arcsinx1?x23. ??dx 22x1?xarcsinx1?x2解. ??dx令x?sint22x1?xt1?sin2t2?costdt?t(csct?1)dt 2?sintcost? ??tcottdt?tdt??tcott?cottdt? ??tcott?ln|sint|????12t?c 212t?c 21?x21 ??arcsinx?ln|x|?(arcsinx)2?c
x24.
arctanx?x2(1?x2)dx
arctanx?x2(1?x2)dx令x?tantt22?tan2tsec2tsectdt??t(csct?1)dt
解.
?tcsctdt?tdt??tdcott? ??tcott?ln|sint|?t?c???2??121t??tcott??cotdt?t2 22122arctanxx1?ln||?(arctanx)2?c x21?x2arctanx1x212?ln?(arctanx)?c ??2x21?x2十三. 求下列不定积分: 1.
?x34?x2dx
解.
32332x4?xdx令x?2sint8sint2cost2costdt?32sintcostdt ??? ?32(1?cost)costdtdcost???223232cos3t?cos5t?c 354122 ??(4?x)2?(4?x)2?c
352.
35??x2?a2 xx2?a2令x?asectxatant1?cos2t?asectasecttantdt?a?cos2tdt
解.
21
?atant?at?c?x2?a2?aarccosa?c x3.
?ex(1?ex)1?e2xdx
t(1?t)dt1?t1?sinu?dt令t?sinu?1?t2t?1?t2?cosucosudu
解.
?ex(1?ex)1?e2xdx令ex?t ?u?cosu?c?arcsinex?1?e2x?c 4.
?xxdx (a > 0)
2a?xxu4解. ?xdx 令u?x 2?du 令u?2asint 8a2?sin4tdt
2a?x2a?u2(1?cos2t)2dt?2a2?(1?2cos2t?cos22t)dt =8a?421?cos4ta422dt?3at?2asin2t?sin4t?c =2at?2asin2t?2a?24222 =3a2t?4a2sintcost?a2sintcost(1?2sin2t)?c =3at?3asintcost?2asintcost?c =3aarcsin22223xx2a?xx?3a2?2a22a2a2a2ax3a?x?x(2a?x)?c 2a2x2a?x?c
2a2a =3aarcsin2十四. 求下列不定积分: 1.
?sinx?sinxdx1?cosxdx
解.
1?cosx??sinxdxsin2x1?cosx???d(1?cosx)sin2x1?cosx??2?d1?cosx
1?cos2x 令1?cosx?u?2dudu??2?1?(u2?1)2?u2(2?u2)
??(
?11112?u?)du??ln||?c 22u22u2?u2?u22
?11?cosx?122ln|2?1?cosx2?1?cosx|?c
2?sinx?2?cosxdx 2?sinx1d(2?cosx)dx?2?dx??解. ?
2?cosx2?cosx2?cosx2dt2x2dt1?t 令tan?t 2??ln|2?cosx|?2?3?t2?ln|2?cosx| 21?t22?1?t22. =
4t41xarctan?ln|2?cosx|?c?arctan(tan)?ln|2?cosx|?c
23333sinxcosx?sinx?cosxdx
sinxcosx11?2sinxcosx?1dx??dx 解. ?sinx?cosx2sinx?cosx3.
1(sinx?cos)2?1111dx??(sinx?cosx)dx??dx =?2sinx?cosx22sinx?cosx)124 =(sinx?cosx)??24sin(x??)4 =
d(x??12x?(sinx?cosx)?ln|tan(?)|?c 2428十五. 求下列不定积分: 1.
??x1?xxxdx
2d(1?t3)4dt?????1?t3?c
331?t31?t32t2解.
1?xxdx令x?t3?4 ??1?x2?c
32.
??ex?1dx ex?1ex?1ex?1xdx?dx令e?sectx?2xe?1e?1sect?1?tanttantdt??(sect?1)dt
解.
23
?ln|sect?tant|?t?c?ln(e?ex2x?1)?arccos1?c ex3.
?x?1arctanx?1dx
x解. 令t?arctanx?1,tant?x?1,x?sec2t,dx?2sec2ttant
?ttantx?1arctanx?11?cos2t22dx??2secttantdt?2?ttantdt?2?tdt 22xsectcostt22dt?2tdt?2tdtant?t?2ttant?2tantdt?t ?cos2t??? ?2 ?2ttant?2ln|cost|?t2?c
?2x?1arctanx?1?ln|x|?(arctanx?1)2?c
第三章 一元函数积分学(定积分)
一．若f(x)在[a，b]上连续, 证明: 对于任意选定的连续函数?(x), 均有
?baf(x)?(x)dx?0,
则f(x) ? 0.
证明: 假设f(?)? 0, a < ? < b, 不妨假设f(?) > 0. 因为f(x)在[a，b]上连续, 所以存在? > 0, 使得在[?－?, ? + ?]上f(x) > 0. 令m =
????x????minf(x). 按以下方法定义[a，b]上?(x): 在[?－?,
22? + ?]上?(x) =??(x??), 其它地方?(x) = 0. 所以
?baf(x)?(x)dx????????f(x)?(x)dx?m??22?0.
和
?baf(x)?(x)dx?0矛盾. 所以f(x) ? 0.
?二. 设?为任意实数, 证明: I???2011?2=. dxdx??01?(cotx)?1?(tanx)?4??证明: 先证:
?20f(sinx)?f(cosx)dx?=?2dx
0f(sinx)?f(cosx)4f(sinx)?f(cosx)令 t =
?2?x, 所以
?20
?f(sinx)dx?f(sinx)?f(coxs)?0?2f(cots)d(?t)
f(cots)?f(sitn) 24
? = 于是
??20f(cost)dt?f(cost)?f(sint)??20f(cosx)dx
f(cosx)?f(sinx)2?20f(sinx)f(sinx)dx??2dx?0f(sinx)?f(cosx)f(sinx)?f(cosx)f(sinx)?f(cosx)?dx??2dx?
0f(sinx)?f(cosx)2????20f(cosx)dx
f(cosx)?f(sinx)??=
?20所以
?20f(sinx)?f(cosx)dx?=?2dx.
0f(sinx)?f(cosx)4f(sinx)?f(cosx)???(cosx)?2 dx????0(cosx)??(sinx)?4?sinx?1????cosx??1所以 I??2dx??2?01?(tanx)01??同理 I??201?. dx??1?(cotx)4
三．已知f(x)在[0，1]上连续, 对任意x, y都有|f(x)－f(y)| < M|x－y|, 证明
1n?k?M ?f(x)dx??f???
0nk?1?n?2n1证明:
?10f(x)dx???knk?1k?1nn1nkf(x)dx, ?f()?nk?1nk??knk?1k?1nnkf()dx nn1nkk?n?f(x)dx?f()?|f(x)?f()?dx|??k?1??0?nk?1nn?k?1n?1 ???knk?1k?1nnnnkkf(x)?f()dx???kn?1M(x?)dxnnk?1nk
?M??
knk?1k?1nM?k??xdx???2?n?1M??22nk?1nn?四. 设In??40tannxdx, n为大于1的正整数, 证明:
?11?In?.
2(n?1)2(n?1)ntdt 证明: 令t =tanx, 则In??4tanxdx??001?t2n1 25