第四章 不定积分习题详细解答20110919
习 题 4-1
,1.已知之一的原函数为,求. fxx()dfx()sin3x,
,, 解:. fxxfxCxCxC()d()(sin3)3cos3,,,,,,,
2,2( 设,求. fx()[ln()]secfxx,
2tanx, 解: 因为,故 (为任意常数), ln()tanfxxC,,[ln()]secfxx,CfxCe(),11
2,x3.若是的原函数,求. xfxx(ln)dfx()e,
2x1,,xx,lnx2,解: = fx()fxe,,,,(),ee,,xfxxC(ln)d,,,(ln),,x2
fx(ln),x4. 若是的原函数,求. fx()edx,x
1,x,x,lnx,,所以, 解:因为 fxe(),fxeC(),,,fxeC,,,,,(ln),00x
Cfx(ln)1fx(ln)10,. ,,,dln||xCxC,,,02,xxxxx
5.求下列不定积分:
135,1222(1)解: ,,,,,,(5)d(5)d22xxxxxxxxC,,x
xxx,4269xx2(2)解:(23)d,x ,,,,C,2ln2ln62ln32111,,xx(3). d[]darctanln||xxxxC,,,,,,,22xxxx(1)1,,
1122(4) 解: (cot)dd(csc1)d,,,,xxxxx,,,22xx,,11
= arcsincotxxxC,,,
x80xx3xxx(5) 解:102dx ,,,,xxC108d80d,,,ln80x1112(6) 解:= sindx,,,,,(1cos)dsinxxxxC,,2222
22cos2xcossinxx,(7) dx,,,,,,,d(cossin)dsincosxxxxxxC,,,cossinxx,cossinxx,
22cos2xcossin11xx,(8) 解: dx,,,d()dxx22,,,2222cossinxxcossinsincosxxxx
,,,,cottanxxC
2sec(sectan)dsecdsectandxxxxxxxxx,,,,(9) 解: tansecxxC,,,,,
,,,xx,1,,(10) 解:设fxx()max{1,},,. 则fxx()1,11,,,,,
,xx,1,,
fx()(,)在上连续,,,,,
1
1,2,,,,xCx,11,2,, 则必存在原函数Fx()又须处处连续,有Fx()FxxCx(),11,,,,,,2,12,xCx,,,132,.112 , xCxC,,,,即,,,,,1,CClim()lim()2121,,xx,,,,1122
112 , xCxC,,,即,,,CC1,lim()lim()3232,,xx,,1122
1 联立并令CC,,可得,CCCC,,,,1.1232
1,2,,,,xCx,1,2,1,故max{1,},11.xdxxCx,,,,,, ,,2,1,2xCx,,,1,1,2,
dy36. 解:设所求曲线方程为,其上任一点处切线的斜率为,从而 yfx,()(,)xy,xdx
134 yxxxC,,,d,4由,得,因此所求曲线方程为 y(0)0,C,0
14 . yx,4
7.解:因为
,,11,,,,22, ,,coscossinxxxsinsincosxxx,,,,,22,,,,
,11,, ,,,cos2sin2sincosxxxx,,42,,
11122所以、 、 都是的原函数. ,cos2xsincosxxsinx,cosx422
习 题 4-2
1.填空.
111(1) = ( + C) (2) = (+ C) ,dxdxddlnx2xxx
xx2(3) = (+ C) (4) = (+ C) tanxexddesecdxxd(5) = (+ C) (6) = (+ C) ,cosxsindxxdcosdxxdsinx
2
x12(7) = (+ C) (8) = (+ C) dxdxdarcsinxd1,x221,x1,x
1 = (+ C) (10) = (+ C) (9)secxtansecdxxxddarctanxdx2x,1
21x(11) = (2+ C) (12) = (+ C) darctanxxxdddx2(1)xx,
2.求下列不定积分:
5x1e55xx(1) exexCdd5,,,,,55
7792222(2) (2)d(2)d(2)(2),,,,,,,,,xxxxxC,,9
d1d(13)1xx,(3) ,,,,,,ln|13|xC,,133133,,xx
xxeed(3),x(4) dln(3)xeC,,,,,,xxee,,33
1123xxx2334xxxxxx(5) (22)deeex,, ,,,,,,,eeeeeeC(22)d()2,,32
26d(3)xx,2(6) d33ln(3)xxC,,,,,,22xx,,33
12,x141x,222 (7) dx,,,,d()(4)d(4)xx,,,2222x,4x,4
1222 ,(4)4xCxC,,,,,
1134444(8) xxxxxxCcosdcosdsin,,,,,44
45lnxlnx4 (9) dx,,,lnd(ln)xxC,,x5
111x1exx (10) ,,,,,eeCdxd(),,2xx
,3xe22,,33xx (11) ,,,,,, dd(3)xexeC,,33x
arctanxarctanarctanxx(12) dx,,2d2d()xx,,,21,xxx(1),1(),x
2,,,2arctand(arctan)(arctan)xxxC ,
3xd()dxd113xx2 (13) ,,,,arcsinC,,,233233xx49,x2221()1(),,22
3
2arcsinxarcsinx (14) dx,,,arcsind(arcsin)xxC,,221,x
arccosxarccosx1010arccosx(15) ,,,,10d(arccos)xx,Cd,,2ln10,x1
2xxsin1,2222(16) tan1dtan1d1d1xxxxx,,,,,,,,,22xx,,1cos1
2dcos1x,2 ,,,,,,ln|cos1|xC,2cos1x,
1334(17) cossindsindsinsinxxxxxxC,,,,,4
1sinx(18) dx,,,,,d(cos)2cosxxC,,cosxcosx
2sincosxx,13 (19) dx,,,,,d(sincos)2(sincos)xxxxC,,33sincosxx,sincosxx,
1ln,x11(20) dx,,,,d(ln)xxC22,,(ln)xx(ln)lnxxxx
111(21) dx,,,,d(ln)d(lnln)lnlnlnxxxC,,,xxxlnlnlnlnlnlnlnlnxxx
21cos212cos2cos2,,,xxx42 (22) cosdxx ,,()ddxx,,,24
2xx,sin21cos4,x1cos2cos2xx ,,,,,()dxdx,,44242
3sin2xx,sin4x ,C,,44
3sinx322(23) cosdxx,coscosdxxx,,(1sin)d(sin)xx ,,,sinxC,,,3
352525sincosdxxx,,,,sincosdcos(1cos)cosdcosxxxxxx(24) ,,,
1186 ,,,coscosxxC86
352424tansecdxxxtansecdsec(sec1)secdsecxxxxxx,,(25) ,,,
1175 ,,,secsecxxxC75
sin9sin11xx,(26) cos5sin4ddcos9cosxxxxxxC,,,,,,,2182
343232tansecdxxx,,,tansecdtantan(tan1)dtanxxxxxx(27) ,,,
4
1165 ,,,tantanxxxC64
,, (28) 设xtt,,tan||)(2
22dsecdcosdxttttdsin11tx, ,,,,,,,,,CC,,,,22222tansecsintttsinsinttxxx,1
, (29) 设, xtt,,sin||)(2
22xxtttdsincosd,2 ,,sindtt,,,2cost1,x
112 ,,,,,,,(sincos)(arcsin1)ttttCxxxC22
,(30) 令,,,则 t,[0,]xt,secdsectandxttt,2
111= dsectanddxtttttC,,,,arccos,C,,,2xsectanttxx,1
, (31) 令,,,则 t,[0,]xt,4secd4sectandxttt,2
2xt,164tan2 d4sectand4(sec1)d4(tan)xtttttttC,,,,,,,,,,xt4sec
2,,x,16442 ,,,,,,,4arccos164arccosCxC,,1,,3xx,,
1d()x,dd111xx2(32) ,,,,,arctan()xC22,,,1445(21)4442xxx,,,,2()1x,,2
32d(217)2dxxx,,,31x,2(33) dx,,,22xxxx,,,,217217
23d(217)d311xxxx,,,2 ,,,,,,,2ln(217)arctanxxC,,2222217(1)4224xxx,,,,
dd11114xxx,(34) ,,,,,()dlnxC2,,,xxxxxxx,,,,,,,34(4)(1)54151
2xxxx,,,11d(56)7d(35) dx,,,,,22xxxxxx,,,,,,562562(2)(3)1711172x,,,22 ,,,,,,,,,,ln(56)dln(56)lnxxxxxC,,,2223223xxx,,,,,32xx114,,22 (36) dd1dxxx,,,,,,,,222xxx,,,42424,,
5
21d(4)1x,222 ,,,,,,d22ln(4)xxxC,,2242x,
22xxxx,arctanarctan(37) dddxxx,,,,,222xxx,,,111
1123 ,,,,ln(1)arctanxxC23
11xx(38) x,,,eeCdd()arctanxx,2,,x,eee,1
2xxx(11),,22(39) dd(1)dxxxxxx,,,,,,,22211(11)(11),,,,,,xxx
33322xxx1(1),2222 ,,,,,,,,,,xxxxxxxCd1d1d(1),,,3233
2axaxaxxx,,,()d(40) ddddxxxax,,,,,,,,,22222222axax,,axaxax,,,
22xaxx1d(),22 ,,,,,,aaaxCarcsinarcsin,22aa2ax,
11111x (41) dd()d()x,,,,,,,22xx1122xx1,x()1()1,,xx
211,x1122 ,,2()1,,2C,,2d(()1),xxx12()1,x
2sin111x,,(42) d1dddxxxx,,,,,,,,,,,22211sin1sinsin,,xxx,,1,2sinx
cotxd()dcot11cotxx,,2 ,,,,,,,xxxCarctan,,22,,2cot,x222,,cotx,,1,,,2,,
习 题 4-3
求下列不定积分
1x1xxxsin2d(1) ,,xxd(cos2),,,cos2cos2dxxx,,,222
x1 ,,,,cos2sin2xxC24
,x,,,,,xxxxxxexd,,,,,,,,,xexeexxeeCdd(2) ,,,
6
3333233xxxxxxx2(3) xxxlnd,,,,,lnd()lnd(ln)lndxxxxx,,,lnxC,,,,3933333
(4)略.
22222(5) xxxcosd,,,,,xxxxxxxxxxxdsinsinsindsin2sind,,,,
22 ,,,,,xxxxxxxxxxsin2dcossin2cos2cosd,,
2 ,,,,xxxxxCsin2cos2sin
(6) arctan31darctan31darctan31xxxxxx,,,,,,,
1d1x ,,,,,,,,xxxxxCarctan31arctan3131,2331x,,x,x,,xx(7)exxsin2d,,sin2dxe,,,exexsin2d(sin2) ,,,,,xx,,,xxx,,,exxesin22cos2d() ,,,,exexexsin22cos22d(cos2) ,,
,,,xxx,,,,exexexxsin22cos24sin2d ,
,,xx,,exexsin22cos2,xexxsin2d ,,C,5
333xxx2(8)xxxarctand ,,,arctandarctandarctanxxx,,,3333333xx1xxxx1,, ,,arctandxx,,arctandxx,,22331,x331,x3x122 ,,,,,arctanln(1)xxxC33
21cos21,xx12(9xxxcosd) ,,xxxcos2d,,,xxxxxxd(cos2)d,,,,4222
22x1x11 ,,xxdsin2,,,xxxxsin2sin2d,,44444
2x11 ,,,,xxxCsin2cos2448
1,,,2arcsind2arcsin2darcsinxxxxxx(10) arcsindxx,,,x
1 ,,,,2arcsin21xxxC,,2arcsindxxx,1,x
2323xx122xexe23x2333xxxxexd(11) ,,,,,xexexxeddd,,,,33339
23xxe2233xx ,,,,xeeC3927
coslndxx,,,,xxxxxxxxcoslndcoslncoslnsinlnd(12)因为 ,,,
,,,xxxxxxcoslnsinlndsinln ,
7
,,,xxxxxxcoslnsinlncoslnd,
xxxxcoslnsinln,于是 coslndxx,,C,2
xxxxcos1xxd,, (13) ,,,ddsindxx,,,,,333222,,,,sinsinsin2sinxxxx2sinsinxx,,
x1122 ,,,,,,,cscd(csccot)xxxxxC,22sin22x
xln(1)e,,,,xxxxxx(14) dln(1)dln(1)dln(1)xeeeeee,,,,,,,,,,,xe
xexd,,,xxxxx ,,,,,,,,eeexeeln(1)dln(1),,xx,,ee11
xxee,,1,xx ,,,,eexln(1)d,xe,1
,xxx ,,,,,,,eexeCln(1)ln(1)ln(1)1ln(1)dxxx,,,,(15) dln(1)dxx,,,,,,2,,,(2)22(2)(1),,,,,xxxxx,,
ln(1)dln(1)111xxx,,,, ,,,,,dx,,,,2(2)(1)2321,,,,,,xxxxxx,,
ln(1)11xx,, ,,,lnC232,,xx
,,,,,,xfxx()d,,,,,,xfxxfxfxxxfxfxCd()()()d()()(16) ,,,
习 题 4-4
求下列不定积分
33xx,,1112(1) dx,,,,,d(1)ddxxxxx,,,,x,1xx,,11
32xx ,,,,,,xxCln132
542xx,,8xx,,82(2) dx,,,,(1)ddxxxx,,,33xx,xx,
8432 ,,,,,,(1)d()dxxxx,,xxx,,11
32xx ,,,,,,,,,xxxxC8ln4ln13ln132
2,,,,xx2342213xx,,1(3) ,dxdx,,ddxx222,,,,22x,2(2)(1)xx,,xx,,1(1)
221d(1)13d(1)4xx,, ,,,,,,ln22ddxxx,,,,2222222112(1)(1)xxxx,,,,
132x2 ,,,,,,,,,ln2ln(1)2arctan2arctanxxxxC2222(1)1xx,,
(上式最后一个积分用积分
公式28)
8
24216114xx,,(4) dx,,,[]dx2,,2xx(1),xxx,,1(1)
112 ,,,,2ln(1)xxC,,,,,4ln2ln1xxCx,1x,1
xx1d11xx,(5) ,dxdx,,dx3222,,,,(1)(1)xx,,xxx,,,12121xx,,
1112 ,,,,,,ln1ln(1)arctanxxxC242
1dudx2dxdu(6) ux,tan,,22,,,,237cos2,x3sin,x34,u21(),u3
12tanx ,,arctanC233
x(7) 令 ,可得 u,tan2
2du2ddxux1,u ,,,,,,,,ln1ln1tanuCC,,,221uu,sincos112xxu,,,,,12211,,uu
2dx133dtt23(8)tx,,1 ,,,3(1)dtt,,,,,tttCln1,,,31,t1,t211,,x
24t1121,x1,x(9) dxt,dt,,,()dt2,,,221,x(1)(1)tt,,ttt,,,111xx1,
t,1 ,,,ln2arctantCt,1
题 4-5 习 利用积分表计算下列不定积分:
dxd(2)x,(1)因为 ,,,2254,,xx1(2),,x
在积分表中查得公式(73)
dx22 ,,,,ln()xxaC,22xa,
现在,,于是 a,1xx,,2
dx2 ,,,,,,ln(542)xxxC,254,,xx(2)在积分表中查得公式(135)
nnn,1lnd(ln)lndxxxxnxx,, ,,
现在,重复利用此公式三次,得 n,3
332lndxx. ,,,,,xxxxxxxCln3ln6ln6,
(3)在积分表中查得公式(28)
9
11dxx dx,,2222,,()2()2baxbaxbbaxb,,,
于是现在,,于是 a,1b,1
xxx1d1 ,,,,arctanxCdx,22222,,2(1)212(1)xxx,,,(1),x
(4)在积分表中查得公式(51)
11a darccosxC,,,2axxxa,于是现在,于是 a,1
dx1 ,,arccosC,2xxx,1(5)令,因为 tx,,1
222222 xxxx,2d,,,xxx(1)1d,,,,(21)1dtttt,,,
由积分表中公式(56)、(55)、(54)
2xa222222222 xxaxxaxaxxaC,,,,,,,,d(2)ln,88
122223 xxaxxaC,,,,d(),3
2xa222222 xaxxaxxaC,,,,,,,dln,22
x,1222222xxxx,2d于是 ,,,,,[2(1))(1)xaxa,8
251a22223. ,,,,,,,,,ln1(1)[(1)]xxaxaC83
(6)在积分表中查得公式(16)、(15)
ddxaxbax, ,,,,,2bxb2xaxbxaxb,,
d2xaxb, ,,arctanC,,bxaxbb,,于是现在,,于是 a,2b,,1
dx21dxx,21x, ,,,,,,2arctan21xC,,2xxxx21,xx21,(7) 在积分表中查得公式(135)
11n,nnn,,12 cosdcossincosdxxxxxx,,,,nn
现在,重复利用此公式三次,得 n,6
15151x653cosdxx. ,,,,,•xxxCcossincossin(sin2)xx,6242442(8)在积分表中查得公式(128)
10
1axax ebxxeabxbbxC,,,sind(sincos)22,ab,
现在,,于是 a,,2b,3
1,2xax exxexxC,,,,sin3d(2sin33cos3),13
1ax . ,,,,exxC(2sin33cos3)13
本章复习题 A 一、填空.
2sinxsinx2(1)已知是的一个原函数,则 = . Fx()d(())Fx2dxxx
2,(2)已知函数的导数为,且时,则此函数为 . yfx,()yx,2y,2yx,,1x,1
xxxx(3fxxxxC()dsin,,,efex(1)d,)已知,则=. sin(1)1eeC,,,,,,
2)如果,则=. (4fx()2x fxxxxC(sin)cosdsin,,,
,,xx,xe(e)dfx(5)设是的一个原函数,则= Fx()fx(),,FC(e),
,fx(ln)(6) =. fxC(ln),dx,x
21222,,,xfxfxx()()d(7) =. fxC(),,,,4
22xfxx()d(8) 设是的一个原函数,则=. fx()cscxxxxCcsccot,,,
二、 选择题 1(, 2(, 3(, 4(, 5(, 6 C 三、求下列不定积分.
2221cos,x1cos,x11cos,x2,,(1sec)dxx(1) dx,dx,dx,,,,221cos2,x2cosx12cos1,,x
,,,xxCtan
,,xxdxexedd(1),x(2) ,,,,ln(1)eC,,,x,,,,,xx1,e11,,ee
3xx2()xx,x,,,2352312x4(3) x,,xx,,,Cd2()d5()d5,,,x,ln3ln4ln2442
x22(arcsin)dxx(4),,,xxxxarcsin2arcsind ,,21,x
22222,,,xxxxarcsin2arcsind1,,,,,xxxxxxarcsin21arcsin21darcsin ,,
22 ,,,,,xxxxxCarcsin21arcsin2
11
2(5)令,则,于是 tx,,1xt,,1
dx2d2d111tttt, ,,,,,,()dlntC22,,,,(1)1111tttttt,,,,,xx,1
3xxxxx(6) ,,,,[]dddxxxdx222222,,,,22(1),x1(1)1(1),,,,xxxx
112 ,,,,ln(1)xC222(1),x
1dx,2(7) ,,,,(arcsin)d(arcsin)xxC,,22arcsinx(arcsin)1xx,
1,x1x(8) ,dxddxx,,,,22294,x9494,,xx
3
121122 ,,,d()d(94)xx,,2338294,x21(),x3
121x2 ,,,,arcsin94xC234
5443223(9)tansecdxxx,,tansecdsecxxx(sec1)secdsecxxx, ,,,
864secsecsecxxx753,,,(sec2secsec)dsecxxxx ,,,,C,834
ππ(10)令,,于是 xt,sint,,(,)22
td()dxcosd1cos1dtttt,,2 ,,,,,,dttt,,,,,2t1cos1cos1cos,,,ttt211,,xcos2
tt2sinsin2tx11,,22 ,,,,,,,,,tCxCxCtanarcsinarcsintt2x2cossin22
222222111113x2222xxxxxxexd) (11,,,,,,xexeexxeeCdd,,,22222
lnlnxlnlndlnlnlnxxxC,,(12) dx,,x
67dd1dxxxx(13) ,,,,,77777xxxxxx(7)(7)7(7),,,
71111x,,7 ,,,,dlnxC,,,777147147xxx,,,,
12
x,1d()dd111xxx,22(14) ,,,,arctanC2,,,22x,1xx,,292(1)(22)222222x,,()1,22
1cosd(sin),,xxx(15) dlnsinxxxC,,,,,,xxxx,,sinsin2xxln(1)11,2222(16) dln(1)dln(1)ln(1)xxxxC,,,,,,,,,2x,124
2t2x(17) 令 , ,, xt,,ln(1)et,,1ddxt,2t,1
xd1211111xtte,,,,, ,,,,,,,,ddlnlnttCC,,2,,,xxttttt,,,,1111,,ee,,,111
2xxee1,,xxxx(18) dd1deln(1e)xeeC,,,,,,,,,,,,xxx1e1e1e,,,,,
xxln,,d,,ln1d(ln)11lnxxxxx,3,,(19) darctanxC,,,,222,,,3(ln)3(ln),,xxxx333xxln,,1,,,3,,
xxxxxxxxx,,sindsinddd(1cos)(20) dx,,,,,,,,,x1cos1cos1cos1cos,,,,xxxx22cos2
xxx ,,,,,,,xxxxxdtanln(1cos)tantandln(1cos),,222
xx ,,,,,xxCtan2lncosln(1cos)122
xxxx2 ,,,,,,xCxCtan2lncosln2costan12222
1,x,0,,fxx()d三、设,求 . fxxx()1,01,,,,,,,x,12,x,
解:,,使得 fx()(,)在上连续,,,,则必存在原函数Fx()
xC,,,1x,0,1,2 , FxxxCx(),01,,,,,,22,x,12,xC,,3,
又须处处连续,有Fx()
.12 ,即 xCxxC,,,,CC,,lim()lim()1212,,xx,,,,002
13
1322 ,即 xCxxC,,,,1,,,CClim()lim()3232,,xx,,1122
1 联立并令CC,,可得,CCCC,,,,1.1232
,,xC,,x,0,1,2,,,,,xxCx,01故 . fxx()d,,2,x,11,2xC,,,,,2
,四(设和都是连续函数,计算不定积分 fx()fx()
2, sin(cos)dcos(cos)dxfxxxfxx,,,
2, 解: sin(cos)dcos(cos)dxfxxxfxx,,,
, ,,,,sin(cos)dcoscos(cos)dxfxxxfxx,,
,,,sind(cos)cos(cos)dxfxxfxx ,,
,,,,sin(cos)(cos)dsincos(cos)dxfxfxxxfxx ,,
,,,,sin(cos)cos(cos)dcos(cos)dxfxxfxxxfxx ,,
,,sin(cos)xfxC
1nn,1五、若Itand,,xx证明:. n,2,3,?,,,xItanInnn,2,n,1
证明:因为
nnn,,2222Itand,xx,,,tantandtan(sec1)dxxxxxx n,,,
nn,,222nn,,22,,tansecdtandxxxxx,,tandtantandxxxx ,,,,1n,1 ,,xtanIn,2n,1
1n,1故 . ,,xItanInn,2n,1
本章复习题B
一、填空.
111,x23(1) ,e; (2) ; (3) xxxxxCcos4sin6cos,,,xxc,,23x
532442,x22(4) (5) (6) (21),,,xecxC,xxcxc,,,12153
14
1lnx1232(7) (8) (9). ,,Cfxx()ln,,,,(1)xCx23
二、求下列不定积分.
x1111arctane,2xxxxx,2(1)= ,,eeex,eedx,arctand[arctand]2,,xx22,x,eee221()
x111e,,2xxxx,2xx==. ,,,,eeeeC,,,(arctanarctan)[arctan()d]eex,2xx,21ee2
2x2t11,t(2)令,则,,.于是 t,tansinx,cosx,xtdxdt,,arctan,22221,t1,t1,t
dxxx11112 ,,,,,()lntantantdtC,,sin22sin44282xxt,
222(3) ln(1)xxdx,,,,,,,,,xxxxdxxln(1)ln(1),,
xdx222. ,,,,,xxxln(1),,,,,,,xxxxCln(1)1,21,x
2u2x(4) 令,则于是有 xu,,ln(1),ue,,1dxdu,,21,ux22xeuuu(1)ln(1)2,,2,,2ln(1)udu dxdu,,,,2xuu1,e,1
24u22 ,,,,,2ln(1)4arctanuuuuC,,,2ln(1)uudu,21,u
xxx. ,,,,,,,2141arctan1xeeeC
sindsin(1sin)dsin(1sin)dxxxxxxxx,,(5) ,,2,,,1sin(1sin)(1sin)cos,,,xxxx
sindd(-cos)xxx22 ,,,,,tand(sec1)dxxxx,,,,22coscosxx
12; ,,,,,tanxxCcosx
(11)d,,,xxx(6) 原式= ,11(11),,,,,,xxxx
1111,,,,xxxx ,,,,ddxxx,,22x2x
111,,,xxx而 dddxxx,,,,,2x11xx,22()()x,,22
15
1111,x2令 ,,xtsecdsecsecdxttt,,,,2221122()()x,,22
1 ,,,,(tanln|sectan|)tttC2
122 ,,,,,,,(2ln|212|)xxxxxC2
2xxx,12所以 原式= xxxxC,,,,,,,ln212224
lnlntanxt2(7)令 xt,tandsecdcoslntandxttttt,,33,,,sect22(1),x
2sect ,,,,sinlntansindsinlntansecdtttttttt,,tant
xxln2; ,,,,,,,,,sintanlnsectanln1ttttCxxC21,x
xxxe(1sin)eesin,xx(8) dd+dxxx,,,,1cos1cos1cos,,,xxx
xxx2esincosxexxxx22 =d+dedtanetandxxx,,,,,,xx22222cos2cos22
xxxxxxxx; etanetandetandetan,,,,,xxC,,2222
22xxx(1)(1)1,,,,(9) ddxx,,,100100(1)(1),,xx
,,,9899100,,,,,,[(1)2(1)(1)]dxxxx ,
111,,,979899; ,,,,,,,(1)(1)(1)xxxC974999
2x1122(10) arctandxx,xxxxCarctanln(1)(arctan),,,,,2221,x
x4xcos11xx22(11). ,,,xCcsccotdx,,3sinx8242
dx2三、设,求( xyxy,,(),xy,3
342uuu,32 解:令 ,得, x,, , uxy,,xxuu,,()ddxu,222(1)u,u,1
16
3uu,3 又 ,故 xyux,,,,3322u,1
422d31xuuuu,, ,,,dduu,,,2232xyuuuu,,,,3(1)31
1122. ,,,,,,,ln|1|ln|()1|uCxyC22
2四、设是的原函数,且当时,有又, Fx()fx()F(0)1,fxFxx()()sin2,,x,0
,求. Fx()0,fx()
,解:因是的原函数,则=.于是 Fx()fx()fx()Fx()
2, sin2()()()()xfxFxFxFx,,
上式两端积分得:
2,sin2d()()dxxFxFxx, ,,
2Fxxx()sin4 ,,,C228
1sin4x又,,得,故,从而 F(0)1,Fx()0,C,Fxx()1,,,24
1cos4,x,=. fx()Fx(),44sin4xx,,
xxe五、设是的一个原函数,且当 时,,已知Fx()fx()x,0fxFx()(),22(1),x
,试求( FFx(0)1,()0,,fx()
,解:因是的原函数,则=.于是 Fx()fx()fx()Fx()
xxe, ,,fxFxFxFx()()()()22(1),x
上式两端积分得:
xxe, d()()dxFxFxx,2,,2(1),x
22xFxxe() ,,C324(1),x
x2xe1sin4x,又,,得,故,从而= F(0)1,Fx()0,C,fx()Fx(),Fxx()1,,,32422(1),x
xx2六、设,求fxx()d( fx(sin),,sinx1,x
2xtsin22解 令 xt,sinfxxftx()d(sin)d,,,21,x1sin,t
17
sintt ,,,,,2sincosd2sind2sincosttxttxtttC,,cossintt
,,,,,21arcsin2xxxC
xxdteedxx七、解 (1)令, et,dx,,,,litlie()(),,,xxlnt222xxxeee (2), dddxxx,,,,,23221xxxx,,,,22(2)2(2)xxx,,eee4442(2)x, , ddd2(2)(),,,,xexexelie,,,,,,222(2)xxx
2xe,22(1)x同理可得 d(),xelie,,1x2xe,,42(2)22(1)xx 所以 . d()(),,xelieelie,2,,32xx
18