt= ct |s|l)2(ta1art4|s|l3rc43. ??dxosin解. ????dxxcosin121c= ? ???? dxdx cosin12)(osin)(s212= ??)4sin()c(si ?x= cx?|82ta|l)os(in21十三. 求下列不定積分:1. dx??1解. cttdttx ???????? 3332 141)(1令c??2342. ??dxe1解. ??? ??????? dtdtttedxexx )1(secan1sec12令ctt xxx ???ro)ln(|asc|ln23. dxx??ar1解. 令 tdxtxtt ansec2,sec,1an,rc 2????12???? ????? dttdtdtttdxx 222 cos1ananseca1arctn??22 tttos tttct ???2|s|lnacxxx ??2)1(artn|l1rct第三章 一元函數(shù)積分學(xué) (定積分)一．若 f(x)在[a，b] 上連續(xù), 證明: 對于任意選定的連續(xù)函數(shù)?(x), 均有 , 0)(???badxf則 f(x) ? 0.證明: 假設(shè) f(?)? 0, a 0. 因?yàn)?f(x)在[a，b] 上連續(xù), 所以存在? > 0, 使得在[? －?, ? + ?]上 f(x) > 0. 令 m = . 按以下方法定義[a，b] 上?(x): 在)(inxfx????[?－? , ? + ?]上 ?(x) = , 其它地方?(x) = 0. 所以22)(?x.02()( ??????????dxfdxfba和 矛盾. 所以 f(x) ? 0.0)(baf二. 設(shè)?為任意實(shí)數(shù), 證明: = .???20)(tan1??dxI 4)(cot120??????dx證明: 先證: =4)(cos)(sin20??xff 20)(s)(si?ff令 t = , 所以x????20)(cos)(sin?dxff ???02 )(sin)(co?tdftf= 20)(in)(?tftf 0)(i)(xfxf于是 ???20)(cos)(sin?dxfxf ??20)(cos)(sin?dxff ?20)(sin)(co?dxfxf13= 2)(cos)(sin020 ?? ????dxfxf所以 = .4)()(i20?ff ?0)(cos)(sindxfxf所以 ???20)(tan1??dxI 4)(in)(cosi12020 ?????? ???????????dx同理 .4)(cot120????dxI三．已知 f(x)在 [0，1]上連續(xù) , 對任意 x, y 都有|f(x) －f(y)| 0, (0 0, 證明: 對于滿足 0 0因?yàn)?f(0) = f(1) = 0?x0 ? (0,1)使 f(x0) = (f(x))1max?所以 > （1）df?10)(' dff?0)(')(在（0，x 0）上用拉格朗日定理0()'ff）??),(0x??16在（x 0, 1）上用拉格朗日定理0)()('xff???)1,(0x??所以 )(4)1( )('')(''' 0010xfxf ffdxfdd??? ????????????（因?yàn)?）ab?2所以 ??104)(')(dxfxf由（1）得??0)('xf十. 設(shè) f(x)在[0, 1]上有一階連續(xù)導(dǎo)數(shù) , 且 f(1)－f(0) = 1, 試證:1)]('[102??dxf證明: ??102)]('[xf 1)0(1()('' 221002 ?????fdxf十一. 設(shè)函數(shù) f(x)在[0, 2]上連續(xù), 且 = 0, = a > 0. 證明: ? ? ? [0, 2], 使20)(xf20)(f|f(?)| ? a.解. 因?yàn)?f(x)在 [0, 2]上連續(xù), 所以|f(x)|在[0, 2]上連續(xù), 所以? ? ? [0, 2], 取? 使|f(?)| = max |f(x)| (0 ? x ? 2)使|f(? )| ? |f(x)|. 所以|)(||1|)(||)(|1|)(1| 202020 fdxfdxfxdfa ??????? ???第三章 一元函數(shù)積分學(xué) (廣義積分)一. 計(jì)算下列廣義積分:(1) (2) (3) ??2031)(dxex???02)4(1dxx????23)1(xd(4) (5) (6) 10sinl?2 ?023)(arctn解. 17(1) 3223102031 )1()(lim)( ?????????edxedxexx?(2) 24li)4(02202 ????????????? bdx(3) ????23)1(xd因?yàn)?, 所以 積分收斂.所以1)(lim233??x ???023)(xd=2 ????23)1(dtxdan)(023???令 2cossec2003?????tdt(4) 21)s(ln)i(l1lim)si(ll)sin(l 0100 ?????????? ??? xxx(5) 3tansec13212 ?????ddx(6) 12cos)(arctn20203032 ???????tt