第六章 不定积分在不定积分的计算中，有很多方法是机械性的：有很多固定的模式和方法，还有一些常用的公式。

在本章里使用的积分公式除了课本161页给出的10个常用公式外，还有6个很有用的式子，罗列如下：22222211.ln ;212.arctan ;3.arcsin ;4.ln ;5.ln ;26.arcsin .2dx x aC x a a x a dx x C x a a a x C a x C a x C a x C a -=+-+=++=+=+=+=+⎰⎰这六个公式在答案中的使用次数很大，使用的时候没有进行说明，敬请读者仔细甄别。

当然答案计算过程中不免有不少错误，敬请原谅并修改。

第一节 不定积分的概念1.求下列不定积分：3353646422112111(1)(.4643*4646x x dx x x x C x x x C +-=+-+=+-+⎰ 3341(2)(5)(5)(5)(5).4x dx x d x x C -=---=--+⎰⎰114211313333222223(3)(32)63.34dx x x x x dx x x x x C --=+++=++++⎰⎰22424242422311111(4)()()(1)1111arctan .3dx x x dx dx dx dx x x dx x x x x x x x x x x C ------=+=+=-+++++=-+++⎰⎰⎰⎰⎰⎰22233(5)(3)33arctan .11x dx dx x x C x x =-=-+++⎰⎰1132222(6)().3x x dx x C -=+=+⎰(7)(2sin 4cos )2cos 4sin .x x dx x x C -=--+⎰ 221(8)(3sec )(3)3tan .cos x dx dx x x C x-=-=-+⎰⎰ 22222sin 3cos 1(9)(tan 3)(2)tan 2.cos cos x x x dx dx dx x x C x x ++==+=++⎰⎰⎰22222sin 3cos (10)3tan .cos cos x x dx dx x x C x x+-==-+⎰⎰222sin tan 11cos (11)(cos ).cos cos cos cos sin 22x x x dx dx d x C x x x x x ==-=+-⎰⎰⎰22cos 2cos sin (12)(cos sin )sin cos .cos sin cos sin x x xdx dx x x dx x x C x x x x-==+=-+--⎰⎰⎰2221(13)tan .1cos 21cos sin 2cos 2dx dx dx x C x x x x ===+++-⎰⎰⎰ 22252(14)(51)(52*51)5.2ln 5ln 5x xxxxdx dx x C +=++=+++⎰⎰ 121(15)(2()).35ln2ln 335xx xxxx e e dx C +-=--+⎰ (16)(1(.x xx x e dx e dx e C -=-=-⎰⎰ 221(17)(cos sin 2arctan arcsin .14x dx x x x C x -=--++⎰113724444(18).7x x dx x dx x C ===+⎰⎰212(19)2312.ln12xx xxdx dx C ==+⎰⎰3(20)sin )sin )arcsin cos .2x dx x dx x x C +=+=-+⎰⎰222.(),(,())2,(2,5).'()2,()'()2.(2)5,45,1,1y f x x f x x f x x f x f x dx C xdx C x C f C C y x ===+=+=+=+===+⎰⎰求一曲线它在点处的切线的斜率为且过点解：设那么令那么于是有因此函数曲线满足条件。

3.(),().f x f x 已知满足给定的关系式试求(1)'()1,(0);1'(),1()ln .xf x x f x xf x dx x C x =>===+⎰解：可得那么 2'()(2)1,(0);'(), ().2f x x xf x x x f x xdx C =>===+⎰解：可得那么有212(3)()'()1,(0);[()'()]1,(),2()f x f x x f x f x dx dx f x C x C f x =>=+=+=⎰⎰解：可知于是有 因此有 12'()(4)1,(()0);()'()1,()ln[()],().x f x f x f x f x dx dx f x f x C x C f x Ce =>=+=+=⎰⎰解：可知于是有 因此有 第二节 换元积分与分部积分法1.用凑微分法求下列不定积分：1(56)1(1)ln 56.565565dx d x x C x x -==-+--⎰⎰ 12(2)()ln ln 21.(12)21dx dx x x C x x x x =-=-++++⎰⎰33223322122(3)[(1)(1)]2331[(1)(1)].3x x Cx x C ==+--+=+--+(4)arcsin).dx dxC===+⎰⎰331(5)).323212dx dxCx x===+++⎰⎰222(6)2()2.2x x xxe dx e d e C---=--=-+⎰⎰222222111(7)()().222x x x xxe dx e d x e d x e C----==--=-+⎰⎰⎰1(8)()ln(1).11xx xx xedx d e e Ce e==++++⎰⎰22()1(9).212(1)1x xx x x x x xdx e dx d eCe e e e e e-===-+++++++⎰⎰⎰22()(10)arctan().11x xxx x x xdx e dx d ee Ce e e e-===++++⎰⎰⎰sin1(11)tan(cos)ln cos.cos cosxxdx dx d x x Cx x==-=-+⎰⎰⎰552562tan1(12)tan sec tan(tan)tan.cos6xx xdx dx xd x x Cx===+⎰⎰⎰1222212sin12sin12 (13)tan2(cos)tan.cos cos cos cos cos x xdx dx dx x d x C x C x x x x x -=-=++=-+⎰⎰⎰⎰2222221(tan)1(tan)cos(14)sin cos tan tan tan1).dxdx d x d xxAA xB x A x B A x B B xBx C===++++==+⎰⎰⎰⎰54222435(15)cos cos (sin )(1sin )(sin )(12sin sin )(sin )21sin sin sin .35xdx xd x x d x x x d x x x x C ==-=-+=-++⎰⎰⎰⎰2222221()2()cos (tan )222(16)2221sin 12sin cos sin cos 2sin cos tan 12tan22222222cos 2(tan 1)22 2.(tan 1)tan 122xd x x x d d dx x x x x x x x x x x x d C x x ===+++++++==-+++⎰⎰⎰⎰⎰cos 2cos 2cos 21(17)2(2)(sin 2)ln sin 2.sin cos sin 2sin 2sin 2x x x dx dx d x d x x C x x x x x ====+⎰⎰⎰⎰22444sin cos sin 111(18)(sin )(sin )arctan(sin ).1sin 1sin 21sin 2x x x dx d x d x x C x x x ===++++⎰⎰⎰ 22222211111(19)()(4)ln(4).424242x dx d x d x x C x x x ==+=+++++⎰⎰⎰ 222244211111(20)()()arctan().42442421()2x x x dx d x d C x x x ===++++⎰⎰⎰175444711473(21)()(3)ln 32.323323333239x dx dx x d x C x x C x x x -=-+=-++=-+-+---⎰⎰⎰ 1111(22)sin 2cos32sin 2cos3(sin 5sin )cos5cos .22102x xdx x xdx x x dx x x C ==-=-++⎰⎰⎰223(ln )1(23)(ln)(ln )ln .3x dx x d x x C x ==+⎰⎰21111(24)sinsin ()cos .dx d C x x x x x=-=+⎰⎰ 2231(25)(arcsin )(arcsin )arcsin .3x d xx C ==+⎰22arctan 1(26)arctan (arctan )arctan .12x dx xdx x C x ==++⎰⎰ (27)22.C ===(28)2.1dCx==++⎰(29)()arcsin().xx xe e C==+(30)2()2()222(sin cos)()2(sin cos).22222x xx x x x xd C===+=-+⎰22222222.(1)((22(ln.2229(7),dx dxaadxx CT=+=-=+-=++⎰⎰⎰用换元积分法求下列不定积分：可参考第四章第一节此题所得可以当做公式使用,即有2ln.2ax C=+2222sin222221224sin(2)(2sin)(2cos)4sin;2cossin,cos,(sin cos);1(cos sin)cos2x t tt t dt tdtxI tdt J tdtI J t t dt dt t CJ I t t dttdt======+=+==+-=-==⎰⎰⎰⎰⎰⎰⎰⎰令可得232sin2.21sin2.242sin22arcsin sin(2arcsin).22t CtI t Cx xt t C C+=-+=-+=-+⎰于是有因此222(2)424arcsin2arcsin)2arcsin.222xx x xC C==-=-=-+=211111 (3)()22111())]222xxxx x-+-==+-=-+-=11)]221arcsin.2x CC-+=1(4)()21921arcsin[()]832921arcsin.83xx CxC=-=-+-=+22222233332222222222222222213222222221222222111(5)()()()()111()2()()11111[321()(2dx x a x x a xdx dx dxa a ax a x a x a x axdx d x aa ax a x adx xda ax a x+-+==-++++=-+++=--++⎰⎰⎰⎰⎰⎰⎰122111122222222222212222])111111[]()()()1()adx x dx Ca a ax a x a x axCax a+=+-++++=++⎰⎰⎰222(6)((1)1ln.22xdx xdx xdxx xxx C-==-=---=-++⎰⎰⎰⎰21(7)(1)22()22222.tt t t t ttt te d t te dt td e te e dt Cte e C C-===-+=-+-+⎰⎰⎰⎰221244(8)(1)(1)2(2)(24)1111 44ln 114ln(1.t t t t dx d t tdt t dt t dtt t t t t t t C C --=-=-=-+++++=-++++-+⎰⎰⎰3866422227537/65/61/21/61/666(9)(6666)11166266arctan 756266arctan().75t t t t dx dt dt t t t dt t t t t t t t t Cx x x x x C ==-+-++++=-+-++-+-++⎰⎰2322323/2222(10)(222)22ln 1111322ln(1.3t t t t dt dt t t dt t t t t Ct t t x x C ==-+-=-+-+++++-+++⎰⎰54222222243535222211(1)(11)(1)(1)(12)222121 (1)(1).3535t t t d t t dt t t dt t t t t C x x C -=-=--=--+=-+-+---+⎰⎰⎰22332322(1)2(1)3(12).(1)(1)(1)111(1)2(1)3(1)1(1)(1)231ln 1.12(1)x x x dx d x x x d x d x d x x x x x C x x ++-++=+++=+-++++++=++-+++⎰⎰⎰⎰⎰3.用分部积分法求下列不定积分：222212121222(1)cos sin sin sin sin 2sin sin 2cos sin 2cos 2cos sin 2cos 2sin .x xdx x d xx x xdx C x x x xdx C x x xd x C x x x x xdx C x x x x x C ==-+=-+=++=+-+=+-+⎰⎰⎰⎰⎰⎰3444143144(2)ln 1ln 41ln (ln )441ln 441ln 416x xdxxdx x x x d x C x x x dx C x x x C ==-+=-+=-+⎰⎰⎰⎰ 11(3)ln ln (ln )ln ln .xdx x x xd x C x x dx C x x x C =-+=-+=-+⎰⎰⎰1122122(4)arc tan arctan (arctan )arctan 111arctan 211arctan ln(1).2xdxx x xd x C xx x dx C x x x dx C x x x x C =-+=-++=-++=-++⎰⎰⎰⎰111(5)2arcsin 2arcsin 2(arcsin )2arcsin 22arcsin 22arcsin .xd x x C x C x C x C =-=-++=-++=-++=-+⎰⎰⎰⎰22212221122211(6)arctan arctan arctan arctan 222111arctan arctan (1)221221arctan arctan .222x x xdx xdx x x d x C x x x x dx C x dx C x x x x xx C ==-+=-+=--+++=-++⎰⎰⎰⎰⎰1132222322ln 11111111(7)ln ()ln (ln )ln 22222ln 1.24x dx xd x d x C x dx C x x x x x x x C x x=-=-++=-++=--+⎰⎰⎰⎰1122(8)cos(ln )cos(ln )[cos(ln )]cos(ln )sin[ln ] cos(ln )sin(ln )[sin(ln )][cos(ln )sin(ln )]cos(ln )[cos(ln )sin(ln )]cos(ln ).2x dx x x xd x C x x x dx C x x x x xd x C x x x x dx C x x x x dx C =-+=++=+-+=+-++=+⎰⎰⎰⎰⎰⎰因此533312331145233515(9)sec sec (tan )sec tan tan (sec )sin sin sec tan 3tan sec tan 3cos cos 1cos sec tan 3sec tan 3sec cos xdx xd x x x xd x C x x x x x dx C x x dx C x x x x x dx C x x xdx x ==-+=-+=-+-=-+=-⎰⎰⎰⎰⎰⎰31335233233233sec ,sec tan 3sec sec ;4sec sec (tan )sec tan tan (sec )sin sin sec tan tan sec tan cos cos xdx C x x xdxxdx C xdx xd x x x xd x C x xx x x dx C x x dx C x x +++=+==-+=-+=-+⎰⎰⎰⎰⎰⎰⎰⎰⎰因此有2333333341cos 1 sec tan sec tan sec cos cos sec tan sec ln sec tan ;sec tan ln sec tan sec .2x x x dx C x x xdx dx C x x x x xdx x x C x x x xxdx C -=-+=--+=--++-+=+⎰⎰⎰⎰⎰于是综上可得353(sec tan ln sec tan )sec tan sec .48x x x x x x xdx C -+=++⎰ 22222212111(10)ln()ln(1)ln(1)ln(1)()ln(1)()12211{ln(1)[ln(1)]}{ln(1)[ln(1)]}2211 ln(2x x dx x x dx x x dx x d x x d x x x x x d x x x x d x C x +=+--=+---=+-+----+=⎰⎰⎰⎰⎰⎰⎰22221122211222111)()ln()121121111111 ln()(1)ln()2112111111 ln()ln(2121x x x x x dx C x dx C x x x x x x x dxx dx C x x C x x x x x xx x x ++-++=-+-+---++=--+=+-+----++=+--⎰⎰⎰⎰).C x+-222212223(11)sin sin ,cos ;21(cos sin )cos 2(sin 2)2sin 21sin 2cos 2 sin 2.2224x xdx I x xdxJ x xdx x I J xdx C J I x x x dx x xdx xd x x x x x xxdx C C ⎧=⎪⎨=⎪⎩+==+-=-===-+=++⎰⎰⎰⎰⎰⎰⎰⎰解：取那么于2222sin 2cos 2sin 448.sin 2cos 2cos 448x x x x I x xdx C x x x x J x xdx C⎧==--+⎪⎪⎨⎪==+++⎪⎩⎰⎰是有222sin 2cos 2(12)cos ,cos .448x x x xx xdx x xdx C =+++⎰⎰依上题可知1(13)[ln(ln )]ln(ln )ln(ln )[ln(ln )]ln ln ln 11 ln(ln )ln(ln )ln ln ln ln ln dx dxx dx x dx x x xd x C x x xx dx dxx x dx C x x dx Cx x x x xx +=+=-++=-++=-++=⎰⎰⎰⎰⎰⎰⎰⎰⎰(ln ).x C +2222222(11)(14)(1)(1)1(1)1(1)1 ()11(1)1 ()1(1)1(1)1x x x x x xx x xx x x xxe x e e e de e dx dx dx dx dx x x x x x x e e e d dx C x x x e e e e d dx C x x x x x +-==-=-++++++=--++++=+-+=+++++⎰⎰⎰⎰⎰⎰⎰⎰⎰⎰.C +arcsin 222221122122(15)(arcsin )(sin )sin sin ()sin 2sin sin 2(cos )sin 2cos 2cos sin 2c x tx dxt d t t t td t C t t t tdt C t t td t C t t t t tdt C t t t ===-+=-+=++=+-+=+⎰⎰⎰⎰⎰⎰arcsin 2os 2sin (arcsin )2.x tt t Cx x x x C =-+=+++1121cos (16)(cot )cot cot cot sin sin (sin )cot cot ln sin .sin x xdx xd x x x xdx C x x dx C xx d x x x dx C x x x C x=-=-++=-++=-++=-++⎰⎰⎰⎰⎰111(17)ln(ln([ln(ln(ln(x dx x x xd x Cx x Cx x C=-++=-+=-+⎰⎰21ln(ln(.x x Cx x C=-+=-333222222213133222222113333222222222(18)ln()ln(ln)3332228ln2ln ln ln()33392882ln ln(ln)ln3993xdx xd x x x x d x Cx x x xdx C x x xd x Cx x x x x d x C x==-+=-+=-+=-++=⎰⎰⎰⎰⎰⎰312222333222288ln992816ln ln.3927x x x x dx C x x x x x C-++ =-++⎰4.求下列不定积分的递推公式：11 11111(1)()().n kx n kx n kx kx n n kx kx n n kx n nn nI x e dx x d e x e e d x x e e x dx x e Ik k k k k k k--===-=-=-⎰⎰⎰⎰11(2)(ln)(ln)(ln)(ln)(ln)(ln).n n n n n nn nI x dx x x xd x x x n x dx x x nI--==-=-=-⎰⎰⎰2222222221222sin(1cos)tan(3)tan tan tan tancos cos costantan(tan)tan.1nn n n n nnn nnx x xI xdx x dx xdx dx xdxx x xxxd x xdx In---------====-=-=--⎰⎰⎰⎰⎰⎰⎰1111(4)(arcsin)(arcsin)(arcsin)(arcsin)(arcsin)(arcsin)(arcsin)(arcsin))[(arcsin)](arcsin)n n n n n nn nn n nI x dx x x xd x x x xx x n x dx x x xx x----==-=-=+=+-=⎰⎰⎰⎰⎰1212)(1)(arcsin)(arcsin))(1).n n nn nnx n n x dxx x x n n I----+--=+--⎰5.求下列有理函数的积分：5423328843(1)(1)118ln 4ln 13ln 1.32x x dx x x dx x x x x x x x x x x x C +-=+++---+-=+++-+--+⎰⎰2222221111111(2)()(1)(1)2121412121111ln 1arctan ln 1.422dx x dx dx dxdx x x x x x x x x x x C -=-+=-+++++++++=-+++++⎰⎰⎰⎰⎰2422111111(3).()ln arctan .1211412x x dx dx x C x x x x +=-=-+--+-⎰⎰1322121112112(4)()ln 1131133113111 ln 1()3321()1111 ln 1()3622 dx x x dx x dx C x x x x x x x x x C d x x x C -+-=+=+-+++-+-+--=+-++-=+-+++⎰⎰⎰22111ln 1ln 1)]36211 ln 1ln 1.36x x x x Cx x x C =+--++-+=+--+++2222737127132222(5)()ln 712ln 7171712222()2422134ln 712ln 2ln 4ln 3.223x x x dx d x x x C x x x x x x x C x x C x -+---=-=-+++-+---+-=-+++=---+-⎰⎰241954352(6)()ln 1ln 1ln 2.(1)(2)6112263x dx dx x x x C x x x x x +-=++=-++-+++-++-+⎰⎰41111(7)11)1)8448x x x x dxdx dx xC+++++ ==++=+++-+=⎰⎰⎰1)1).44C+++-+222232(1)55(8)ln(21).21(1)1x xdx dx x x Cx x x x-+-==++++++++⎰⎰22(2)(9).42(2)2dx d xCx x x+==++++-⎰⎰2112(10)()ln.826(2)(4)62464dx dx dx dx xCx x x x x x x-=-=--=-+ ---+-++⎰⎰⎰⎰6.求下列三角函数有理式积分：222222222()()22(1)2245cos4cos4sin5cos5sin9cos sin2222221()2cos(tan)tan3122222ln.39tan9tan tan3222x xd ddxx x x x x xxxdx x xdCx x x==+++---===-+--+⎰⎰⎰⎰⎰222222221cos(2)54sin25sin5cos8sin cos5tan58tan(tan)1(tan)1(tan)8435tan58tan55tan tan1(tan)()55515arcta53dxdx dx xx x x x x x xd x d x d xx x x x x==+++++===++++++=⎰⎰⎰⎰⎰⎰45(tan)1545n arctan(tan).3333xC x C++=++222(tan )1(tan )(3))22sin 3tan 23tan 3).dx d x d x x Cx x x x C ===++++=+⎰⎰⎰22222221sec 1cos cos (4)11(1sec )(cos 1)(1)(cos 1)(1)cos cos ()()1122 22cos 1(cos 1)2cos (2cos )22tan 2x x x dx dx dx dx x x x x xx x d d dx dx x x x x x ===+++++=-=-++=-⎰⎰⎰⎰⎰⎰⎰⎰11222231222(tan )tan 1122tan tan ()2222cos 2cos cos222tan tan tan 11222 tan tan tan .222322cos cos 2cos 222x x d x x C d C x x x x x x x x x dx C C x x x +=-++=-++=-++⎰⎰⎰cos (5),1tan cos sin cos cos sin ,sin cos sin ;cos sin (cos sin )ln cos sin cos sin cos sin dx xdx x x xx I dx x x x J dx x x I J dx x Cx x d x x I J dx x x C x x x x =++⎧=⎪⎪+⎨⎪=⎪+⎩⎧+==+⎪⎨-+-===++⎪++⎩⎰⎰⎰⎰⎰⎰⎰取那么有于是ln cos sin cos cos sin 2.ln cos sin sin cos sin 2x x x x I dx C x x x x x x J dx C x x ⎧++==+⎪⎪+⎨-+⎪==+⎪+⎩⎰⎰121sin 2cos 1(cos )21(sin )(6)()(2cos )sin 32cos sin 32cos 3sin 3sin 12sin 1ln 2cos ln sin 331cos 312 ln 3dx x x d x dx d x dx x x x x x x xxdx x x C x --=+=+-+++=++-+-=⎰⎰⎰⎰⎰⎰cos 11cos ln .sin 31cos x xC x x++-+-22sin cos 112sin cos (7)()sin cos 2sin cos sin cos 12cos 1(sin cos )2 (2)2sin cos 2tan 1tan 2211 sin cos 224x x x x dx dx dx x x x x x xxd x x x dx x x x x x x +=-++++=-++-=--⎰⎰⎰⎰⎰.C +3222222sin sin 1cos 2(8)(cos )(cos )(1)(cos )1cos 1cos 1cos 1cos cos 2arctan(cos )x x x dx d x d x d x xx x x x x C-=-=-=-++++=-+⎰⎰⎰⎰4(3)322111(9)tan tan tan tan ln cos .22T xdxx xdx C x x C =-+=++⎰⎰参见sin 222222222sin 1cos sin 11(10)()sin cos sin cos sin (1sin )(1)111111sin 1 ln ln 21sin 2sin 1x t x t x d x dt dx dx dt x x x x x x t t t t t x C Ct t x x ======+-----=--+=--+++⎰⎰⎰⎰⎰cos (11)sin 2cos cos sin 2cos ,sin sin 2cos 2;cos 2sin (sin 2cos )2ln sin 2cos sin 2cos sin 2cos xdxx xx I dx x x x J dx x x I J dx x Cx x d x x I J dx x x C x x x x +⎧=⎪⎪+⎨⎪=⎪+⎩⎧+==+⎪⎨-+-===++⎪++⎩⎰⎰⎰⎰⎰⎰取那么于是cos 21ln sin 2cos .sin 2cos 55x I dx x x x C x x ==++++⎰221222212sin sin 2(12)(1)21cos 2sin 2sin 2sin (tan ) 2tan ).tan 22x x dxdx dx dx x C x x x x d x x C x x C x ==-+=-+++---=-++=-+++⎰⎰⎰⎰⎰1/6563222127.634161(1)6()(12)(1)(21)2121116()3342 6ln ln 11724()416x t t dt dt t dx t t t t t t t t t t t t t t dt C t =-===--+++-++-+-+=-+-+-+⎰⎰⎰⎰求下列无理函数的积分：1/621/61/61/31/639116ln ln 1ln )]242243911 ln ln 1ln )]24224x tt t t t t Cx x x x x C==-+--+-+=-+--+--+2(2)1 (ln .22x x dx x C ====-+⎰⎰111()(3)221()122211.22txtxdd tt CCx====-=-+=-=-+++=-+++⎰2322(4)[(1(1)211(22)ln1.32x dxxx x x C =-=-+=-+-⎰⎰⎰2(5)11()()11ln.2d dCx====-+=+1(6)ln.2x C==++⎰22213() (7)1112()2()5132arcsin ln1arcsin2822122xx x xx x Cx-+ =-+=-+---=--+---=--⎰⎰212()111ln1.82xx x C-++-+13(8)ln.8x C==++222242441111(9)22441(1)arctan11ln4(1)(1)2412x t t syyd y y dy y yCy y y y=======--==++--++⎰⎰22114241ln.4t sxC CC===++ =++2442341()41(10)(1)18[))8111[8(8ttd t dttt ttCtxxx+===-+=--+++++--⎰⎰3/4)C-++++8.(1)2()22arcsin arcsin.a bx x a bC Ca b a b==+---=+=+--⎰求下列不定积分：221()()11(2).22d dCββ==-=-=(3)sin (cos )cos cos ()cos cos ()sin sin sin (sin )sin (sin cos )sin cos cos cos x x x x x x x x x x x x x x x x xe xdx xe d x xe x xd xe xe x x e xe dx xe xdx x xde x xe e d x x x xe e x x x dx xe xdx xe xdx xe x xe =-=-+=-++==-=-+-=-+⎰⎰⎰⎰⎰⎰⎰⎰⎰⎰由于有于是有;sin cos sin sin (sin cos )cos 2,(sin cos )cos 2(sin cos )sin sin 22cos x x x x xx xxx x x xx dxxe xdx xe xdx xe x xe dxe x x xe dx e x x xe dx xe x x e x xe xdx C xe xd ⎧⎪⎨+=-⎪⎩⎧-=⎪⎪⎨+⎪=⎪⎩-=-++⎰⎰⎰⎰⎰⎰⎰可以求得于是有.(sin cos )cos 22x xxe x x e x x C ⎧⎪⎪⎨+⎪=-+⎪⎩⎰(4)cos x xe xdx ⎰ 可见上题！sin()sin()(5)sin()sin()sin()sin()sin()sin()sin()sin()sin()cos()sin()cos()sin()sin()sin() dx b a dx x b x a dx x a x b b a x a x b b a x a x b x b x a x a x b dxb a x a x b -+--==++-++-++++-++=-++⎰⎰⎰⎰cos()cos() sin()sin()sin()sin()11ln sin()ln sin()sin()sin() x a dx x b dxb a x a b a x b x a x b Cb a b a ++=--+-+=+-++--⎰⎰1sin()ln .sin()sin()x a C b a x b +=+-+112sin sin()cos cos()sin sin()(6)tan tan()[1]cos cos()cos cos()cos cos cos cos()cos cos sin cos sin x x a x x a x x a x x a dx dx dxx x a x x a a adxx dx C x C x x a x a x x a +++++==-++=-++=-+++-⎰⎰⎰⎰⎰1cos (tan )cos ln sin tan cos .cos tan sin sin ad x ax C x a x a C a x a a=-++=---++-⎰32222222111(7)()()()22211 (1)(1)2212 arcco 23x x x xd x x ==-+=---=⎰3223322221s (1)arccos 11 arccos (1)(1)(arccos )33arccos (arccos )1 arcco 3xd x xd x x x d x x x C --=----++=⎰⎰⎰322213232321s (1)(1)arccos 3111 arccos (1)()arccos 33321 2)arccos .39x x x dx x dx C x x x x x x Cx x x x C -+---+=-+---+=----+⎰⎰222122tan sin cos sin cos (8)1tan tan cos sin cos sin 1sin cos 1sin cos 1sin cos 1sin cos cos sin cos sin x x x x x dx dx dx x xx x x x x x x x dx dxdx x C x x x x x x x x==++++++=-=-+++++⎰⎰⎰⎰⎰⎰1122(tan )(tan )131tan tan (tan )241(tan )].332d x d x x C x C x x x x x C =-+=-+++++=-++⎰⎰11112(9)22222()222122242ln 11xx x e t x x yxd C d e C dt C e ty y y dy C y Cy y ====+=+=+-+=-+-+⎰⎰⎰⎰2.yC +1sin (10)1cos 1sin 1cos ,'(),'()1cos sin 1cos '()ln sin .sin 1(), 4.3(())','()1s '()x xdxxx x xf x f x x x xxf x dx dx x x C x xf x f x f x x dx f x -++++==+++==+++=+=⎰⎰⎰⎰ 取即那么有由于单调递增由定理可以知道因此有1in (),1cos ()ln sin .xdx f x x f x x x C -=+=++⎰其中22222222sin (10)1cos 2sin 1,1cos 122arctan sin 221(2arctan )11cos 1111 x xdxx t x t t x t tt x x t t dx dt t dt t x t t t ++⎧=⎪⎪+⎨-⎪=⎪+⎩+++==+-+++++⎰⎰⎰⎰ 令于是有2122122tan2()2arctan 2(arctan )1()2arctan 2(arctan )11 2arctan 2tan arctan(tan )22x t d t t t td t C tt d t t t d t C t t x x t t C ==-+++=-++++=+=+⎰⎰⎰⎰tan .2xC x C =+1tan cos sin (cos sin )(11)ln cos sin .1tan cos sin cos sin x x x d x x dx dx x x C x x x x x--+===+++++⎰⎰⎰sin sin sin sin sin sin sin sin 1(12)sin 22sin cos 2sin (sin )2sin ()2sin 2(sin )2sin 2.x x x x x x x x e xdx e x xdx e xd x xd e xe e d x C xe e C ====-+=-+⎰⎰⎰⎰⎰111(13)arctan(1arctan(1(arctan(12arctan(11arctan(12dxx xd Cxxx dx Cx C+=-++=+-+=+-+⎰⎰⎰122121212arctan(1222arctan(12222arctan(1(1)22t t tx dt Ct ttx dt Ct ttx dt Ct t+-+++=+-++++=+--+++⎰⎰⎰2122(2)arctan(122arctan(1ln22arctan(1ln2.td t tx t Ct tx t t t Cx x C+=+-++++=+-+++++-++⎰2444432423ln221(2)1(14)(12)ln22(12)ln22(12)ln2(1)11(1)(1)(1)1()ln2(1)1111[ln ln1]ln212(1)3(1)xx x tx x x x xdx dx d dtt tt t tdtt tt tt t t====++++++++++=-+=-++++++++⎰⎰⎰⎰⎰2231111[ln2ln(21)].ln2212(21)3(21)x tx xx x xCC ==-++++++++此文只供参考，写作请独立思考，不要人云亦云，本文并不针对某个人（单位），祝您工作愉快！一是主要精力要放在自身专业能力的提升上，二是业余时间坚持写作总结，这是一个长期的积累过程，剩下的，不用过于浮躁，交给时间就好了。