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1、精品 第五章 习题 5-1 1.求下列不定积分：(1)2(5)x x dx;(2)2(1)xxdx;(3)3 exxdx；(4)2cos2xdx；(5)2 35 23xxx dx；(6)22cos2dcossinxxxx.解 5151732222222210(1)(5)(5)573ddddx xxxxxxxxxxC 113222221132223522(1)1 2(2)(2)242235ddddddxxxxxxxxxxxxxxxxxxxxC 213(3)3(3)(3)ln(3)1 ln31 cos1111(4)coscossin2222222 35 222(5)25()25()333125 2
2、25()223(ln2ln3)3ln()3eede deedddddddd xxxxxxxxxxxxxxxxCCxxxxxx xxxCxxxxxCxC 2222222222cos2cossin(6)(cscsec)cossincossincscseccottanddd ddxxxxxxxxxxxxx xx xxxC 2.解答下列各题：(1)一平面曲线经过点(1，0)，且曲线上任一点(x,y)处的切线斜率为 2x-2,求该曲线方程；(2)设 sinx为f(x)的一个原函数，求()fxdx；(3)已知f(x)的导数是 sinx，求f(x)的一个原函数；(4)某商品的需求量Q是价格P的函数，该商品的
3、最大需求量为 1000(即P=0 时，Q=1000),精品 已知需求量的变化率(边际需求)为Q(P)=-10001()3Pln3,求需求量与价格的函数关系.解 (1)设所求曲线方程为y=f(x),由题设有f(x)=2x-2,2()(22)2df xxxxxC 又曲线过点(1,0),故f(1)=0 代入上式有 1-2+C=0 得C=1,所以,所求曲线方程为 2()21f xxx.(2)由题意有(sin)()xf x,即()cosf xx,故 ()sinfxx,所以 ()sinsincosdddfxxx xx xxC.(3)由题意有()sinfxx,则1()sincosdf xx xxC 于是 1
4、2()(cos)sinddf xxxCxxC xC.其中12,C C为任意常数,取120CC,得()f x的一个原函数为sin x.注意 此题答案不唯一.如若取121,0CC得()f x的一个原函数为sin xx.(4)由1()1000()ln33PQ P 得 111()1000()ln31000 ln3()1000().333ddPPPQ PxxC 将P=0 时,Q=1000 代入上式得C=0 所以需求量与价格的函数关系是1()1000()3PQ P.习题 5-2 1.在下列各式等号右端的空白处填入适当的系数，使等式成立：(1)dx=d(ax+b)(a0)；(2)dx=d(7x-3)；(3)
5、xdx=d(52x)；(4)xdx=d(1-2x)；(5)3xdx=d(3x4-2);(6)2exdx=d(2ex)；(7)2exdx=d(1+2ex);(8)dxx=d(5lnx);(9)2d1xx=d(1-arcsinx);(10)2d1x xx=d21x;精品(11)2d1 9xx=d(arctan3x);(12)2d12xx=d(arctan2x);(13)(32x-2)dx=d(2x-3x)；(14)cos(23x-1)dx=dsin(23x-1).解 1(1)()(0)()ddddaxba x axaxba 22224334222221(2)(73)7(73)71(3)(5)10(
6、5)101(4)(1)2(1)21(5)(32)12(32)121(6)()2()2(7)(1)dd dddd dddd dddd dddeed eddedeexxxxxxxxxxxx xx xxxx xx xxxxxxxxxx 222222222221()2(1)251(8)(5ln)(5ln)51(9)(1 arcsin)(1 arcsin)111(10)1(2)12 1113(11)(arctan3)1 9d e ddeddd dddd dddd dddd xxxxxxxxxxxxxxxxxxxxxxxxxxxxxx 222322231(arctan3)1 9321(12)(arctan
7、2)(arctan2)12122(13)(2)(23)(32)(32)(2)222232(14)sin(1)cos(1)cos(1)sin(1)333323ddddd ddd dddd dxxxxxxxxxxxxxxxxxxxxxxxx 2.求下列不定积分：(1)5e dtt;(2)3(32)xdx;(3)d1 2xx;(4)3d23xx;(5)sindttt;(6)dlnlnlnxxxx;(7)102tansecdxx x;(8)2edxxx;精品(9)dsincosxxx;(10)22dtan 11x xxx;(11)deexxx;(12)2d23xxx;(13)343d1xxx;(14)
8、3sindcosxxx;(15)21d94xxx;(16)32d9xxx;(17)2d21xx;(18)d(1)(2)xxx;(19 2cos()dtt);(20)2cos()sin()dttt;(21)sin2 cos3 dxx x;(22)cos cosd2xxx;(23)sin5 sin7 dxx x;(24)3tansec dxx x;(25)arctand(1)xxxx;(26)22d(arcsin)1xxx;(27)lntandcos sinxxxx;(28)21 lnd(ln)xxxx;(29)222d,0 xx aax;(30)23d(1)xx;(31)29dxxx;(32)2
9、d1xxx;(33)2d11xx;(34)d,0axx aax.解 5555111(1)5(5)555edededetttttttC 33411(2)(32)(32)(32)(32)28ddxxxxx 精品 1223333111(3)(1 2)ln1 21 22 1 2211 31(4)(23)(23)()(23)(23)33 2223sin1(5)2 sin2 sin()2cos2111(6)(lnln)lnlnlln lnlnlnlnlnlnlnddddddddddxxCxxxxxxxCxCxtttttttCttxxxxxxx xxx 222210210112n1(7)tansectan(
10、tan)tan11111(8)(2)222(9)22csc22sincos2sincossin2lnlncsc2cot2tansin cdde dede d(-edddd d 或xxxxCxxx xxxxCxxxxxCxxxxxxxxxxCCxxxxx 2cos1tanlntanossincostanddxxxCxxxxx 222222222222234(10)tan 1tan 11lncos 111(11)()arctan11()161(23)(12)6323232 231123233333(13)14dddeddeee eeeddd ddxxxxxxxx xxxxCxxxxCxxxxxxx
11、xxxxCxxx 344443233222224313(1)ln114 14sinsin1(14)coscoscoscoscos211(15)94949411211(8)2382941()31121dddddddd dd xxxCxxxxxxxxxxCxxxxxxxxxxxxxxx 1222221(94)(94)382()3d()d xxxx 2121arcsin94234xxC 精品 3322222222999(16)()9999119(9)ln(9)29221111111(17)()21222212121211111(21)(21)2 2212 221xxxxxxxxxxxxx xxxxC
12、xxxxxxxxxxxxxxddd dddddd dd 211121lnlnln21212 22 22 2211111111(18)()(2)(1)(1)(2)32132311112lnlnln2133311 cos(22)11(19)cos()cos(22224xCCxxxxxxxxxxxxxxCCxxxtttttt dddd ddd223)(2)11cos(22)(22)2411sin(22)241(20)cos()sin()cos()cos()1cos()3(21)sin2 cos3ttttttCttttttCxx d d dd 111(sin5sin)sin55sin210211cos
13、5cos10213133(22)cos cos(coscos)cos()cos()22223222213sinsin3221(23)sin5 sin7(cos12xxxxxxx xxxCxxxxxxxxxxxxCxx x dddd dddd d2cos2)11cos12(12)cos2(2)24411sin12sin2244xxxxxxxxxC d dd 精品 32232222(24)tansectan(sec)(sec1)sec1secsec3arctan1(25)2 arctan2 arctan(arctan)(1)2(1)(arctan)1(26)(arcsin)(arcsin)1ddd
14、 ddd ddxx xxxxxxxCxxxxxxxxxxxCxxxx1(arcsin)arcsinxCx 22222222222222222lntan1(27)lntanseccos sintan1lntan(lntan)(lntan)21 ln111(28)(1 ln)(ln)(ln)ln(ln)ln(29)()dd ddddddddxxxx xxxxxxxCxxxxxxCxxxxxxxxxxaxaxxaaxaxaxax x 利用教材5.2 例 16 及公式(20)可得:原式=222222211arcsinarcsinarcsin2222xaxaxax axCx axCaaa.(30)令ta
15、n,(,)2 2xt t,则2secddxt t.所以223231seccossinsec(1)(tan1)ddddxtt tt ttCtxt 22tan,sin11 原式xxxttCxx.(31)令3sec,(0,)2xt t,可求得被积函数在x3 上的不定积分,此时 23sectan93tanddxtt txt 故 22293tan3sectan3 tan3(sec1)3secddddxtxtt tt tttxt 3tan3ttC.由3sec,(0,)2xt t得29tan3xt,又由3secxt得33sec,cos,arccos3xtttxx,精品 22293393arccos93arc
16、cos)dxxxCxCxxx 又令x=3sect,类似地可得被积函数在x-3 上的不定积分.22211293393arccos93(arccos)393arccosd xxxCxCxxxxCx 综上所述有 229393arccosdxxxCxx.(32)令sin,(,)2 2xt t,则cosddxt t.2211cossincossincossincos2sincos111111(sincos)lnsincos22sincos2211arcsinln.122dddddxttttt ttttttxxttttCttttxCxx(33)令sin,(,)2 2xt t,则cos,ddxt t 222
17、cos1(1)sec()1cos1cos221111tanarcsin.2dddd xttttttttxtxtCxCx (34)222222211()2ddddaxxxxaxxaaxaxaxax 22arcsinxaaxCa.习题 5-3 1.求下列不定积分：(1)sin dxx x;(2)e dxxx;(3)arcsin dx x;(4)ecos dxx x;(5)2esin d2xxx;(6)2tandxx x;精品(7)2edttt;(8)2(arcsin)dxx;(9)2e sindxx x;(10)3e dxx;(11)cos(ln)dxx;(12)2(1)sin 2 dxx x;(
18、13)ln(1)dxxx;(14)22cosd2xxx;(15)32lndxxx;(16)sincos dxxx x.解 (1)sincoscoscoscossindddxx xxxxxx xxxxC (2)()(1)eddeeedeed eeexxxxxxxxxxxxxxxxxCxC 2222111(3)arcsinarcsinarcsin(1)211arcsin1ddd x xxxxxxxxxxxxxC(4)coscoscos(sin)cossincossincoseddeeedede eeedxxxxxxxxxx xxxxxxxxxx x 12cos(sincos)(sincos)cos
19、2edeeedxxxxx xxxCxxx xC 22221111(5)sinsinsincos22222222eddeeedxxxxxxxxxx 2222222211sincos22821111sincos(sin)2282822111sincossin2282162edeeeedeeedxxxxxxxxxxxxxxxxxx 2221221711sinsincos16222822sin(cos4sin)21722edeeedexxxxxxxxxCxxxxC 精品 222222222222221(6)tan(sec)sec211(tan)tantan221tanlncos2111(7)22211
20、11(2)2424ddd dd e ddeee d ee detttttttxx xxxxxxx xxxxxxxx xxxxxCxtttttttt 2222222222221(8)(arcsin)(arcsin)2arcsin1(arcsin)2 arcsin11(arcsin)2 1arcsin211(arcsin)2 1arcsin2edd d d d tCxxxxxxxxxxxxxxxxxxxxxxxx222(arcsin)2 1arcsin21 cos211(9)sincos222211cos222 edededed eedxxxxxxxxxxxCxx xxxx xx x 而 cos2
21、cos2cos22sin2cos22 sin2edde eededexxxxxxx xxxx xxx cos22sin24cos2eeedxxxxxx x 11cos2(cos22sin2),511111(cos22sin2)(sin2cos2).2102510ede原式eeexxxxxx xxxCxxCxxC(10)令3xt,则32,3ddxtxtt 3322222223233336363663663(22)3(22)e de ddeee dedeeee deeeeextttttttttttttxxtttttttttttttCttCxxC(11)令 lnx=t,则,ede dttxxt,精品
22、 cos(ln)coscos dee cose sine cossinee cose sine coscos(ln)sin(ln)cos(ln)cos(ln)cos(ln)sin(ln)2dedd dd ddtttttttttxxt tttt tttttt txxxxxxxxxxxC 22222211(12)(1)sin2sin2sin2cos2sin2(2)2211cos2cos2cos222111cos2cos2sin222211cos2cos2sin222ddddd d d xx xxx xx xxxxxxxxx xxxxxxxxxxx 2212sin22111cos2cos2sin2c
23、os2222413()cos2sin2222d xx xxxxxxxCxxxxC 2222222221(13)ln(1)ln(1)()ln(1)222111 1111ln(1)ln(1)(1)22122211 11ln(1)()ln122 221(1)ln(1)2ddd ddd xxxxxxxxxxxxxxxxxxxxxxxxxCxxx 211.42xxC 2222232321cos11(14)coscos22221111sinsinsin6262dddd ddxxxxxxxxxx xxxxxxxxx x 3232321111sincossincoscos626211sincossin.62d
24、dxxxxxxxxxxx xxxxxxxC 精品 333222323223232232ln111(15)ln()ln3ln11131ln3 ln()lnln6ln131lnln6 ln()1361lnlnln613lnlnddd dd d d xxxxx xxxxxxxxxx xxxxxxxxxxxxxxxxxxxxxxx 3266ln1(ln3ln6ln6)xxCxxxxxCx 11(16)sincossin2cos22411cos2cos2cos2cos2244481cos2sin248ddd dd xxx xxx xxxxxxx xxxxxxxC 习题 5-4 求下列不定积分：(1)31
25、d1xx;()5438dxxxxx;()sind1 sinxxx;(4)cotdsincos1xxxx.解 (1)令 322111(1)(1)11ABxCxxxxxxx 则 2331()()()11AB xBCA xACxx 从而 001ABBCAAC 解得 131323ABC 于是 精品 2322222123(1)3(1)1112111331612()2411121lnlnarctan1136331(1)121lnarctan6133dddd d xxxxxxxxxxxxxxxxCxxxxxCxx 542233323323288(2)(1)11832111111ln8()13221218ln3ln4ln1132dd dd d xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxCxxx 222sinsin(1 sin)1(3)cos(sec1)1 sincoscos1tansectancosdddd xxxxxxxxxxxxxCxxxCx 注 本题亦可用万能代换法(4)令tan2xt,则 222222112sin,cos,cot,2arctan,1121ddtttxxxxtxttttt 则 222221cot211 11221sincos112221111111lnlntantan222222ddddd txttxttttttxxtttttxxtCCt