ax,b(一)含有的积分(1~9)???????????????????????????????????????????????????????1
ax,b(二)含有的积分(10~18)???????????????????????????????????????????????????5
22x,a(三)含有的积分(19~21)????????????????????????????????????????????????????9
2ax,b (a,0)(四)含有的积分(22~28)????????????????????????????????????????????11
2ax,bx,c (a,0)(五)含有的积分(29~30)????????????????????????????????????????14
22x,a (a,0)(六)含有的积分(31~44)?????????????????????????????????????????15
22x,a (a,0)的积分(45~58)?????????????????????????????????????????24 (七)含有
22a,x (a,0)(八)含有的积分(59~72)?????????????????????????????????????????37
2,a,bx,c (a,0)(九)含有的积分(73~78)????????????????????????????????????48
x,a
,(x,a)(b,x)(十)含有 或的积分(79~82)???????????????????????????51 x,b
(十一)含有三角
数的积分(83~112)???????????????????????????????????????????55
a,0(十二)含有反三角函数的积分(其中)(113~121)???????????????????????68 (十三)含有指数函数的积分(122~131)??????????????????????????????????????????73 (十四)含有对数函数的积分(132~136)??????????????????????????????????????????78 (十五)含有双曲函数的积分(137~141)??????????????????????????????????????????80 (十六)定积分(142~147)????????????????????????????????????????????????????????????81
附录:常数和基本初等函数导数
?????????????????????????????????????????85
?????????????????????????????????????????????????????????????????????????????????????86
团队人员??????????????????????????????????????????????????????????????????????????????87
ax,b(一)含有的积分(1~9)
dx11. ,,ln ax,b ,C,ax,ba
1b
: 被积函数 f(x),的定义域为 {x|x,,}ax,ba
1 令 ax,b,t (t,0) ,则dt,adx ,? dx,dta dx11 ? ,dt,,ax,bat
1 ,,ln t ,Ca
dx1 将 t,ax,b 代入上式得: ,,ln ax,b ,C,ax,ba
1μμ,12. (ax,b)dx,,(ax,b),C (μ,,1),a (μ,1)
1 证明: 令 ax,b,t ,则dt,adx ,? dx,dta
1μμ ? (ax,b)dx,tdt,,a
1μ,1 ,,t,C a (μ,1)
1μμ,1 将t,ax,b代入上式得:(ax,b)dx,,(ax,b),C,a (μ,1)
x1.dx,ax,b,b,ln ax,b ,C3 ,2ax,ba,,
xbf(x),x|x,, 证明: 被积函数 的定义域为{}ax,ba
11ax,b,t t,,x,t,b , dx,dt 令 (0) 则aa,,
1t,bxb11,,,,a ?dx,?dt,1,dt ,,,,,2ax,btata,,
b11 ,dt,dt 2,,2taa
tb ,,,ln t ,C 22aa
1 ,t,b,ln t ,C ,,2a
x1,,t,ax,bdx,ax,b,b,ln ax,b ,C 将代入上式得:2,ax,ba
- 1 -
2x11,,224. dx,(ax,b),2b (ax,b),b,ln ax,b ,C,3,,ax,b2a,,
222x1(ax,b),2abx,b) 证明: dx,dx,,2ax,bax,ba
2112abx1b ,(ax,b)dx,dx,dx,,,222ax,bax,baaa112 ? (ax,b)dx,(ax,b),C1,23a2a
12abx2bax,b,b dx,d(ax) ,,23ax,bax,baa
22b2b1 ,dx,d(ax,b),,33ax,baa
22b2b ,x,ln ax,b ,C233aa
2221bb1b dx,d(ax,b),ln ax,b ,C3233,,ax,bax,baaa
2x11,,22 由以上各式整理得: dx,(ax,b),2b (ax,b),b,ln ax,b ,C3,,,ax,b2a,,
dx1ax,b5. ,,,ln ,C,x (ax,b)bx
1b 证明: 被积函数 f(x),的定义域为 {x|x,,}x,(ax,b)a
1AB 设,, , 则 1,A(ax,b),Bx,(Aa,B)x,Abx,(ax,b)xax,b
1,A,,Aa,B,0,,b ? 有 , ,,Ab,1a,,B,,,b,
dx1a11a1 于是 ,[,]dx,dx,dx,,,,x (ax,b)bxb,(ax,b)bxbax,b
1111 ,dx,d(ax,b),,bxbax,b
11 ,,ln x ,,ln ax,b ,Cbb
1x,1 ,,ln ,C提示: logb,,logbaabax,b
1ax,b ,,,ln ,Cbx
- 2 -
dx1aax,b 6. ,,,,ln ,C22,bxxx (ax,b)b
1b f(x), {x|x,,}证明:被积函数的定义域为2 ax,(ax,b)
1ABC2 ,,, , 1,Ax (ax,b),B(ax,b),Cx设则 22xax,bx,(ax,b)x 2 x(Aa,C),x (Ab,aB),Bb,1即
a, A,,2, bAa,C,0,, 1,, ? Ab,aB,0 , B,,,有 b,,Bb,12, ,aC,, 2b, 2dxa111a1 ,,dx,dx,dx 于是2222,,,,xbax,bx (ax,b)bxb
a111a1 ,,dx,dx,d(ax,b) 222,,,xbax,bbxb
a1a ,,,ln x ,,,ln ax,b ,C22 bxbb
1aax,b ,,,,ln ,C2 bxxb
x1b,, 7. dx,ln ax,b ,,C,,22,ax,b (ax,b)a,,
xb f(x), {x|x,,}证明:被积函数的定义域为2a (ax,b)
xAB , ,, x,A(ax,b),B设则 22ax,b (ax,b) (ax,b)
x,Aa,(Ab,B),x即
1,A,,Aa,1,,a ? , ,,有Ab,B,0b,,B,,,a,
x11b1 dx,dx ,dx 于是22,,,aax,ba (ax,b)(ax,b)
11b1 ,d(ax,b) , d(ax,b) 222,,ax,baa(ax,b)
1b ,,ln ax,b ,,C22aa(ax,b)
1b,, ,ln ax,b ,,C,,2ax,ba,,
- 3 -
22,,x1b ,,8. dx,ax,b,2b,ln ax,b , ,C,,,23ax,b(ax,b)a ,,
2xb 证明: 被积函数 f(x),的定义域为{x|x,,} 2a(ax,b) 11 令ax,b,t (t,0), 则 x,t,b , dx,dt,,aa 2222x(b,t)b,t,2bt ? ,,22222(ax,b)atat
2222 xb,t,2btb112b1 ? dx,dt,dt,dt,dt ,,,,,2323233t(ax,b)atataa 2b12b ,,,,t,,ln t ,C333ataa 21b ,(t,2b,ln t ,),C3 ta
22 ,,x1b,, 将t,ax,b代入上式得:dx,ax,b,2b,ln ax,b , ,C ,23,,ax,b(ax,b)a,,
dxax,b11.,,ln||,C 9 ?22,bax,bxxax,bb()()
b1f(x),x|x,, 证明:被积函数 的定义域为{}2axax,b()
ABD1,,, 设: 2xax,bxax,bax,b()()2
1,Aax,b,Bxax,b,Dx 则 ()()2
,Aax,Ab,2Aabx,Bax,Bbx,Dx2222
2,xAa,Ba,x2Aab,Bb,D,Ab ()()22
1,A,2,b,Aa,Ba,0,2a,,? 2Aab,Bb,D,0 , B,, 有 ,,b2,,Ab,1,2a,D,,,b,
dxaa1111,dx,dx,dx 于是 222,,,,xax,bxax,bbbbax,b()()
1111 ,,ln|x|,,ln|ax,b|,?,C22bax,bbb
ax,b11 ,,?ln||,C2bax,bxb()
- 4 -
ax,b(二)含有的积分(10~18)
2310. ax,b dx,,(ax,b),C ,3a
111,11122 证明:ax,b dx,(ax,b)d(ax,b),,,(ax,b),C ,,1aa1, 2 23 ,,(ax,b),C3a
2311. xax,b dx,,(3ax,2b),(ax,b),C,2 15a
22t,b2tt,b 证明: 令ax,b,t (t,0) , 则x, , dx,dt , xax,b,,t aaa 2t,b2t242 ? xax,b dx,,t,dt,(t,bt)dt,,,2aaa
22b22b5353 ,dt,dt,,t,,t,C,,22225a3a5a3a 32t2 ,(3t,5b),C215a
2 3 将t,ax,b代入上式得:xax,b dx,[3(ax,b),5b],(ax,b),C2, 15a
23 ,,(3ax,2b),(ax,b),C2 15a
22222312. xax,b dx,,(15ax,12abx,8b),(ax,b),C3, 105a
2t,b2t 证明: 令ax,b,t (t,0) , 则x, , dx,dt , aa
22523 (t,b)t,bt,2bt2 xax,b,,t,22aa 22523 ? xax,b dx,t,(t,bt,2bt)dt3,,a 222b4b624 ,tdt,tdt,tdt333,,,aaa 2212b14b1 6,11,24,1 ,,,t,,,t,,,t,C333 1,61,21,4aaa
2 22b4b735 ,,t,,t,,t,C333 7a3a5a
3 2t422 ,,(15t,35b,42bt),C 3105a 将t,ax,b代入上式得:
2232222 ,, xax,b dx,,(ax,b) 15ax,15b,30abx,35b,42b,(ax,b)3,105a
22223 ,,(15ax,12abx,8b),(ax,b),C3105a
- 5 -
x2 13. dx,,(ax,2b),(ax,b),C2, 3aax,b
2t,b2t ax,b,t (t,0) , x, , dx,dt , 证明:令则 aa 2xt,b2t ? dx ,, dt ,,ataax,b
222 ,tdt,bdt22,,aa
212b,21 ,,,t,,t,C221,2aa
22b 3 ,,t,,t,C223aa
x22b t,ax,b dx,,(ax,b),(ax,b),,(ax,b),C将代入上式得:,22 3aaax,b
2 ,,(ax,2b),(ax,b),C2 3a
2x2 22214. dx,,(3ax,4abx,8b),(ax,b),C3,15a ax,b
2 t,b2t 证明: 令ax,b,t (t,0) , 则x, , dx,dt , aa
22xt,b12t2 ? dx ,(),, dt ,,ataax,b 2422 ,(t,b,2bt)dt3,a 224b422 ,tdt,bdt,tdt333,,,aaa
212b523 ,(t,bt,t),C353a
2t422 ,,(3t,15b,10bt),C315a
将代入上式得: t,ax,b
2x22222,, dx,,(ax,b),3(ax,b,2abx),15b,10b,(ax,b),(ax,b),C3,15aax,b
2222 ,,(3ax,4abx,8b),(ax,b),C315a
- 6 -
,1ax,b,bln C (b0),,,,bax,b,bdx,15. ,,,xaxb,2ax,b,,arctan ,C (b,0),,bb,,
2t,b2t 证明: 令ax,b,t (t,0) , 则 x, , dx,dt , aa
dx12t ? ,, dt,,2atb,xax,b,ta
2 ,dt,2t,b
21dx1x,a 1. 当 b,0 时 , dt,2dt,,222公式 21 : ,,ln ,C22,t,bt,(b)2ax,ax,a
1t,b ,,ln ,C
bt,b
dx1ax,b,b 将t,ax,b代入上式得: , ,ln ,C,xax,bbax,b,b
21 2. 当 b,0 时 , dt,2dt2,,22dxx1t,bt,(,b)公式:,,arctan,C 19 22,aax,a2t ,,arctan ,C
,b,b
dx2ax,b 将t,ax,b代入上式得: , ,arctan ,C ,,bxax,b,b
,1axbb,,,ln ,C (b,0),bax,b,bdx, 综合讨论 1 , 2 得: ,,,xax,b2ax,b,arctan C (b0),,,,,b,b,
- 7 -
dxax,badx 16. ,,,,,2bx2bx ax,bx ax,b
1ABax,b ,, , 1,Ax ,B(ax,b)证明:设则 22xx,ax,bxax,b a, A,,,A,Ba,0,,b ? , ,,有Bb,11,, B,,b,
dxa11ax,b ,,dx,dx,,,于是22 bbxx ax,bxax,b
a111 ,,dx,ax,b d ,,bbxxax,b
a1ax,b11 ,,dx,,dax,b ,,bbxbxxax,b
1 a1ax,b11a,2 ,,dx,,,(ax,b)dx ,, bbxbx2xax,b
a1ax,ba1 ,,dx,, dx,,bbx2bxax,bxax,b
ax,badx ,,,, bx2bx ax,b
ax,bdx17. dx,2ax,b,b ,,xxax,b
2t,b2t (0) , , 证明:令ax,b,tt,则x,dx,dtaa
2ax,bat2tt ? dx,, dt,2 dt,,,22xat,bt,b
222t,b,b1 2 22 ,dt,dt,bdt,,,22t,bt,b
1 ,2t,2b dt,2t,b
1 ? b取值为R , 符号可正可负 ? dt 不能明确积分,2t,b
ax,b1 ? dx,2t,2b dt,,2xt,b
1a 22,t,b,dx2,2tt,b
ax,b1a 将t,ax,b代入上式得: dx,2(ax,b), 2b,dx,,xax,b,b2ax,b
dx 2 ,ax,b,b,xax,b
- 8 -
ax,bax,badxdx,,,18. ,,2x2xxax,b
ax,b1 证明:dx,,ax,bd ,,2xx
ax,b1,,,dax,b , xx
1 ,ax,ba12,,,,ax,b,dx () ,xx2 ax,badx ,,, ,x2xax,b
22x,a(三)含有的积分(19~21)
dx1x 19. ,,arctan,C,22 aax,a
ππ2 证明: 令 x,a,tant (,,t,) , 则 dx,d(a,tant),a,sect dt 22
1dx1 ,, 222222 x,aa,(1,tant)asect
dx12 ? , ,a,sect dt 22,,22x,aasect 1 ,dt ,a 1 ,,t,C a x ? x,a,tant ? t,arctana 1xdxx 将t,arctan代入上式得: ,,arctan,C22,aaax,a
- 9 -
dxx2n,3dx20. ,, ,,22n222n,1222n,1(x,a)2(n,1),a,(x,a)2(n,1),a(x,a)
dxx1 证明: ,,x d,,22n22n22n(x,a)(x,a)(x,a)
x 22,n,1 ,,x,(,n),(x,a),2x dx,22n (x,a)
2 xx ,,2ndx ,22n22n,1(x,a)(x,a) 222xx,a,a ,,2ndx,22n22n,1(x,a)(x,a)
x11 2 ,,2ndx,2nadx,,22n22n22n,1(x,a)(x,a)(x,a)
dxx12 移项并整理得: (1,2n) ,,2nadx,,22n22n22n,1 (x,a)(x,a)(x,a)
,,11xdx ? dx,,(2n,1) ,,,22n,1222n,22n(x,a)2na(x,a)(x,a),,
,,dx1xdx 令n,1,n , 则 ,,(2n,3),,,22n222n,1,22n,1 (x,a)2(n,1),a(x,a)(x,a),,
x2n,3dx ,, 222n,1222n,1,2(n,1),a,(x,a)2(n,1),a(x,a)
dx1x,a21. ,,ln ,C,222ax,ax,a
dx111证明: ,[,] dx22,,2ax,ax,ax,a
1111 ,dx,dx,,2ax,a2ax,a
11 ,,ln x,a ,,ln x,a ,C2a2a
1x,a ,,ln ,C2ax,a
- 10 -
2ax,b (a,0)(四)含有的积分(22~28)
,1a ,arctan ,x,C (b,0),babdx,22. , (a,0),2, ax,b1a,x,,b,,ln ,C (b,0),2abaxb,,,, ,
证明:
11111 1. b0 当,时,,,,, 2baaaxb,b222x, x,()aa
dx11 ?,dx2,,aax,bb 22()x,a
1aa ,,,arctan,x,C abb
1a ,,arctan,x,C bab 11111 2. 当 b,0 时 , ,,,,2baaax,bb222 ()x,,x,(,)aa
dx11 dx?,2,, aax,bb22()x,, a
,bx, 11a ,,,ln ,C a,b,b 2x,aa
1a,x,,b ,,ln ,C 2,aba,x,,b
,1a ,arctan ,x,C (b,0),babdx, 1 , 2 综合讨论得:,,2,ax,b 1a,x,,b,ln C (b0),,,, 2,aba,x,,b,
x1223. dx,,ln ax,b ,C (a,0) ,22aax,b x112 dx,dx证明: ,,222ax,bax,b 112 ,d(ax,b)2,2aax,b
12 ,,ln ax,b ,C2a
- 11 -
2xxbdx24. dx,, (a,0) ,,22aaax,bax,b 22xbax1 证明: dx,,dx ,,22abax,bax,b b11 ,(,)dx,2abax,b
b1b1 ,dx,dx,,2abaax,b
xbdx ,,2,aaax,b
2 dxx1lna025. ,,,C (,) 2b2xaxb,2(,)axb,
dxxdx ,222xaxbxaxb,,证明:(,)(,)
112dx ,22xaxb,2(,)
AB1 ,,2222xaxbxaxb设:(,),
222AaxbBx xAaBAb 1,(,),,(,),
则
1,A,,AaB0,,,,b ? ,,,Aba,1有,,B,,,b,
dxa112dx ,[,]222,,xaxbbxbaxb于是2(,)(,)
a11122 dxdx,,2bb2,,xaxb22,
111122 dxdaxb,,(,)22bb,,xaxb22,
1122?lnx? ln axbC , ,,,
bb22
2x1?ln ,,C 2b2axb,
- 12 -
dxadx1 ,,,a,26. (0),,222bxbxax,bax,b()
AB1,, 222证明:设:2xax,bxax,b() 222,Aax,b,Bx ,xAa,B,Ab 1()()则
,1A, ,Aa,B,0,,b ? , ,,有Ab,a1 ,,B,,, b,
dxa1,,dx [] ,,22于是22xax,bbxbax,b()()
a 111 ,dx,dx ,,2 2bbxax,b
adx1 ,,, 2,bxbax,b
2ax,b1dxa 27. (),ln,,Ca,03222,2()22 xax,bbxbx
dxx 证明:,dx 34,,22()()xax,bxax,b
11 2 ,dx4 ,22()xax,b
1ABC 设: ,,,44222()xax,bxxax,b
24 22 则 1()(),Axax,b,Bax,b,Cx 42 ()(),Aa,Cx,Ab,Bax,Bb
1,B, ,b0Aa,C, ,,
a,, 有 0?Ab,Ba, , A,,,,2 b,,1Bb,2, ,aC, ,2b, 21111dxaa 222 于是 ,,dx,dx,dx32242 222,()2,,2,bxax,bbxxbax,b
1aa22 ,,?lnx,,? ln ax,b,C222222bbxb
2ax,b1a ,ln,,C22222bxbx
- 13 -
dxxdx1.,,a,028 (),,222bax,bbax,bax,b2()2()
dx111111,,d,,,,d ,,,2222axaxaxax,bax,bax,bax,b222()2证明:1111 ,,,,,dx ,22axax,bax,bax222 AB122,,,Aax,b,Bax,(Aa,Ba)x,Ab 1()2 2222axax,baxax,b2()2 2设:,则,1A, ,2Aa,Ba,,,20b2? , ,,Ab,11,, B,,,有b,2
111,,,,dx (),222axax,babxbax,b2()22()
于是上式11111,,,dx,dx ,,222abbaxax,bxbax,b222()2()
2ax,b,b111111,,,,dx,,dx ,,2222 abxbbaxax,bbax,babxax,bbax,b2222()2()2()2()
xdx1 ,, 22,bbax,bax,b22()
2ax,bx,c (a,0)(五)含有的积分(29~30)
22axb,, 2,arctan ,C (b,4ac),22 4ac,b4ac,bdx,29. , (a,0), ,22ax,bx,c12ax,b,b,4ac,2,ln ,C (b,4ac) ,22b,4ac2ax,b,b,4ac,
1 222,,证明:? ax,bx,c,(2ax,b),(4ac,b) 4a
dx1 ? ,4adx,,222 ax,bx,c(2ax,b),(4ac,b)
dx12当时, 1. b,4ac ,4adx ,,2222ax,bx,c(2ax,b),(4ac,b)
4a1dxx1 ,d(2ax,b)公式:,,arctan ,C 19 ,22,2222aaax,a(2ax,b),(4ac,b)
22ax,b arctan C,,, 224ac,b4ac,b
dx12 2. 当 b,4ac 时 , ,4adx ,,222ax,bx,c(2ax,b),(4ac,b) 1 4adx,,22(2ax,b),(b,4ac)
4a1 dxxa1, ,d(2ax,b)公式:ln,21 ,, ,C22222,2a (2axb)(b4ac),,,axaxa2,,
212ax,b,b,4ac ln C,,,22b,4ac2ax,b,b,4ac
22axb,,2,arctan ,C (b,4ac),224ac,b4ac,bdx,综合讨论得: 1 , 2 ,,,22ax,bx,c12ax,b,b,4ac2,ln C (b4ac),,,22, - 14 - b,4ac2ax,b,b,4ac,
x1bdx230. dx,,ln ax,bx,c , (a,0)22,, 2a2aax,bx,cax,bx,c
x12ax,b,b 证明: dx, ,dx22,, 2aax,bx,cax,bx,c
12ax,b1,b ,dx,dx22,, 2a2aax,bx,cax,bx,c
11b12 ,d(ax,bx,c),dx 22,,2a2aax,bx,cax,bx,c 1bdx2 ,,ln ax,bx,c , 2,22aaax,bx,c
22x,a (a,0)(六)含有的积分(31~44)
dxx22 31. ,arsh,C,ln (x,x,a),C (a,0),122a x,a
1 f(x),{x|x,R}证明:被积函数的定义域为 22x,a ππ22 x,a tant (,,t,), dx,d(a tant),a sectdt , x,a,|a sect|可令则222
ππ122 ,,t, , sect,,0 , ? x,a,a sect?22cost
dx1 ? ,,a sect dt2,,22a sect 公式 87 : sectdt,ln|sect,tant|,Cx,a, ,sect dt,
,ln sect,tant ,C 2
22 RtΔABC,B,t , |BC|,a ,|AC|,x , |AB|,x,a在中,设则 221x,ax ? sect,, , tant, costaa
dx ? ,ln sect,tant ,C2,22 x,a
22x,a,x ,ln ,C2a
22 ,ln x,a,x ,lna,C2
22 ,ln x,a,x ,C3
22 x,a,x,0 ? dx22 ? ,ln (x,x,a),C,22x,a
- 15 -
dxx32. ,,C (a,0) ,223222(x,a)ax,a
1 f(x),{x|x,R}证明:被积函数的定义域为223 (x,a)
ππ223233 x,a tant (,,t,), dx,d(a tant),a sectdt , (x,a),|a sect| 可令则22 ππ122333 ,,t, , sect,,0 , ? (x,a),a sect?22cost
dx111 2 ? ,,a sect dt ,dt2,,,22333secta secta (x,a)
11 ,costdt,sint,C 22,aa 22 RtΔABC,B,t , |BC|,a ,|AC|,x , |AB|,x,a在中,设则
|AC|x ? sint,,22 |AB|x,a
dx1x ? ,,sint,C,,C 2,223222a(x,a)ax,a
x22 33. dx,x,a,C (a,0),22x,a
2222 证明: 令x,a,t (t,0) , 则x,t,a
1,1t222 ? dx,(t,a),2tdt,dt222t,a
22 xt,at ? dx,,dt,,2222 tx,at,a
,dt,t,C,
x 2222 将t,x,a代入上式得: dx,x,a,C,22 x,a
x1 34. dx,,,C (a,0),22322 (x,a)x,a
33,,x12222222 证明: dx,x,(x,a)dx,(x,a)dx ,,,2232(x,a) 3, 122222 ,(x,a)d(x,a), 2
3 ,111222 ,,,(x,a),C 321, 2
1 ,,,C22x,a
- 16 -
22 xxa222235. dx,,x,a,ln (x,x,a),C (a,0) ,2222x,a 2222xx,a,a 证明: dx, dx,,2222 x,ax,a
1222 ,x,adx,adx ,,22x,a 2xa222222 ? x,adx,,x,a,,ln (x,x,a),C (公式39),22
122 dx,ln (x,x,a),C (公式31),22 x,a
22 xxa2222222 ? dx,,x,a,ln (x,x,a),a,ln (x,x,a),C ,2222x,a 2xa2222 ,,x,a,,ln (x,x,a),C 22
2 xx2236. dx,,,ln (x,x,a),C (a,0), 22322(x,a)x,a 2x f(x),{x|x,R}证明:被积函数的定义域为223 (x,a)
222ππxatant2 x,a tant (,,t,), dx,d(a tant),a sectdt ,可令则, 3322322|a sect|(x,a)
22 ππ1xtant ,,t, , sect,,0 , ? ,? 322322costa sect(x,a) 2222xtanttantsect,1 2 ? dx,, a sectdt ,dt,dt,,,,3223sectsecta sect (x,a)
1 ,sectdt,dt,sectdt,costdt ,,,,sect ,ln sect,tant ,sint,C公式 87 : sec t dt,ln|sect,tant |,C1,
22 RtΔABC,B,t , |BC|,a ,|AC|,x , |AB|,x,a在中,设则
22xx1x,a ? sint, , tant, , sect,,22acostax,a
2 x ? dx,ln sect,tant ,sint,C,1 223(x,a)
22x,a,xx ,ln ,,C122 ax,a
x22 ,ln x,a,x ,,lna,C122x,a
22 x,a,x,0 ?
2xx22 ? dx,,,ln (x,x,a),C ,22322(x,a)x,a
- 17 -
22dx1x,a,a 37. ,,ln,C (a,0),22a x x,x,a
2222 证明:令则 x,a,t (t,0) , x,t,a
1,1t222 ? dx,(t,a),2tdt,dt 222t,a
dx1t ? ,,dt,,222222x,x,at,t,at,a
dxxa 1,1公式:ln21 ,, ,C ,dt22,,22axaxa 2,,t,a
1t,a ,,ln ,C 2at,a
21(t,a) ,,ln ,C222at,a
222 dx1(x,a,a)22将代入上式得: t,x,a ,,ln ,C,222 222ax,a,ax,x,a
2221(x,a,a) n提示: logb,nlogb ,,ln ,Caa2 2ax
221x,a,a ,,ln,C a x
22dxx,a 38. ,,,C (a,0)2,222axx,x,a
dx11 证明: ,,d,,22222x x,x,ax,a
11 令t , (t,0) , 则x, xt
111t ? ,d,,dt,,dt,,, 2222x1x,a1,at2,a 2t 212at ,,dt2,222a1,at
11 22 ,,d(1,at)2,22 2a1,at
1 ,111222 ,,,(1,at),C 212a1, 2 122 ,,,1,at,C 2a
221dxx,a 将t,代入上式得: ,,,C 2,222xaxx,x,a
- 18 -
2xa22222239. x,adx,x,a,,ln (x,x,a),C (a,0), 22
222222 证法1: ? x,adx,xx,a,x dx,a,,
2x22 ,xx,a,dx ,22x,a
2 x2222 ? x,adx,dx,xx,a ?,, 22x,a 22xa 22 又 x,adx,dx,dx ?,,,2222 x,ax,a
222122 ,a,ln (x,x,a),C公式31:dx,ln(x,x,a),C (a,0) ,221x,a 2222222 由?,?得, 2x,adx,xx,a,a,ln (x,x,a),
2 xa222222 即 x,adx,x,a,,ln (x,x,a),C ,2 22xa22222239. x,adx,,x,a,,ln (x,x,a),C (a,0) ,22 ππ222 证法 2: 令x,a,tant (,,t,) ,则x,a,a1,tant,a?sect,
22
ππ122 ? ,,t, , sect,,0 , ? x,a,a?sect
22cost
222 ?x,adx,a?sectd(a?tant),asectdtant,,, 2222提示: 1,tant,sect tantdsect ,asect,tant,a ? ,
2sint 又tantdsect ,tant,sect,tantdt,dt,,,3cost 21,cost111 ,dt,,dt,dt,,,32costcost costcost
,sectdtant,sectdt ?,,
1222 联立??有asectdtant,(asect?tant,asectdt) ?,,2
又sectdt,ln|sect,tant|,C (公式 87 ) ?1, 11222 联立??有asectdtant,asect,tant,aln|sect,tant|,C ?2,22
?x,a,tant , ? 在RtΔABC中,可设 ,B,t,|BC|,a ,
22 则|AC|,a?tant,x,|AB|,a,x
221a,xx ?sect,,,tant, costaa
22211xax,x,a2222 ?asect?tant,aln|sect,tant| ,,x,a,,ln|| 2222a 22xaa2222 ,,x,a,,ln (x,x,a),,lna
222
2xa222222 综合?????得 x,adx,,x,a,,ln (x,x,a),C ,22
- 19 -
x3 40. (x,a) dx,,(2x,5a)x,a,,a,ln (x,x,a),C (a,0)2232222422,88
f(x),(x,a){x|x,R}223 证明:被积函数的定义域为
ππ223 x,a tant (,,t,), (x,a),|a sect|33可令则 22
ππ1223 ,,t, , sect,,0 , ? (x,a),a,sect33 ?22cost
2234 ? (x,a) dx,a,sect d (a tant),asect d tant333 ,,,
44 ,asec3t,tant,atant d sec3t,
4432 ,asect,tant,atant,3,sect,sect tant dt ,
44323 ,asect,tant,3atant,sect dt,
4432 ,asect,tant,3atant,sect d tant,
4432 ,asect,tant,3a(sect,1),sect d tant ,
44433 ,asect,tant,3asect d tant,3asect d tant,,
1 44433 asect d tant,(asect tant,3asect d tant) 移项并整理的:?,,4
sect d tant,sect ,tant,tant d sect?,, 2 ,sect ,tant,tant,sectdt ,
2 ,sect ,tant,(sect,1),sectdt,
3 ,sect ,tant,sect dt,sect dt ?,,
3 sect d tant,sect dt 又??,,
11 43公式 87 : sec t dt,ln|sect,tant |,C asect d tant,,sect ,tant,sect dt ,联立??得:,, 22
11 ,,sect ,tant,ln sect,tant ,C ?1 22
133444433 asect d tant,asect,tant,asect,tant,a,ln sect,tant ,C联立??得1, 488
22 RtΔABC,B,t , |BC|,a ,|AC|,x , |AB|,x,a在中,设则
22 x1x,a ? tant, , sect,, acosta
42242222axx,a3ax,ax3x,a,x42243 ? asect d tant,,,,x,a,,,,a,ln ,C1,34a8aa8aa
24 x3a,x3a22222222 ,(x,a)x,a,,x,a,,ln x,a,x ,C488
x32232222422 ? (x,a) dx,,(2x,5a)x,a,,a,ln (x,x,a),C,88
- 20 -
12222341. x,x,adx,(x,a),C (a,0) ,3 1122222 2 证明: x,x,adx,(x,a)dx,,2 11 22222 ,(x,a)d(x,a), 2
1 ,111222 ,,,(x,a),C 121, 2
1223 ,(x,a),C3
- 21 -
xa4 42. x,x,a dx,,(2x,a)x,a,,ln (x,x,a),C (a,0)222222222, 88
f(x),x,x,a{x|x,R}222证明:被积函数的定义域为 ππ x,a tant (,,t,), x,(x,a),atant|a sect|2222可令则222
ππ1 ,,t, , sect,,0 , ? x,(x,a),atant,sect 2223?222cost
222344 ? x,(x,a) dx,atant,sect d (a tant),atant,sect d tant,atant,sect dt2223,,,,
4 ,atant,sect dsect2 ,
444 ,atant,(1,tant) dsect,atant dsect,atant dsect23,,,
444 ,atant dsect,a,tant,sect,asect dtant 33,,
4443 ,atant dsect,a,tant,sect,3asect tant dt32,,
4442 ,atant dsect,a,tant,sect,3asect tant dsect3,, 1444 atant,sect dsect,(atant dsect,a,tant,sect) 移项并整理的:23,,4 44aa ,tant dsect,,tant,sect ?3,44
tant d sect,sect ,tant,sect dtant ,sect ,tant,sectdt?3,,, ,sect ,tant,(1,tant),sectdt2,
,sect ,tant,sect dt, tant sect dt2,,
,sect ,tant,sect dt, tant d sect ,,
11 tant d sect ,,sect ,tant,sect dt 移项并整理得:公式 87 : sec t dt,ln|sect,tant |,C ,,,22
11 ,,sect ,tant,ln sect,tant ,C ? 122 444aaa4 23 atant,sect dsect,sect,tant,ln sect,tant ,tant,sect,C联立??得:,1884
22 RtΔABC,B,t , |BC|,a ,|AC|,x , |AB|,x,a在中,设则
22x1x,a ? tant, , sect,, acosta
4224224322axx,aax,a,xaxx,a42 ? atant,sect dsect,,,,,ln ,,,,C1,38aa8a4aa
443axax222222 ,,x,a,,ln x,a,x ,,x,a,C2884
4xa222222 ,,(2x,a)x,a,,ln x,a,x ,C88
44aa222222 x,a,x,0 ? ,ln x,a,x , ,ln (x,x,a)?,88
4xa222222222 ? x,x,a dx,,(2x,a)x,a,,ln (x,x,a),C,88
- 22 -
2222x,ax,a,a2243. dx,x,a,a,ln,C (a,0)x x ,
22 x,a f(x),{x|x,0} x证明:被积函数的定义域为
2222 x,a,t (t,0t,a) , x,t,a令且则 11t 222 ? dx,(t,a),2tdt,dt222t,a ,
222x,attt ? dx,,dt ,dt 22x2222t,at,at,a,,, 222t,a,a12 ,dt ,dt,adtdxxa1,2222t,at,a公式:ln21 ,, ,C ,,,22,axaxa2,,21t,aa(t,a) 2 ,t,a,,ln ,C ,t,,ln ,C222at,a2 t,a
22222x,aa(x,a,a)2222 t,x,a dx,x,a,,ln ,C222x2将代入上式得:x,a,a,
22(x,a,a)222222 ,x,a,a,ln ,Cx,ax,a 22x44. dx,,,ln (x,x,a),C (a,0),2 xx
22 x,a,a2222x,a ,x,a,a,ln,C 证明: 被积函数 f(x),的定义域为{x|x,0} x 2x π 1. 当x,0时 ,可令x,a tant (0,t,), 则dx,d(a tant),a sectdt ,22
2222 a sect x,aπ1x,asect , , ? 0,t, , sect,,0 , ? ,22 2costxa tantxa tant
222 22x,asectsect ? dx,, a sectdt ,,(1,tant)dt ,,,2xa tanttant22 22sect1cost ,sectdt,,dt ,sectdt,,dt2,,,,costtantsint 22cost1 ,sectdt,dt,sectdt,dsint,,,,sintsint 221 ,ln sect,tant ,,C sint1 22公式 87 : sec t dt,ln|sect,tant |,C 在RtΔABC中,设,B,t , |BC|,a ,则|AC|,x , |AB|,x,a, 22xx1x,a ? sint, , tant, , sect,,22acosta x,a
222222x,ax,a,xx,a ? dx,ln ,,C,2axx 1
22x,a22 ,,,ln x,a,x ,lna,C x 12222x,ax,a 2222 ? x,a,x,0 ? dx,,,ln (x,a,x),C,2xx
2222 x,ax,a22 2. 当x,0时 , 同理可证得: dx,,,ln (x,a,x),C ,2xx 2222x,ax,a22 综合讨论1 , 2 得: dx,,,ln (x,a,x),C,2x x- 23 -
22x,a (a,0)(七)含有的积分(45~58)
dxx|x|2245. ,,arsh ,C,ln|x,x,a|,C (a,0) 1,22|x|ax,a 1 证法1 : 被积函数 f(x),的定义域为 {x|x,a或x,,a}22x,a
π 1 . 当 x,a时 , 可设x,a,sect (0,t,) , 则dx,a,sect,tantdt 2
π22222 x,a,asect,1,a,tant ?0,t, , x,a,a,tant 2
dxa,sect,tant ?,dt,sectdt,,, 公式 87 : sectdt,ln|sect,tant|,C22a,tant,x,a ,ln|sect,tant| ,C2
22 在Rt ΔABC中,可设 ,B,t ,|BC|,a , 则|AB|,x ,|AC|,x,a
221x|AC|x,a ? sect,,, tant,, costa|BC|a 22dxx,x,a ?,ln|sect,tant|,ln||,22a x,a
22 ,ln|x,x,a|,C3 2 . 当 x,,a ,即,x,a时,令 μ,,x ,即x,,μ
dxdμ22 由讨论 1可知 ,,,,ln|μ,μ,a|,C4,,2222 x,aμ,a
122 ,,ln|,x,x,a|,C ,ln,C 4422|,x,x,a|
22|,x,x,a| ,ln,C42 a
22 ,ln|,x,x,a|,C5 dxx|x|22 综合讨论 1 , 2 ,可写成 ,,arsh ,C,ln|x,x,a|,C 1,22|x|ax,a
- 24 -
dxx|x|2245. ,,arsh ,C,ln|x,x,a|,C (a,0),1 22|x|ax,a 1 证法2 :被积函数 f(x),的定义域为 {x|x,a或x,,a} 22x,a
x 1 . 当x,a时 , 可设x,a,cht (t,0) , 则t,archa
22222 x,a,acht,a,a,sht , dx,a,shtdt
dxa,sht ?,dt,dt,t,C,,,1 22a,shtx,a 2,,xxx ,,,, ,arch,C,ln ,,1,C,,2 aaa,,,,,, 22 ,ln|x,x,a|,C 3
2 . 当x,,a ,即,x,a时,令 μ,,x ,即x,,μ
dxdμ22 由讨论 1可知 ,,,,ln|μ,μ,a|,C4,, 2222x,aμ,a 122 ,,ln(,x,x,a),C ,ln,C4422|,x,x,a|
22 |,x,x,a| ,ln,C4 2a
22 ,ln|,x,x,a|,C5
dxx|x|22 综合讨论 1 , 2 ,可写成 ,,arsh ,C,ln|x,x,a|,C 1,22|x|a x,a
- 25 -
dxx 46 (0).,,,C a,,223222()x,aa,x,a
1 证明 : 被积函数 的定义域为 {或}f(x),x|x,ax,,a 223()x,a
π 1 . 当 时 可设 (0) , 则 x,a,x,a,sect,t,dx,a,sect,tantdt2
π2233322333 () ?0 , 0 , () x,a,a,tant ,t,tant,x,a,a,tant2
1dxa,sect,tantsect ?,dt,dt,,,3323223a,tantatant ()x,a
2111costcost ,,dt,dt ,,2222costasintasint 11 ,dsint,22asint 1 ,,,C2asint
22 在Rt 中,可设 , 则 ΔABC,B,t,|BC|,a|AB|,x,|AC|,x,a
22x,a ?sint, x dxx ?,,,C,223222()x,aa,x,a
2 . 当 即时,令 即x,,a,,x,aμ,,x,x,,μ
dxdμ ?,,,, 223223()()x,aμ,a
dμμ 由讨论 1可知 ,,,C,223222()()μ,aa,μ,a
dxx 将代入得: μ,,x,,,C, 223222()x,aa,x,a
dxx 综合讨论 1 2 得: ,,,,C,223222()x,aa,x,a
x 2247. dx,x,a,C (a,0),22 x,a
1,x12222 dx,(x,a)dx 证明:,,222x,a 1,1 22222 ,(x,a)d(x,a),2
1 1,11222 ,,(x,a),C 121, 2
22 ,x,a,C
- 26 -
x1 48. dx,,,C (a,0),22322(x,a)x,a
x 证明: 被积函数 f(x),的定义域为 {x|x,a或x,,a} 223(x,a)
π 1 . 当 x,a时 , 可设x,a,sect (0,t,) , 则dx,a,sect,tantdt 2
xa,sect πxsect , ? 0,t, , , 23332232232a,tanta,tant (x,a)(x,a)
xsect ? dx,,a,sect,tant dt,,23 223a,tant(x,a) 21sect 11 ,dt,dt,,22aatantsint
11 2 ,,,csctdt,,,cott,C, aa
22 在Rt ΔABC中,可设 ,B,t ,|BC|,a , 则|AB|,x ,|AC|,x,a
a ? cott, 22x,a
x1a1 ? dx,,,,C,,,C,2232222a (x,a)x,ax,a
2 .当 x,,a ,即,x,a时,令 μ,,x ,即x,,μ
xμ ? dx,dμ ,,223223(x,a)(μ,a)
μ1 由讨论 1可知 dμ,,,C,22322 (μ,a)μ,a
x1 将μ,,x代入得: dx,,,C ,22322(x,a)x,a
x1 综合讨论 1 , 2 得: dx,,,C ,22322 (x,a)x,a22xxa2222 49. dx,x,a,,ln x,x,a ,C (a,0),2222x,a
2222 xx,a,a 证明: dx,dx,, 2222x,ax,a
2a22 ,(x,a,)dx,22x,a
1222 ,x,adx,adx,,22x,a
2xa222222 ? x,a dx,,x,a,,ln x,x,a ,C ? (公式53),22
dx2222 a ,a,ln x,x,a ,C ? (公式45),22x,a
22xxa2222 ? 由?,?得: dx,x,a,,ln x,x,a ,C,2222x,a
- 27 -
2xx2250. dx,,,ln x,x,a ,C (a,0),32222(x,a)x,a
2x 证明: 被积函数 f(x),的定义域为 {x|x,a或x,,a} 322(x,a)
π 1 . 当 x,a时 , 可设x,a,sect (0,t,) , 则dx,a,sect,tantdt 2
22222 xa,sect πxsect , ? 0,t, , ? , 3 333322222a,tanta,tant (x,a)(x,a)
2232xsect sect 1cost1 ? dx, ,a,sect,tant dt,dt,,dt,dt,,,,,32322322a,tanttantcostsintsint,cost(x,a)
cost111 ,dt,dsint,(,)dsint,,, 222222sint,costsint (1,sint)sint 1,sint
1111 ,dsint,dsint,dsint,dsint,,,, 2222sint 1,sintsint sint,1 1111 ,dsint,(,)dsint ,,22sint,1sint,1sint 11111 ,dsint,d(sint,1),d(sint,1),,,22sint,12sint,1sint
111 ,,,ln sint,1 ,ln sint,1 ,C1sint22
211sint,111(sint,1) ,,,ln ,C,,,ln ,C112 sint2sint,1sint2sint,1
211(sint,1)1 ,,,ln ,(,1) ,C,,,ln tant,sect ,C 122sint2sintcost
22 在Rt ΔABC中,可设 ,B,t ,|BC|,a, 则|AB|,x ,|AC|,x,a
2222x,ax,ax ? sint, , tant, , sect, xaa 222xxx,x,ax 22 ? dx,,,ln ,C ,,,ln x,x,a ,C,23222222a (x,a)x,ax,a
2 .当 x,,a ,即,x,a时,令 μ,,x ,即x,,μ
22xμ ? dx,,dμ,,332222(x,a)(μ,a)
2μμ22 由讨论 1可知 ,dμ,,ln μ,μ,a ,C,32222(μ,a)μ,a
2xx22 将μ,,x代入得: dx,,,ln ,x,x,a ,C,32222(x,a)x,a
22222xx(x,a,x)(x,a,x) ,dx,,,ln ,C,3222222(x,a)x,ax,a,x
222xx,a,xx1 ,,,ln ,C,,,2lna,ln ,C22222222x,a(x,a,x)x,ax,a,xn提示:logb,nlogbaax22 ,,,ln x,x,a ,C22x,a
2xx22 综合讨论 1 , 2 得: dx,,,ln x,x,a ,C,32222(x,a)x,a - 28 -
dx1a51. ,,arccos ,C (a,0) ,22a|x|xx,a
1 证法 1 :被积函数 f(x),的定义域为 {x|x,ax或,,a}22 xx,a
π 1 . x当,a , 时可设x,a,sect (0,t,) , 则
2 22222 xx,a,a,sectsect,1,asect,tant , dx,a,sect,tant d t
dxa,sect,tant1 ? ,dt,dt 2,,,22aasect,tantxx,a
1 ,t,C1 a
aa x?,a,sect, ? cost,, ? t,arccos xx
dx1a ? ,,arccos ,C,22ax xx,a
2 . x当,,a ,,即x,a时,令 μ,,x ,x即,,μ
dxdμ1a 由讨论 1可知 ,,,arccos ,C2,,2222aμ xx,aμμ,a
1a ,,arccos ,C a,x
dx1a 综合讨论 1 , 2 ,可写成 ,,arccos ,C ,22a|x| xx,a
- 29 -
dx1a51. ,?arccos ,C (a,0), 22a|x|xx,a
1 证法2 :被积函数 f(x),的定义域为 {x|x,a或x,,a}22 xx,a
1 . 当x,a 时, 可设x,a,cht (0,t) , 则
222 xx,a,a,cht,a,sht,acht,sht , dx,a,sht dt
dxa,sht11 ? ,dt,,dt,,, 22a,cht,shtachtxx,a
1cht11 ,dt,dsht22,,aacht1,sht
1dx1x ,,arctan (sht ),C公式 19 : ,arctan ,C22, aax,aa
22xx,a2 ?x,a,cht, ? cht,, ? sht,1,cht,
aa
22 x,a 在RtΔABC中,设 tany,sht,,,B,y ,|BC|,a a 2222 ? y,arctan (sht ),|AC|,x,a,|AB|,|AC|,|BC|,x
|BC|a ? cosy,, |AB|x
a 即cosy,cos arctan (sht ),
x
a ? arctan(sht ),arccos ,C
x
dx11a ? ,,arctan (sht ),C,,arccos ,C, 22aaxxx,a
2 . 当x,,a ,即,x,a时,令 μ,,x ,即x,,μ
dxdμ1a 由讨论 1可知 ,,,arccos ,C2,,2222 aμxx,aμμ,a
1a ,,arccos ,C
a,x
dx1a 综合讨论 1 , 2 ,可写成 ,,arccos ,C ,22 a|x|xx,a
- 30 -
22 dxx,a52. ,,C (a,0), 2222axxx,a 1 证明 : 被积函数 f(x),的定义域为 {x|x,a或x,,a} 222xx,a 31111t 1 . 当 x,a时 , 可设x, (0,t,) , 则 dx,,dt , ,222222tat xx,a1,at
3 dxt1 ? ,, (,)dt,,2 22222txx,a1,at 1,t12222 ,,dt ,,(1,at)dt,,2221,at 11,,1111 22222222 ,(1,at)d(1,at) ,,,(1,at),C,221 2a2a1,2
22 1,at ,,C2 a
2211dx111x,a22 将x, , 即t,代入上式得: ,,1,a(),C ,,,C ,222222txxaaxxx,a 221x,a ,,,C2 x a
22dxx,a ? x,a,0 ? , ,C ,2222axxx,a
2 . 当 x,,a ,即,x,a时,令 μ,,x ,即x,,μ
22 μ,adxdμ 由讨论 1可知 ,,, ,,C,,2 222222aμxx,aμμ,a
22dxx,a 将μ,,x代入上式得: , ,C,2222axxx,a
22 dxx,a 综合讨论 1 , 2 得: , ,C 2, 222axxx,a
- 31 -
2xa 22222253. x,a dx,x,a,,ln x,x,a ,C (a,0),22
22 证明: 被积函数 f(x),x,a的定义域为 {x|x,a或x,,a}
π22 1 . 当 x,a时 , 可设x,a,sect (0,t,) , 则x,a, a,tant 2 π22 ? 0,t, , ? x,a,a,tant 2 222 ? x,a dx,a,tant d (a,sect),atant dsect,,,
22 ,a,tant,sect,asect dtant, 223 ,a,tant,sect,asect dt ,
222 ,a,tant,sect,asect (1,tant) dt,
2222 ,a,tant,sect,asect dt,asect tant dt ,,
222 ,a,tant,sect,asect dt,atant dsect,,
222 ,a,tant,sect,a,ln sect,tant ,atant dsect,
22aa2 移项并整理得: atant dsect , ,tant,sect,,ln sect,tant ,C,122
22 在Rt ΔABC中,可设 ,B,t ,|BC|,a , 则|AB|,x ,|AC|,x,a
22x,ax ? tant, , sect,aa
222 ? x,a dx, atant dsect,,
222222ax,axax,a,x ,,,,,ln ,C12aa2a
2xa2222 ,x,a,,ln x,x,a ,C 22
π 2 .当 x,,a时,可设x,a,sect (,,t,0) 同理可证2
2xa222222 综合讨论 1 , 2 得: x,a dx,,x,a,,ln x,x,a ,C,22
- 32 -
x3223222242254. (x,a) dx,,(2x,5a)x,a,,a,ln x,x,a ,C (a,0), 88
33 223222222 (x,a) dx,x,(x,a),xd (x,a) 证明:,,
31 3222222 ,x,(x,a),x,,(2x),(x,a)d x ,2 312222222 ,x,(x,a),3x(x,a)d x, 312222222 22 ,x,(x,a),3(x,a,a)(x,a)d x, 331 2222222222 ,x,(x,a),3(x,a)d x,3a(x,a)d x,, 312x3a 223222222 (x,a) dx,,(x,a),(x,a)d x 移项并整理得:?,, 44
12 xa2222222 (x,a)d x,x,a,,ln x,x,a ,C ( 53) 又公式? ,22 联立??得: 3x3x322322222422 2 (x,a) dx,(x,a),,a,x,a,,a,ln x,x,a ,C,488
32 xax3x322222422 ,(,)x,a,,a,x,a,,a,ln x,x,a ,C 4488
x32222422 ,,(2x,5a)x,a,,a,ln x,x,a ,C 88
1 2222355. xx,a dx,(x,a) ,C (a,0),3
122222 证明: xx,a dx,x,a dx,, 2
1 122222 ,(x,a) d (x,a) ,2 11,11222 ,,,(x,a),C12 1,2
1223 ,(x,a) ,C3
- 33 -
4222222222xa56. xx,a dx,,(2x,a)x,a,,ln x,x,a ,C (a,0),88
222 证明: 被积函数 f(x),xx,a的定义域为{x|x,a或x,,a}
2222π 1.当 x,a时, 可令x,a, sect (0,t,), 则 xx,a,asect|a tant|22
π2223 ? 0,t, , tant,0 , ? xx,a,asect,tant 2 222234 ? xx,a dx,asect,tant d (a sect),asect d tant dt ? ,,,
232 44aa23 ,sect,3,sect,tant dt,sect dtant,,332 44aa33 ,,sect,tant,tant dsect,33 44 aa32 ,,sect,tant,tant (sect,1)dsect, 33
444 aaa32 ,,sect,tant,tant ,sectdsect,tant dsect,, 333
444aaa323 ,,sect,tant,tant ,sect dt,tant dsect ,,333 44aa43 移项并整理得: asect d tant dt ,,sect,tant,tant dsect ?,,44 32
又 tant d sect,sect ,tant,sect dtant ,sect ,tant,sectdt ,,,
3 ,sect ,tant,(1,tant),sectdt, 2 ,sect ,tant,sect dt, tant sect dt,,
2
,sect ,tant,sect dt, tant d sect ,,
11 移项并整理得: tant d sect ,,sect ,tant,sect dt ,, 22
11 ,,sect ,tant,ln sect,tant ,C ?1 22
444 aaa4 将?式代入?式得:asect d tant dt,tant,sect,sect,tant,ln sect,tant ,C 1,488323 22 在RtΔABC中,设,B,t , |BC|,a ,则|AB|,x , |AC|,x,a
22 x,a1x ? tant, , sect,, acosta
422422422axx,aax,axax,x,a422 ? asect d tant dt,,,,x,a,,,,,ln ,C1,34a8aa8aa 32
24xaxa22222222 ,,(x,a),x,a,,x,a,,ln x,x,a ,C2488
4xa222222 ,,(2x,a)x,a,,ln x,x,a ,C88
2. 当x,,a,即,x,a时,令μ,,x,则x,,μ
222222 由讨论1得: xx,a dx,,μμ,a dμ,,
4,μa222222 ,,(2μ,a)μ,a,,ln ,μ,μ,a ,C88
4xa222222222 将μ,,x代入上式得: xx,a dx,,(2x,a)x,a,,ln x,x,a ,C,88
4xa222222222 综合讨论 1,2得: xx,a dx,,(2x,a)x,a,,ln x,x,a ,C, - 34 - 88
22x,aa22 57 (0).dx,x,a,a,arccos,C a,,||xx
22x,a 证法1 : 被积函数 的定义域为 {或}f(x),x|x,ax,,ax
π 1 . 当 时 可设,x,a,x,a,sect (0,t,) 2
22x,aa,tant则 , , dx,a,sect,tant d txa,sect
22x,aa,tant,a,sect,tant2 ? dx,dt,a,tantdt,,,xa,sect
2211sint,cost ,adt,adt,adt,dt,,,,222costcostcost ,a,tant,a,t,C
aa? x,a,sect, ? cost,, ? t,arccosxx
22 在Rt中,设B|BC| ,则 ,ΔABC,,t,,a|AB|,x|AC|,x,a
22|AC|x,a ? tant,,|BC|a
22x,a ? dx,a,tant,a,t,C,x
a22 ,x,a,a,arccos ,C x
2 . 当 即时,令 即x,,a,,x,aμ,,x,x,,μ
2222μ,ax,aa22 由讨论 1可知 dx,dμ,μ,a,a,arccos ,C,,xμμ
a22 ,x,a,a,arccos ,C,x
22x,aa22 综合讨论 1 2 ,可写成: ,dx,x,a,a,arccos,C ,||xx
- 35 -
22x,aa2257. dx,x,a,a,arccos,C (a,0),x|x|
22x,a 证法 2 : 被积函数 f(x),的定义域为 {x|x,a或x,,a}x
1 . 当 x,a时 , 可设x,a,cht (0,t) ,
22x,aa,shtsht 则 , , , dx,a,sht dtxa,chtcht
222x,ashtsht ? dx,,a,sht dt,adt,,,xchtcht
2cht,1cht ,adt,achtdt,adt,,,2chtcht
1提示 : cht,sht,122 ,achtdt,adsht,,2 1,sht, (cht),sht
dx1x ,a,sht,a,arctan(sht ),C公式 19 : ,,arctan ,C, (sht),cht22,aax,a22xx,a2 ? x,a,cht, ? cht,, ? sht,1,cht,aa
22x,a 在RtΔABC中,设 tany,sht,,,B,y ,|BC|,aa
2222 ? y,arctan (sht ),|AC|,x,a,|AB|,|AC|,|BC|,x
|BC|a ? cosy,,|AB|x
a 即cosy,cos arctan(sht ),x
a ? arctan(sht ),arccos x
22x,aa22 ? dx,x,a,a,arccos,C,xx
2 . 当 x,,a ,即,x,a时,令 μ,,x ,即x,,μ
2222μ,ax,aa22 由讨论 1可知 dx,dμ,μ,a,a,arccos,C,,xμμ
a22 ,x,a,a,arccos,C,x
22x,aa22 综合讨论 1 , 2 ,可写成: dx,x,a,a,arccos,C ,x|x|
- 36 -
2222x,ax,a22 58. dx,,,ln x,x,a ,C (a,0)2,xx
22 x,a122 dx,,x,ad2证明:,, xx
22x,a122 ,,,dx,a ,xx 122, 2x,a1122 ,,,,,2x,(x,a)dx, xx2
dx 2222公式45:,ln x,x,a ,Cx,a1,22 ,,,dxx,a,22xx,a
22 x,a22 ,,,ln x,x,a ,C x
22a,x (a,0)的积分(59~72) (八)含有
dxx 59. ,arcsin,C (a,0),22aa,x
1 证明 : 被积函数 f(x),的定义域为 {x|,a,x,a}22 a,x
ππ11 ? 可设x,a,sint (,,t,) , 则dx ,a,costdt , , 2222a,costa,x
ππ11 ? ,,t, , cost,0 ? , 2222a,costa,x dx1 ? ,,a ,costdt ,,22a,costa,x
,dt ,
,t,C
x ? x,a,sint ? t,arcsin a dxx ? ,arcsin,C ,22aa,x
- 37 -
dxx60. ,,C (a,0) ,223222(a,x)a,a,x
1 证明 : 被积函数 f(x),的定义域为 {x|,a,x,a}223 (a,x)
ππ11 ? 可设x,a,sint (,,t,) , 则 dx,a,cost dt , , 3322322a,cost(a,x) ππ11 ? ,,t, , cost,0 ? ,3322322a,costa,x()
dx1 ? ,, a,cost dt ,,33223a,cost a,x()
1,dt ,22a,cost
12 ,,sect dt ,2a
1 ,,tant,C 2a
22 在Rt ΔABC中,设 ,B,t ,|AB|,a , 则|AC|,x ,|BC|,a,x x ? tant,22a,x
dxx ?,,C ,223222 (a,x)a,a,x
x22 61. dx,,a,x ,C (a,0),22 a,x
1 x1,2222 证明: dx,(a,x) dx ,,222a,x 1,1 22222 ,,(a,x) d (a,x),2
1 1,1122 ,,,,(a,x),C 121, 2
22 ,,a,x ,C
- 38 -
x1dxa62. , ,C (,0) 22322,axax(,),
3 x1,2222证明:dxaxdx ,(,) 223,,2ax(,) 31, 22222axdax ,,(,) (,), 2
3 1,11222axC ,,,,(,), 321, 2 1 , ,C 22ax, 22xxax2263. ()dx,,a,x,,arcsin,Ca,0 ,2222aa,x 2x () {|}fx,x,a,x,a证明:被积函数的定义域为22 a,x
2 22ππxa,sint () cos , ?x,a,sint ,,t,dx,a,tdt, 可设,则2222 a,cost a,x 22ππxa,sint , cos0 ,,t, t,?,?2222costa,x
22 xa,sint cos ?dx,,a,tdt,, 22costa,x 2222 ,asintdt提示: cos2t,cost,sint, 2 ,1,2sint1cos2,t 2 ,adt, sin2t,2,sint,cost2
22 aa cos2 (2),dt,tdt,, 24
22aa ,,t,,sin2t,C 24 22aa ,,t,,sint,cost,C22 22 Rt ,|| || ,||ΔABC,B,tAB,aAC,xBC,a,x 在中,设,则
22xa,x ?sint, , cost, aa 22xxax22 ?dx,,a,x,,arcsin,C ,2222aa,x
- 39 -
2xxx64. dx,,arcsin,C (a,0) ,22322a(a,x)a,x
2 x 证明 : 被积函数 f(x), 的定义域为 {x|,a,x,a} 223(a,x)
222ππxa,sint ? 可设x,a,sint (,,t,) , 则 dx,a,cost dt , ,3322322 a,cost (a,x)
2 2ππxsint ? ,,t, , cost,0 ? , 322322a,cost(a,x) 22xsint ? dx,, a,cost dt ,,322 a,costa,x
2sint , dt ,2cost 21,cost , dt,2cost 1 , dt,dt,,2cost
,d tant,dt ,,
,tant,t,C
22 在Rt ΔABC中,设 ,B,t ,|AB|,a , 则|AC|,x ,|BC|,a,x
x ? tant,22 a,x
2 xxx ? dx,,arcsin,C ,22322a(a,x)a,x
- 40 -
22dx1a,a,x65. ,,ln,C (a,0),a x 22xa,x
1 证明 : 被积函数f(x), 的定义域为 {x|,a,x,a且x,0} 22xa,x
π 1. 当,a,x,0 时, 可设x,a,sint (,,t,0) , 则 dx,a,cost dt 2
π22222 xa,x,a,sint,|a,cost| ? ,,t,0 , cost,0 ? xa,x,a,sint,cost2
dx1 ? ,,a,cost dt,,2a,sint,cost22xa,x
11 , dt,asint
1sint , dt,2asint
11 ,, dcost,a1,cost2
111 ,,( ,)dcost,2a1,cost1,cost
1111 ,, d(cost,1), d(1,cost),,2a1,cost2a1,cost
11 ,,,ln 1,cost ,,ln cost,1 ,C12a2a
1cost,1 ,,ln ,C12a1,cost
21(cost,1) ,,ln ,(,1) ,C122a1,cost
21(cost,1) ,,ln ,C22asint2
1cost,1 ,,ln ,C2asint
1 ,,ln cott,csct ,C2a
在Rt ΔABC中,设 ,B,t ,|AB|,a , 则|AC|,x ,|BC|,a,x22
a,x1a22 ? cott, , csct,,xsintxdx1a,x,a1a,a,x2222 ? ,,ln ,C,,ln ,(,1) ,C,2222axaxxa,x
1a,a,x22 ,,ln ,C3ax
? a,a,x,022
22dx1a,a,x ? ,,ln,C ,22a x xa,x
π 2. 当0,x,a 时, 可设x,a,sint (0,t,),同理可证2
22dx1a,a,x 综合讨论 1 , 2 得:,,ln,C ,22a x xa,x
- 41 -
22dxa,x66. ,,,C (a,0) ,2222axxa,x
1 证明 : 被积函数 f(x), 的定义域为 {x|,a,x,a且x,0}222xa,x
π 1. 当,a,x,0 时, 可设x,a,sint (,,t,0) , 则 dx,a,cost dt , 2
111 ,, 2 222 a,cost a,sintxa,x
2ππ11 ? ,,t, , cost,0 ? , 22222a,sint,costxa,x 32dx1 ? , , a,cost dt ,,222a,sint,cost xa,x
32 11 , dt,2 asint
2 12 ,,,csct dt,2 a
1 ,,,cott,C 2a 22 在Rt ΔABC中,设 ,B,t ,|AB|,a , 则|AC|,x ,|BC|,a,x
22 a,x ? cott, x
22dxa,x ? ,,,C ,2222axxa,x
π 2. 当0,x,a 时, 可设x,a,sint (0,t,),同理可证2
22 dxa,x 综合讨论 1 , 2 得:,,,C2, 222axxa,x
- 42 -
2xax222267 ().a,xdx,a,x,,arcsin,C a,0,22a
22 {}f(x),a,xx|,a,x,a证明:被积函数的定义域为
ππ22 () , ?x,a,sint ,,t,dx,a,costdta,x,a,cost可设,则22
ππ22 0 ,,t,, cost, ? a,x,a,cost?22
22 ? a,xdx,a,cost,a,costdt,,
22 ,acostdt,
22 1 ,a(,sint)dt,
222 ,adt,asintdt,,?2222 a,xdx,acostdt,,又
2 ,acostdsint,
22 ,a,sint,cost,asint d cost,
222 ,a,sint,cost,asintdt,?
222222 2 ,a,xdx,adt,a,sint,cost,at,a,sint,cost,,由??得:
22aa22 ?a,xdx,t,,sint,cost,C,22
22 Rt ΔABC,B,t,|AB|,a|AC|,x,|BC|,a,x在中,设,则
22xa,x ?sint, , cost,aa
2222axaa,xx22 ?a,xdx,,arcsin,,,,C,22aaa
2xax22 ,a,x,,arcsin,C22a
- 43 -
3xx2232222468. () (52) (0)a,xdx,,a,xa,x,,a,arcsin,Ca, ,88a 33223222222 证明: () () ()a,xdx,x,a,x,xda,x,, 313222222 ()(2)() ,x,a,x,x,,,x,a,xdx,2 31 2222222 ()3() ,x,a,x,xa,xdx, 31 222222222 ()3()() ,x,a,x,x,a,aa,xdx,
331 2222222222 ()3() 3() ,x,a,x,a,xdx,aa,xdx,,
312 3xa223222222 移项并整理得: () ()() ? a,xdx,,a,x,a,xdx ,,44
12xax22222 又() (公式 67) ?a,xdx,a,x,,arcsin,C ,22a 联立??得:
333xxx2232222242 () ()a,xdx,a,x,,a,a,x,,a,arcsin,C,488a
2333axxxx222224 (),,a,x,,a,a,x,,a,arcsin,C4488a
3xx22224 (52),,a,xa,x,,a,arcsin,C88a
122223 69. xa,xdx,,(a,x),C (a,0),3
22 证明 : 被积函数 f(x),xa,x 的定义域为 {x|,a,x,a}
ππ22 ? 可设x,a,sint (,,t,) , 则 dx,a,cost dt , xa,x,a,sint,|a,cost|22
ππ222 ? ,,t, , cost,0 ? xa,x,a,sint,cost 22
22232 ? xa,xdx,a,sint,cost,a,cost dt ,acost,sint dt,,,
3 a323 ,,acost dcost,,cost,C,3
33a22 ,,(1,sint),C3
ππx ? x,a,sint (,,t,) , ? sint,22a
3322322(a,x)a,x222 ? (1,sint),(),23aa
33a2222 ? xa,xdx,, (1,sint),C,3
1223 ,,(a,x),C3
- 44 -
4xax222222270. (2) (0)xa,xdx,,x,aa,x,,arcsin,Ca,,88a
222 (){|}fx,xa,xx,a,x,a证明:被积函数的定义域为 ππ2222 (), ||?x,a,sint,,t, xa,x,a,sinta, cost22可令则2
ππ2223 , cos0 , ,,t,t,?xa,x,a,sint,cost22?222234 () ?xa,xdx,asint,cost d a,sint,asint,costdt ,,,2224a 3 ,,sint,cost,costdt ,324a ,costdsint,3344aa3 cos,,t,sint,sint d cost,33344aa3 cos(1) ,,t,sint,sint,,costd cost,332444aaa3 cos ,,t,sint,sintd cost,sint,costd cost,,3332444aaa32 cos ,,t,sint,sintd cost,sint,costdt,,33322224 xa,xdx,asint,costdt ,,移项并整理得:2244aa3 cos ,,t,sint,sintd cost,44?
sintdcost,sint,cost,costdsint,,?
cos ,sint,cost,tdt,2 (1 ),sint,cost, ,sintdt ,2 ,sint,cost,dt,sintdt ,,?2 sintdcost,,sintdt ,,又??211 sintdcost,,sint,cost,dt,,22联立??得:
11 C ,,sint,cost,,t,122?
444aaa2223 cos Cxa,xdx,,t,sint,,sint,cost,,t,,488联立??得:
22 RtB , |A| ,|AC| , |BC|ΔABC,,tB,a,x,a,x在中,设则22a,xx ? cost, , sint,aa
42234224aa,xxaa,xxax222 ?xa,xdx,,,,,,,,arcsin,C3,488aaaaa
4xax2222 (2),,x,aa,x,,arcsin,C88a
- 45 -
2222a,xa,a,x2271. dx,a,x,a,ln,C (a,0),x x
1 证明 : 被积函数 f(x), 的定义域为 {x|,a,x,a且x,0}
22xa,x
π 1. 当,a,x,0 时, 可设x,a,sint (,,t,0) , 则 dx,a,cost dt 2
2222a,x|a,cost|πa,xcost , ? ,,t,0 , cost,0 ? ,xa,sint2xsint
222a,xcostcost ? dx,,a,cost dt,a dt,,,xsintsint
1,sint1 ,a dt,a dt,asint dt,,, sintsint2sint1 ,a dt,asint dt,,a dcost,asint dt,,,,2sint1,cost
a11 ,,( ,)dcost,asint dt2,,21,cost1,cost
a1a1 ,, d(cost,1), d(cost,1),asint dt,,,21,cost2cost,1
aa ,,,ln 1,cost ,,ln cost,1 ,a,cost,C122
acost,1 ,,ln ,a,cost,C121,cost
2a(cost,1) ,,ln ,(,1) ,a,cost,C1221,cost
2a(cost,1) ,,ln ,a,cost,C22sint
cost,12 ,a,ln ,a,cost,C2 sint
,a,ln cott,csct ,a,cost,C2
22 在Rt ΔABC中,设 ,B,t ,|AB|,a , 则|AC|,x ,|BC|,a,x
2222a,x1aa,x ? cott, , csct,, , cost,xsintxa
222222a,xa,x,aa,x ? dx,a,ln ,a,,C,2xxa
2222a,a,xa,x ,a,ln ,(,1) ,a,,C2xa
22221a,a,xa,x ,,ln ,a,,C3axa
22 ? a,a,x,0
2222a,xa,a,x22 ? dx,a,x,a,ln,C ,x x
π 2. 当0,x,a 时, 可设x,a,sint (0,t,),同理可证2
2222a,xa,a,x22 综合讨论 1 , 2 得: dx,a,x,a,ln,C ,x x
- 46 -
2222a,xa,xx72. dx,,,arcsin,C (a,0),2 xax
22a,x 证明 : 被积函数 f(x), 的定义域为 {x|,a,x,a且x,0}2x
22 a,cost πa,x 1. 当,a,x,0 时, 可设x,a,sint (,,t,0) , 则 dx,a,cost dt , , 222xa,sint222πa,xcost ? ,,t,0 , cost,0 ? ,22xa,sint222a,xcost ? dx, , a,cost dt ,,2xa,sint2cost , dt2,sint21,sint , dt2,sint2 ,csctdt,dt,,2
,,cott,t,C
22 在Rt ΔABC中,设 ,B,t ,|AB|,a , 则|AC|,x ,|BC|,a,x
22a,x ? cott,x
2222a,xa,xx ? dx,,,arcsin,C,2xax
π 2. 当0,x,a 时, 可设x,a,sint (0,t,),同理可证2
2222a,xa,xx 综合讨论 1 , 2 得:dx,,,arcsin,C,2xax
- 47 -
2,a,bx,c (a,0)(九)含有的积分(73~78)
dx1273 . ,,ln 2ax,b,2aax,bx,c ,C (a,0) ,2aax,bx,c 12 f(x),ax,bx,c,0 证明:若被积函数成立,则恒成立2ax,bx,c 2 a,0 ? Δ,b,4ac,0?
1222 ax,bx,c,[(2ax,b),4ac,b]?4a
122 2 ,[(2ax,b),(b,4ac)]4a
dx1 ? ,2adx,,2222 ax,bx,c(2ax,b),(b,4ac)
2a1 ,d(2ax,b) ,222a2(2ax,b),(b,4ac)
11 dx22 ,d(2ax,b)公式45: ,ln|x,x,a|,C,,22 222ax,a(2ax,b),(b,4ac)
1222 ,,ln 2ax,b,(2ax,b),(b,4ac) ,C a 12 ,,ln 2ax,b,4a,(ax,bx,c) ,C
a
12 ,,ln 2ax,b,2aax,bx,c ,C
a
22ax,b4ac,b22274. ax,bx,c dx,ax,bx,c,,ln 2ax,b,2aax,bx,c ,C (a,0) ,34a8a 22 证明: 若被积函数 f(x),ax,bx,c成立,则ax,bx,c,0恒成立
2 ? a,0 ? Δ,b,4ac,0
1222 ?ax,bx,c,[(2ax,b),4ac,b] 4a
12222xa222222 ,[(2ax,b),(b,4ac)]公式 53 :x,a dx,x,a,,ln x,x,a ,C ,224a
12222 ? ax,bx,c dx,(2ax,b),(b,4ac)dx,,2a
1222 ,(2ax,b),(b,4ac)d(2ax,b),2a,2a
2,,12ax,bb,4ac222222 ,,(2ax,b),(b,4ac),,ln 2ax,b,(2ax,b),(b,4ac) ,,224a,a,,
212ax,b4ac,b2222,,,2aax,bx,c,,ln 2ax,b,(2ax,b),(b,4ac) ,C3324a8a
212ax,b4ac,b22,,,2aax,bx,c,,ln 2ax,b,4a,(ax,bx,c) ,C3324a8a
22ax,b4ac,b22 ,,ax,bx,c,,ln 2ax,b,2aax,bx,c ,C34a8a
- 48 -
x1b2275. dx,ax,bx,c,,ln 2ax,b,2aax,bx,c ,C (a,0),23aax,bx,c2a
2 证明: ?d(ax,bx,c),(2ax,b)dx
,,x12ax,bb,, ? 可将dx 变换成,, dx,,,,,,222a2a,,ax,bx,cax,bx,c,,
11b1,, ? 上式,,2ax,bdx,dx,,222a2aax,bx,cax,bx,c
1,1b1222,, ,ax,bx,cd(ax,bx,c),dx,,2 2a2aax,bx,c
1b12 ,ax,bx,c,dx,2a2aax,bx,c
b1b12 又dx,,,ln 2ax,b,2aax,bx,c ,C (公式 73)1,22a2aaax,bx,c
b2 ,,ln 2ax,b,2aax,bx,c ,C132a
x1b22 ? dx,ax,bx,c,,ln 2ax,b,2aax,bx,c ,C,23aax,bx,c2a
12,dxaxb76. ,,,C (,0)arcsina,22a,,,cbxaxb4ac
12证明:若被积函数成立,则有解 ,,,,0f(x)cbxax2,,cbxax
2? ,0 ? Δ,,,0ab4ac
1222? ,,,[,(2,)],cbxaxbaxbc4a
22,(2,)b4acaxb ,,44aa
1dx ? ,2adx,,2222,,(,),(2,)cbxaxb4acaxb
12,axb ,,,Carcsin2a,b4ac
dxaxb12,arcisn原
:,,,,C 有误,22acbxaxb4ac,,,
- 49 -
22,,2,axbb4acaxb2277. ,,,,,,,,C (,0)cbxaxdxcbxaxarcsina ,328a8,ab4ac 22证明:若被积函数成立,则有解 ,,, ,,,0f(x)cbxaxcbxax
2 ? ,0 ? Δ,,,0ab4ac
1222? ,,,[,(2,)],cbxaxbaxbc 4a 22,(2,)b4acaxb2xax ,,2222 公式67: a,x dx,a,x,,arcsin,C ,44aa22a 12222 ? ,, ,(,),(2,) cbxaxdxb4acaxbdx,,2a
1222 ,(,),(2,) (2,)b4acaxbdaxb,2,2aa
2 ,,12,,2,axbb4acaxb222 ,(,),(2,),,,b4acaxbarcsinC,, 32224,ab4ac,, 22,,2,axbb4acaxb2 ,4,(,,),,,acbxaxarcsinC33288,aab4ac 22,,2,axbb4acaxb 2 ,,,,,,Ccbxaxarcsin328 a8,ab4ac
12,xbaxb278. ,,,,,,,C (,0)dxcbxaxarcsina,232a,,2,cbxaxab4ac
x2证明:若被积函数成立,则有解 , ,,,0f(x)cbxax2,,cbxax
2? ,0 ? Δ,,,0ab4ac
1222? ,,,[,(2,)],cbxaxbaxbc4a
122,, ,,,(2,)b4acaxb4a
xxx22公式61 : dx,,a,x ,C ? ,2 dxadx,,,222222a,xc,bx,ax(,),(2,)b4acaxb
112,,axbb ,2,, (2,)adaxb,22222aa(,),(2,)b4acaxb
12,1axbb , (2,), (2,)daxbdaxb,,3322222222aa(,),(2,)(,),(2,)b4acaxbb4acaxb
12,baxb222dxx ,,(,),(2,),,,b4acaxbarcsinC公式59 : ,arcsin,C,33222aa,x22,aab4ac
12,baxb2 ,,4,(,,),,,acbxaxarcsinC33222,aab4ac
12,baxb2 ,,,,,,,Ccbxaxarcsin32a2ab,4ac
- 50 -
x,a
,(x,a)(b,x)(十)含有 或的积分(79~82) x,bx,ax,a79. dx,(x,b),(b,a),ln (x,a,x,b) ,C,x,bx,b 2x,ax,aa,bt2t,(a,b) 证明: ? ,0 可 令 t, (t,0) , 则x, , dx,dt222x,bx,b1,t(1,t)
2x,a2t,(a,b)t ? dx,t,dt,2(a,b)dt2222,,,x,b(1,t)(1,t)
21,t,111 ,2(b,a)dt,2(b,a)[,]dt22222,,(1,t)1,t(1,t)
1111 ,2(b,a)dt,2(b,a)dt ,2(a,b)dt,2(a,b)dt222222,,,,1,t(1,t)t,1(1,t)
1t,11t,11 ,2(a,b),,ln ,2(a,b)dt ,(a,b),ln ,2(a,b)dt 2222,,2t,1t,1(1,t)(1,t)
11 对于dt ,dt (t,0)2222,,(1,t)(t,1)
π224 ? 可令 t,seck (0,k,),则 (t,1),tank , dseck,seck,tankdk2
211seckcosk ? dt,,seck,tankdk,dk,dk22443,,,,(t,1)tanktanksink
21,sink111cosk111 ,dk,dk,dk ,,,,dk,dk332,,,,,sink22sinksinksinksinksink
1cosk1111cosk ,,,,dk,,,ln csck,cotk ,,22,22sink22sinksink
2 在 RtΔABC中,, B,k , |BC| ,1 则 |AC|,t,1 , |AB|,t
21t11t,1 ? csck,, , cotk, , cosk, , sink,22sinkttt,1t,1
x,at,11t,1t ? dx,(a,b),ln ,2(a,b)[,,ln ,],C12,2x,bt,122(t,1)t,1
t,1t,1(a,b),t ,(a,b),ln ,(a,b),ln ,,C122t,1t,1t,1
2t,1(a,b),t ,(a,b),ln ,,C12t,1(t,1)
b,a
x,bx,ax,ax,ax,b 将t,代入上式得: ? dx,(a,b),ln ,(a,b),,C1,x,bx,bx,bb,ax,a,x,b
x,b
x,ab,a ,(x,b),(a,b)ln ,C1x,bx,a,x,b
x,a ,(x,b),(a,b)ln b,a ,(b,a)ln x,a,x,b ,C1x,b
x,a ,(x,b),(b,a),ln (x,a,x,b) ,C - 51 - x,b
xaxaxa,,,.dxxbbaarcsin80 ,(,),(,), ,C,bxbxba,,,
2xaxaabttba,,,2,(,)证明?可令tt,则x, dxdt : ,0 , (,0) , ,222bxbx,,tt1,(1,)
2xatbat,2,(,)dxtdtbadt ? ,,,2(,),,,2222bx,tt(1,)(1,)
2t1,,111badtbadt ,2(,),2(,)[,],,22222ttt(1,)1,(1,)
111badtbadtbaarcsintabdt ,2(,),2(,) ,2(,),2(,),,,22222ttt1,(1,)(1,)
tt1,11,11ablnabdtablnabdt ,2(,),, ,2(,) ,(,), ,2(,) ,,2222tt2,1,1tt(1,)(1,)
1对于dtt (,0),22t(1,)
π2242可令ttankk,则tseckdtseckdk ? , (0,,) (,1), , ,2
11122dtseckdkdkcoskdk ? ,,,,,,,,2242tseckseck(1,)
111coskdkdkcoskdk ,(1,2),,2,,,222
k1sinkC ,,,2,124
xak,1dxbakbasinkC? ,2(,),2(,)[,,2],1,bx,24
bakbasinkcoskC ,(,),(,),,1
2在ΔABC中,k ,则 ,t Rt, B, |BC| ,1 |AC|,1 |AB|,,1
t1cosksink ? , , ,22tt,1,1
xatt,1dxbaarcsinba ? ,(,),(,),,,C1,222bx,ttt,1,1,1
ttbaarcsinba ,(,),(,),,C122t,1t,1
,,xaxabxxaxabx,,,,,,,,将t代入上式得:dxbaarcsinba, ? ,(,) ,(,),,C1,,,bxbxbabxbxba,,,,,,,,
xaxa,,baarcsinbx ,(,),(,),,C1babx,,
xaxa,,baarcsinxb ,(,),(,),,Cbabx,,
- 52 -
dxx,a81. ,2arcsin,C (a,b),b,a(x,a)(b,x)
dx1x,a : ,, dx证明,,|x,a|b,x(x,a)(b,x)
22x,aa,bt(b,a)t2t(b,a) t, x, |x,a| , dx,dt令,则,,2222b,x1,t1,t(1,t)
2t b,a , ? |x,a|,(b,a),?21,t
21x,a11,t2t,(b,a) , dx,,,t,dt于是,,222|x,a|b,xb,at(1,t)
1 ,2dt,2arctant,C ( 19)公式,21,t
x,a ,2arctan,Cb,x
x,ax,a tanμ, μ,arctan令,则b,xb,x
RtΔABC, B,μ |AC| ,x,a在中,,
22 ? |BC|,b,x |AB|,|AC|,|BC|,b,a,
x,ax,a ? sinμ, , ? μ,arcsinb,ab,a
dxx,a ? ,2arcsin,C ,b,a(x,a)(b,x)
- 53 -
2xabbaxa2,,(,),.xabxdxxabxarcsinab82 (,)(,),(,)(,),, ,C (,),bx44,
bx,证明xabxdxxadx : (,)(,),,,,xa,
222bxbxbatatttatbtab,,,2,(1,),2(,)2(,)?可令tt,则x, dxdtdt ,0 , (,0) , ,,22222xaxa,,ttt1,(1,)(1,)
22atbaatba,,,,xa ,,, 22tt1,1,
ba,?abxa , ? ,, 2t1,
batab,2(,)xabxdxtdt ? (,)(,),,, ,,222tt1,(1,)
2t2abdt ,,2(,) ,23t(1,)
2tπ2362对于dtt可令ttankk,则tseckdtseckdk (,0) ? , (0,,) (,1), , ,,232t(1,)
222ttanktank222dtseckdkdksinkcoskdk ? ,,,,,,,,,2364tseckseck(1,)
2112sinkcoskdksinkdk ,(2,),2,,,44
k121,,sinkC ,,,4,,,824,,
k1sinkC ,,,4,832
k133sinkcosksinkcoskC ,,,(4,,4,),832
k1133sinkcosksinkcoskC ,,,,,,,888
1233联立以上两式得:xabxdxbaksinkcosksinkcosk (,)(,),,2(,),,(,,,,),C,8
2ba(,)33ksinkcosksinkcosk ,,,(,,,,),C4
2在ΔABC中,k ,则t ,t Rt, B, |BC| ,1 |AC|, |AB|,,1
t1cosksink ? , , ,22tt,1,1
22battt(,)1xabxdxarcsin ? (,)(,),,,(,,,,22224t,1ttt,1,1,1
t11 ,,)222t,1tt,1,1
23battt(,)arcsin ,,,(,,),C222224tt(,1)(,1)t,1
22battt(,)(,1)arcsin ,,,(,),C2224t(,1)t,1
2bxxabbaxa,2,,(,),将t代入上式得:xabxdxxabxarcsin, (,)(,),(,)(,),, ,C,xabx,44,
- 54 -
(十一)含有三角函数的积分(83~112)
83. sinx dx,,cosx,C,
证明: sinx dx,,(,sinx) dx ,,
, ? (cosx),,sinx 即 cosx为,sinx的原函数
? sinx dx,,dcosx,,
,,cosx,C
84. cosx dx,sinx,C,
, 证明: ? (sinx),cosx 即 sinx为cosx的原函数
? cosx dx,dsinx,,
,sinx,C
85. tanx dx,,ln cosx ,C,
sinx 证明: tanx dx, dx,,cosx
1 ,, dcosx,cosx
,,ln cosx ,C
86. cotx dx,ln sinx ,C,
cosx 证明: cotx dx, dx,, sinx
1 , dsinx, sinx
,ln sinx ,C
πx87. secxdx,ln|tan (,)|,C,ln|secx,tanx|,C,42
1cosx 证明 : secxdx,dx,dx2,,,cosxcosx
11111 ,dsinx,dsinx,dsinx2,,,21,sinx21,sinx1,sinx
11 ,,ln|1,sinx|,,ln|1,sinx|,C22 2,,11,sinx11,sinx ,,ln||,C,,ln ,C221,sinx21,sinx
2,,11,sinx1,sinx ,,ln ,C,ln||,C22cosxcosx
1sinx ,ln ,,Ccosxcox
,ln|secx,tanx|,C
- 55 -
x88. cscx dx,ln tan ,C,ln cscx,cotx ,C, 2
xxx222sin,cos1,tan 11222 证法1 : ? cscx,,,, xxxxxsinx2,sin,cos2,sin,cos2tan 22222
x11 又? dtan,,dx x222cos 2
xx2 ? dx,2,cosdtan 22
1xx2 ? cscx dx,,2,cosdtan ,,xx222,sin,cos 22 1x , dtan ,x2tan2
x ,ln tan ,C2
xxx22sinsin2sinx1,cosx222 ? tan,,,,,cscx,cotxxxxxx2sinxcossin,cos2sin,cos22222
x ? cscx dx,ln tan ,C,ln cscx,cotx ,C ,2
1证法: 2 cscx dx , dt,,sint
sint , dt2,sint
1 ,, dcost2,1,cost
111 ,,( ,)dcost,21,cost1,cost
1111 ,, d(cost,1), d(1,cost),,21,cost21,cost
11 ,,,ln 1,cost ,,ln cost,1 ,C122
1cost,1 ,,ln ,C121,cost
21(1,cost) ,,ln , (,1),C1221,cost
21(1,cost) ,,ln ,C222sint
1,cost ,ln ,C2sint
,ln cscx,cotx ,C
- 56 -
2 89. secx dx,tanx,C, 22, (tanx),secx tanxsecx证明:?即为的原函数
2 ? secx dx,dtant,,
,tanx,C
290. cscx dx,,cotx,C,
22 证明: cscx dx,,(,cscx) dx,, 22,?即为的原函数 (cotx),,cscx cotx,cscx 2 ? cscx dx,,dcotx ,,
,,cotx,C
91. secx,tanx dx,secx,C,
, 证明: ? (secx),secx,tanx 即 secx为secx,tanx的原函数
? secx,tanx dx,dsecx,,
,secx,C
92. cscx,cotx dx,,cscx,C,
证明: cscx,cotx dx,,(,cscx,cotx) dx,, , ? (cscx),,cscx,cotx 即 cscx为,cscx,cotx的原函数
? cscx,cotx dx,,dcscx ,,
,,cscx,C
x1293. sinx dx,,,sin2x,C, 24
112 sinx dx,(,,cos2x) dx证明:,, 22
12,cosx112 提示:sinx, ,dx,cos2x d2x,, 224
x1 ,,sin2x,C 24
x1294. cosx dx,,,sin2x,C,24
112 cosx dx,(,,cos2x) dx证明:,,22
12,cosx112 提示:cosx, ,dx,cos2x d2x,,224
x1 ,,sin2x,C24
- 57 -
1n,1nn,1n,2.sinxdx,,,sinx,cosx,sinxdx95 ,,nn nn,1证明:sinxdx,sinx,sinxdx ,,
n,1 ,,sinx dcosx,
n,1n,1 ,,cosx,sinx,cosx dsinx,
n,1n,2 ,,cosx,sinx,cosx, (n,1),sinx,cosx dx, n,12n,2 ,,cosx,sinx,(n,1)cosx,sinx dx,
n,12n,2 ,,cosx,sinx,(n,1)(1,sinx),sinx dx, n,1n,2n ,,cosx,sinx,(n,1)sinx dx,(n,1)sinx dx ,,
nn,1n,2 移项并整理得: nsinx dx,,cosx,sinx,(n,1)sinx dx,,
n,11nn,1n,2?sinxdx,,,sinx,cosx,sinxdx ,,nn
1n,1nn,1n,2.cosxdx,,cosx,sinx,cosxdx96 ,,nn
nn,1证明:cosxdx,cosx,cosxdx ,,
n,1 ,cosx dsinx,
n,1n,1 ,sinx,cosx,sinx dcosx ,
n,1n,2 ,sinx,cosx,sinx, (n,1),cosx,sinx dx,
n,12n,2 ,sinx,cosx,(n,1)sinx,scosx dx,
n,12n,2 ,sinx,cosx,(n,1)(1,cosx),cosx dx,
n,1n,2n ,sinx,cosx,(n,1)cosx dx,(n,1)cosx dx,,
nn,1n,2移项并整理得: ncosx dx,sinx,cosx,(n,1)cosx dx,,
n,11nn,1n,2?sinxdx,,sinx,cosx,cosxdx ,,nn
- 58 -
dx1cosxn,2dx97. dx,,,,,,nn,1n,2 n,1n,1sinxsinxsinx
dx11 证明: dx,,,dx,,nn,22 sinxsinx,sinx
1 ,,dcotx ,n,2sinx cotx1 ,,,cotx d ,n,2n,2sinxsinx cotx1,n ,,,cotx, (2,n),sinx,cosx dx ,n,2sinx 2cotxcosx ,,,(2,n) dx,n,2nsinxsinx 2cotx1,sinx ,,,(2,n) dx,n,2nsinxsinx
cotxdx1 ,,,(2,n) dx,(2,n) dx,,n,2nn,2sinxsinxsinx
dxcotx1 移项并整理得: (n,1) dx,,,(2,n) dx,,nn,2n,2 sinxsinxsinx
cosx1 ,,,(n,2) dxn,1,n,2 sinxsinx
dx1cosxn,2dx ? dx,,,,nn,1n,2,, n,1n,1sinxsinxsinx
dx1sinxn,2dx98. ,,,,,,nn,1n,2 n,1n,1cosxcosxcosx
dx11 证明: ,,dx,,nn,22 cosxcosxcosx
1 ,dtanx,n,2 cosx
tanx1 ,,tanx d, n,2n,2cosxcosx
tanx1,n ,,tanx, (2,n),cosx,sinx dx ,n,2cosx 2tanxsinx ,,(n,2) dx,n,2ncosxcosx 2tanx1,cosx ,,(n,2) dx,n,2ncosxcosx
sinxdx1 ,,(n,2) dx,(n,2) dx,,n,1nn,2cosxcosxcosx
dxsinx1 移项并整理得: (n,1) ,,(n,2) dx,,nn,1n,2cosxcosxcosx
sinx1 ,,(n,2) dxn,1,n,2 cosxcosx
dx1sinxn,2dx ? ,,,,nn,1n,2,, n,1n,1cosxcosxcosx
- 59 -
1m,1mnm,1n,1m,2n?99. cosx,sinxdx,,cosx,sinx,cosx,sinxdx ,,m,nm,n
1n,1m,1n,1mn,2? ,,,cosx,sinx,cosx,sinxdx ,m,nm,n
m,nm,n,1证明?:? dsinxdx,(m,n),sinx,cosxdx
1mnm,11,mm,n ? cosx,sinxdx,cosx,sinxdsinx,,m,n
11m,1n,1m,nm,11,m ,,cosx,sinx,sinxd(cosx,sinx),m,nm,n
m,11,mm,21,m1,m,1m,1? d(cosx,sinx),[,(m,1),cosx,sinx,sinx,(1,m),sinx,cosx,cosx]dx
,mm2,2 ,[(1,m),sinx,cosx,(sinx,cosx,1)]dx
22sinx,cosx,mm ,[(1,m),sinx,cosx,()]dx2cosx
,mm,2 ,[(1,m),sinx,cosx]dx
1m,1m,nm,11,mm,2n ? ,sinxd(cosx,sinx),cosx,sinxdx,,m,nm,n
1m,1mnm,1n,1m,2n ? cosx,sinxdx,,cosx,sinx,cosx,sinxdx ,,m,nm,n
m,nm,n,1证明?:? dcosx,,(m,n),cosx,sinxdx
,1mn1,nn,1m,n ? cosx,sinxdx,cosx,sinxdcosx,,m,n
,11n,1m,1m,nn,11,n ,,sinx,cosx,cosxd(sinx,cosx),m,nm,n
n,11,nn,21,n1,n,1n,1? d(sinx,cosx),[(n,1),sinx,cosx,cosx,(1,n),cosx,sinx,sinx]dx
,nn,22 ,[(n,1),cosx,sinx,(sinx,cosx,1)]dx
22sinx,cosx,nn ,[(n,1),cosx,sinx,()]dx2sinx
,nn,2 ,[(n,1),cosx,sinx]dx
1n,1m,nn,11,nmn,2 ? cosxd(sinx,cosx),cosx,sinxdx,,m,nm,n
1n,1mnm,1n,1mn,2 ? cosx,sinxdx,,,cosx,sinx,cosx,sinxdx ,,m,nm,n
- 60 -
11100. sinax,cosbx dx,,,cos(a,b)x ,,cos(a,b)x,C, 2(a,b)2(a,b)
11提示: sinαcosβ,[sin (α,β),sin (α,β)] sinax,cosbx dx,[sin (a,b)x ,sin(a,b)x]dx证明: ,,22 11 ,sin (a,b)x dxsin(a,b)xdx ,,22 11 ,sin (a,b)x d(a,b)x,sin (a,b)x d(a,b)x ,,2(a,b)2(a,b)
11 ,,,cos(a,b)x ,,cos(a,b)x2(a,b)2(a,b)
11 101. sinax,sinbx dx,,,sin (a,b)x ,,sin (a,b)x,C, 2(a,b)2(a,b)
11提示: sinαsinβ,,[cos(α,β),cos (α,β)] 证明: sinax,sinbx dx,[cos (a,b)x ,cos(a,b)x]dx,, 22
11 ,cos (a,b)x dx,cos (a,b)x dx ,,22
11 ,cos (a,b)x d(a,b)x,cos (a,b)x d(a,b)x ,,2(a,b)2(a,b) 11 ,,sin (a,b)x,,sin (a,b)x ,C2(a,b)2(a,b)
11 102. cosax,cosbx dx,,sin (a,b)x ,,sin (a,b)x,C, 2(a,b)2(a,b)
11 证明: cosax,cosbx dx,[cos (a,b)x ,cos(a,b)x]dx提示: cosαcosβ,[cos (α,β),cos (α,β)],, 22
11 ,cos (a,b)x dx,cos (a,b)x dx ,,22
11 ,cos (a,b)x d(a,b)x,cos (a,b)x d(a,b)x ,,2(a,b)2(a,b) 11 ,,sin (a,b)x,,sin (a,b)x ,C2(a,b)2(a,b)
- 61 -
xa,tan,b dx2222103. ,,arctan,C (a,b),2222 a,b,sinxa,ba,b
x2,tan xxx2t2 证明: 令t,tan , 则 sinx,2,sin,cos,, 2x2221,t21,tan 2
x1x1x1222 dt,(tan )dx,,secdx,(1,tan) dx,(1,t) dx 222222 222bta (1,t),2bt ? dx,dt , a,b,sinx,a,, 2221,t1,t1,t 2dx1,t2 ? ,,dt 22,,a,b,sinxa (1,t),2bt1,t
1 ,2 dt2,a t,2bt,a
1 ,2 dt2, bb2a (t,),,a aa
1 ,2a dt222, (at,b), (a,b)
1 ,2 d(at,b) 222, (at,b), (a,b) 2222 当a,b , 即a,b,0 时
11 2 d(at,b) ,2 d(at,b)222,,2222 (at,b), (a,b) (at,b), (a,b)
2at,bdx1x ,,arctan,C公式 19 : ,,arctan ,C 22,2222aax,aa,ba,b x a,tan,bxdx22 , ,,,C将ttan代入上式得:arctan ,22222,absinxa,ba,b
- 62 -
x22,,,,atanbbadx1222104. ,,ln ,C (a,b),x22a,bsinx22,ba,,,,atanbba2
x2,tan2xxxt2 , , ,2,,,,证明:令ttan则sinxsincos2x2221,t21,tan2
111xxx222 dt,(tan )dx,,secdx,(1,tan) dx,(1,t) dx222222
222bta (1,t),2bt ? dx,dt , a,bsinx,a,,2221,1,1,ttt
21,2dxt ? ,,dt ,,22a,bsinxa (1,t),2bt1,t
1 ,2 dt,2 ,2,atbta
1 ,2 dt,2bb2 (,),,ataaa
1 ,2a dt,222 (at,b), (a,b)
1 ,2 d(at,b),222 (at,b), (a,b)
2222 当a,b , 即a,b,0 时
11 2 d(at,b),2 d(at,b),,222222 (at,b), (a,b) (at,b), (b,a)
1 ,2 (,)datb,2222 (,), (,)atbbadxxa1,公式:ln21 ,, ,C2222,1at,b,b,aaxaxa2,, ,2,,ln ,C22222b,aat,b,b,a
x22a,tan,b,b,a1xdx2 , ,, ,C将ttan代入上式得:ln,22x2,absinx22,ba,,,,atanbba2
- 63 -
,,dx2a,ba,bx22,,105. ,,arctan ,tan,C (a,b) ,,,a,b,cosxa,ba,ba,b2,,
x2 1,tan2x1,t2 证明:令t,tan,则cosx,, 2x21,t2 1,tan2 221,t(a,b),t(a,b) ? a,b,cosx,a,b,,221,t1,t
2 x1x111,t2 ? dt,dtan,,secdx,dx,dx,dx x2221,cosx222cos 2
212,cosθ2 ? dx,dt θ提示:cos,2 1,t2
dx2 ? ,dt2 ,,a,b,cosx(a,b),t(a,b)
22 当 |a|,|b| , 即 a,b 时
221dx1x dt,dt公式 19 : ,,arctan ,C2222,,,aax,aa,b(a,b),t(a,b),, a,b2,,,t,, a,b,, ,,2a,ba,b ,, ,,,arctan ,t,C,,a,ba,ba,b ,,
,,2a,ba,b,, ,,,arctan ,t,C ,,a,ba,ba,b,,
,,1a,b ,, ,2,arctan ,t,C,,,,,,a,b,a,ba,b ,,
,,2a,ba,b,, ,,,arctan ,t,C ,,a,ba,ba,b,,
,,xdx2a,ba,bx,,将t,tan代入上式得:,,arctan,tan,C ,,,2a,b,cosxa,ba,ba,b2,,
- 64 -
xa,btan,dx1a,b2b,a22106. ,,,ln ,C (a,b),a,b,cosxa,bb,axa,btan,2b,a
x21,tan2x1,t2 证明: 令t,tan , 则 cosx,,2x21,t21,tan2
221,t(a,b),t(a,b) ? a,b,cosx,a,b,,221,t1,t
2x1x111,t2 ? dt,dtan,,secdx,dx,dx,dxx2221,cosx222cos2
212,cosθ ? dx,dt22 θ提示:cos,1,t2
dx2 ? ,dt,,2a,b,cosx(a,b),t(a,b)
22 当 a,b , 即 |a|,|b| , ? b,a,0
22 dt,dt,,22(a,b),t(a,b)(a,b),t(b,a)
2121 ,dt ,dt22,,b,aa,b,,,,a,ba,b22,,,,,tt,,,,,b,ab,a,,,,
a,ba,bt,t,21b,a1b,ab,ab,a ,,,,ln ,C,,,ln ,Ca,b2a,ba,ba,ba,ba,bt,t,dxxab,ab,a1,公式:ln21 ,, ,C22,axaxa2,,a,ba,bt,t,11a,bb,ab,a ,(,1),ln ,C ,,,,ln ,C,,,,a,b,b,aa,bb,aa,ba,bt,t,b,ab,a
a,bt,1a,bb,a ,,,ln ,C a,bb,aa,bt,b,a
xa,btan,xdx1a,b2b,a 将t,tan代入上式得: ,,,ln ,C ,2a,b,cosxa,bb,axa,btan,2b,a
- 65 -
dx1b,,107. ,,arctan ,tanx,C,,2222,abaacosx,bsinx,,
dx11 ,,dx证明:22222222,,acosx,bsinxcosxa,btanx
1 , dtanx222,a,btanx
11 , dtanx22,ba2(,tanx)2b
11dx1x , dtanx,2公式 19 : ,,arctan ,Ca22,b22aax,a(),tanx)b
1bb,, ,,,arctan ,tanx,C,,2aab,,
1b,, ,,arctan ,tanx,C,,aba,,dx1b,tanx,a108. ,,ln ,C,22222abb,tanx,aacosx,bsinx
dx11证明: ,,dx,,22222222acosx,bsinxcosxa,btanx
1 , dtanx,222a,btanx
11 , d (b,tanx),22ba,(b,tanx)
11 ,, d (b,tanx)22,b(b,tanx),adxxa1,公式:ln21 ,, ,C22,11b,tanx,aaxaxa2,, ,,,,ln ,Cb2ab,tanx,a
1b,tanx,a ,,,ln ,C,12abb,tanx,a提示: logb,,logbaa
1b,tanx,a ,,ln ,C2abb,tanx,a
- 66 -
11109. x,sinax dx,,sinax,,x,cosax,C2, aa
1x,sinaxdx,,xdcosax 证明: ,, a
11 ,,,x,cosax,cosax dx ,aa 11,,,x,cosax,cosaxdax 2,aa 11 ,,,x,cosax,,sinax,C 2aa
122 22110 .x,sinaxdx,,,x,cosax,,x,sinax,,cosax,C,23 aaa
122 证明: x,sinaxdx,,xdcosax,, a
1122 ,,,x,cosax,cosax dx, aa
122 ,,,x,cosax,x,cosax dx, aa
122 ,,,x,cosax,,x dsinax ,2aa 1222 ,,,x,cosax,,x,sinax,,sinaxdax 23,aaa 1222 ,,,x,cosax,,x, sinax,,cosax 23aaa
11 111. x,cosax dx,,cosax,,x,sinax,C2, aa
1x,cosaxdx,xdsinax 证明: ,, a
11 ,,x,sinax,sinax dx, aa
11,,x,sinax,sinaxdax 2, aa
11 ,,x,sinax,,cosax,C2 aa
12222 112 .x,cosaxdx,,x,sinax,,x,cosax,,sinax,C23,aaa
122 证明: x,cosaxdx,xdsinax,,a
1122 ,,x,sinax,sinax dx,aa
122 ,,x,sinax,x,sinax dx,aa
122 ,,x,sinax,,x dcosax2,aa
1222 ,,x,sinax,,x,cosax,,cosaxdax23,aaa
1222 ,,x,sinax,,x, cosax,,sinax,C23aaa
- 67 -
a,0(十二)含有反三角函数的积分(其中)(113~121)
xx22 113. arcsin dx,x,arcsin ,a,x,C (a,0),aa
xxx 证明: arcsin dx,x,arcsin ,x d arcsin ,,aaa
x11 ,x,arcsin ,x,,dx,aax21,()a
xx ,x,arcsin ,dx,22aa,x
x112 ,x,arcsin ,dx,22a2a,x
1,x122222 ,x,arcsin ,(a,x)d(a,x),a2
1,1x11222 ,x,arcsin ,,,(a,x),C1a21,2
x22 ,x,arcsin ,a,x,Ca22xxaxx22114. x,arcsin dx,(,),arcsin ,a,x,C (a,0),a24a4
x 证明: 令t,arcsin , 则 x,a,sinta
x2 ? x,arcsin dx,a,sint,t d(a,sint) ,at,sint, cost dt,,,a
22aa ,t,sin2t dt,,t dcos2t,,24
22aa ,,,t,cos2t,cos2t dt,44
22aa ,,,t,cos2t,cos2t d2t,48
22aa ,,,t,cos2t,,sin2t,C48提示: sin2x,2,sinx,cosx2222aa2 cos2x,cosx,sinx ,,,t,(2cost,1),,sint,cost,C442 ,2cosx,1222aaa2 ,,,t,cost,,t,,sint,cost,C244
22在中,可设,则 Rt ΔABC ,B,t ,|AB|,a |AC|,x ,|BC|,a,x
22a,xx ? cost, , sint,aa
2222222xaxa,xaxaxa,x ? x,arcsin dx,,,arcsin ,,,arcsin ,,,,C,2a2a4a4aaa
222x,axaxx22 ,,arcsin ,,arcsin ,,a,x,C2a4a4
22xaxx22 ,(,),arcsin ,a,x,C - 68 - 24a4
3xxx122222115. x,arcsin dx,,arcsin ,(x,2a)a,x,C (a,0),a3a9
x 证明: 令t,arcsin , 则 x,a,sinta
x22232 ? x,arcsin dx,a,sint,t d(a,sint) ,at,sint, cost dt,,,a
3a3 ,t dsint,3
33aa33 ,,t,sint,sint dt,33
33aa32 ,,t,sint,sint (1,cost) dt,33
333aaa32 ,,t,sint,sint dt,sint,cost dt,,333
333aaa32 ,,t,sint,,cost,cost dcost,333
333aaa133 ,,t,sint,,cost,,,cost,C3331,2
333aaa33 ,,t,sint,,cost,,cost,C339
22 在Rt ΔABC中,可设 ,B,t ,|AB|,a , 则|AC|,x ,|BC|,a,x
22a,xx ? cost, , sint,aa
22223333xaxxaa,xaa,x222 ? x,arcsin dx,,arcsin ,,,,,,a,x,C,33a3a3a9aa
2232xxaa,x2222 ,,arcsin ,,a,x,,a,x,C3a39
3xx12222 ,,arcsin ,(x,2a)a,x,C3a9
- 69 -
xx 22116. arccos dx,x,arccos ,a,x,C (a,0), aa
xxx 证明: arccos dx,x,arccos ,x d arccos ,, aaa
x11 ,x,arccos ,x,,dx, aax21,() a xx ,x,arccos ,dx ,22aa,x
x112 ,x,arccos ,dx,22a2 a,x
1 ,x122222 ,x,arccos ,(a,x)d(a,x) ,a2
1,1x11222 ,x,arccos ,,,(a,x),C 1a21, 2 x22 ,x,arccos ,a,x,Ca22 xxaxx22117. x,arccos dx,(,),arccos ,a,x,C (a,0) ,a24a4 x t,arccos , x,a,cost 证明:令则a
x2 ? x,arccos dx,a,cost,t d(a,cost) ,,at,cost, sint dt,,,a
22aa ,,t,sin2t dt,t dcos2t ,,24
22aa ,,t,cos2t,cos2t dt ,44 22aa ,,t,cos2t,cos2t d2t,48提示: sin2x,2,sinx,cosx 2222aa cos2x,cosx,sinx ,,t,cos2t,,sin2t,C
482 ,2cosx,122aa2 ,,t,(2cost,1),,sint,cost,C
44
222aaa2 ,,t,cost,,t,,sint,cost,C
244
22 Rt ΔABC ,B,t ,|AB|,a |BC|,x ,|AC|,a,x在中,可设,则
22a,xx ? sint, , cost,
aa
222222xaxxaxaxa,x ? x,arccos dx,,arcsin ,,,arcsin ,,,,C2,a2a4a4aaa
22xxaxx22 ,,arcsin ,,arcsin ,,a,x,C
2a4a4
22xaxx22 ,(,),arcsin ,a,x,C
24a4
- 70 -
3 xxx122222118. x,arccos dx,,arccos ,(x,2a)a,x,C (a,0), a3a9
x 证明: 令t,arccos , 则 x,a,cost a
x22232 ? x,arccos dx,a,cost,t d(a,cost) ,,at,cost, sint dt ,,,a
3a3 ,t dcost ,3 33aa33 ,,t,cost,cost dt,33 33aa 32 ,,t,cost,cost (1,sint) dt,33
333aaa 32 ,,t,cost,cost dt,cost,sint dt,, 333
333 aaa32 ,,t,cost,,sint,sint dsint, 333
333aaa133 ,,t,cost,,sint,,,sint,C 3331,2
333aaa33 ,,t,cost,,sint,,sint,C 339 22 在Rt ΔABC中,可设 ,B,t ,|AB|,a , 则|BC|,x ,|AC|,a,x
22 a,xx ? sint, , cost, aa
22223333xaxxaa,xaa,x222 ? x,arccos dx,,arcsin ,,,,,,a,x,C ,33a3a3a9aa 2232xxaa,x2222 ,,arcsin ,,a,x,,a,x,C3a39 3xx12222 ,,arcsin ,(x,2a)a,x,C3a9
xxa22 119. arctan dx,x,arctan ,,ln (a,x),C (a,0),aa2
xxx arctan dx,x,arctan ,x d x,arctan 证明:,,aaa
x11 ,x,arctan ,x,,dx,xaa21,()a
xx ,x,arctan ,adx,22aa,x
xa12 ,x,arctan ,dx,22a2a,x
xa122 ,x,arctan ,d(a,x),22a2a,x
xa22 ,x,arctan ,,ln a,x ,Ca2
22 ? a,x,0
xxa22 ? arctan dx,x,arctan ,,ln (a,x),C,aa2 - 71 -
x1xa22120. x,arctan dx,(a,x),arctan ,,x,C (a,0), a2a2
x t,arctan , x,a,tant证明:令则 a
x22 ? x,arctan dx,a,tant,t d(a,tant) ,at,sect, tant dt ,,,a
2a2 ,t dsect ,2 22aa22 ,,t,sect,sect dt,22 22aa2 ,,t,sect,,tant,C22
22 在Rt ΔABC中,可设 ,B,t ,|BC|,a , 则|AC|,x ,|AB|,a,x
221a,xx ? sect,, , tant, costaa 2222xaxa,xax ? x,arctan dx,,arctan ,,,,C2,a2a2aa
1xa22 ,(a,x),arctan ,,x,C2a2
33xxxaa2222 121. x,arctan dx,,arctan ,,x,ln (a,x),C (a,0),a3a66
x1x23 证明: ? x,arctan dx,arctan dx,, a3a3 xx1113 ,,arctan ,x,,dx, x23a3a1,() a
33 xxax ,,arctan ,dx22, 3a3a,x 32xxax2 ,,arctan ,dx 22,3a6a,x 3222xxax,a,a2 ,,arctan ,dx22,3a6a,x 32xxaaa22 ,,arctan ,dx,dx22,,3a66a,x
33xxaa1222 ,,arctan ,dx,d(x,a)22,,3a66a,x
33xxaa222 ,,arctan ,,x,ln a,x,C3a66
22? a,x,0
33xxxaa2222 ? x,arctan dx,,arctan ,,x,ln (a,x),C,a3a66
- 72 -
(十三)含有指数函数的积分(122~131)
1xx122. adx,,a,C ,lna 1xx 证明: adx,lna,adx ,,lna xxxx, ? (a),alna , 即alna的原函数为a
1xx ? adx,da,,lna
1x ,,a,Clna
1axax123. edx,,e,C ,a
μ1 ax,μ x, , dx,dμ 证明: 令, 则 aa
11axμμ ? edx,edμ,,e,C,,aa
1ax,,e,C a
1axax124. x,edx,(ax,1)e,C2,a
1axax证明: x,edx,x de,,a
11axax ,,x, e,edx,aa
11axax ,,x, e,edax2,aa
11axax ,,x, e,e,C2aa
1ax ,(ax,1)e,C2a
1nnaxnaxn,1ax125. x,edx,,x,e,x,edx,,aa
1naxnax 证明: x,edx,x de,,a
11naxaxn ,,x, e,edx,aa
1nnaxn,1ax ,,x, e,x,edx,aa
- 73 -
1xxxx126 .x,adx,,a,,a,C,2 lna(lna)
1xx 证明: x,adx,xda ,,lna 111xxxxadx,,a,C公式122 : ,,x, a,adx ,,lnalnalna 11xx ,,x, a,,a,C2lna()lna
1nnxnxn,1x127. x,adx,,x,a,x,adx ,,lnalna 1nxnx x,adx,xda 证明: ,,lna 11nxxn ,,x, a,adx,lnalna
1nnxn,1x ,,x,a,x,adx ,lnalna
1 axax128. e,sinbx dx,,e(a,sinbx,b,cosbx),C,22 a,b
1axax 证明: e,sinbx dx,,edcosbx,, b
11axax ,,,e,cosbx,cosbxde, bb
1aaaxaxax ,,,e,cosbx,,e,sinbx,sinbx de,22 bbb
1aaaxaxax ,,,e,cosbx,,e,sinbx,sinbx de, 22bbb
22a,b1aaxaxax移项并整理得: e,sinbx dx,,,e,cosbx,,e,sinbx,C ,22bbb baaxaxax ? e,sinbx dx,,,e,cosbx,,e,sinbx,C ,2222a,ba,b 1ax ,,e(a,sinbx,b,cosbx),C22a,b
- 74 -
1axax 129. e,cosbxdx,,e(b,sinbx,a,cosbx),C,22a,b 1axax 证明: e,cosbxdx,edsinbx,,b
11axax ,,e,sinbx,sinbxde,bb
1aaxax ,,e,sinbx,sinbx,edx,bb
1aaxax ,,e,sinbx,edcosbx,2bb
1aa axaxax ,,e,sinbx,,e,cosbx,cosbxde,22b bb
2 1aaaxaxax ,,e,sinbx,,e,cosbx,e,cosbxdx,22 bbb
222 aa,b1aaxaxaxax ? (1,)e,cosbxdx,e,cosbxdx,,e,sinbx,,e,cosbx 2,2,2bbbb 1axax ? e,cosbxdx,,e(b,sinbx,a,cosbx),C 22,a,b
- 75 -
1axnaxn,1130. e,sinbx dx,,e,sinbx(a,sinbx,nb,cosbx)222,a,bn
2n,(n,1)baxn,2 , e,sinbx dx222,a,bn
axnaxn,22axn,22证明: e,sinbx dx,e,sinbx,sinbx dx,e,sinbx,(1,cosbx) dx,,,
axn,2axn,22? ,e,sinbx dx,e,sinbx,cosbx dx ,,
1axn,22axn,1又 e,sinbx,cosbx dx,e,cosbx dsinbx,,b,(n,1)
11axn,1n,1ax? ,,e,cosbx,sinbx, sinbx d(e,cosbx) ,b,(n,1)b,(n,1)
n,1axn,1axax又 sinbx d(e,cosbx),sinbx (a,e,cosbx,b,sinbx,e)dx,,
axn,1nax? ,ae,sinbx ,cosbx dx,bsinbx,edx ,,
1axn,1axn,1又 e,sinbx ,cosbx dx,e,sinbx dsinbx,,b
11axnaxn,1 ,,e,sinbx,sinbx d(e,sinbx),bb
11axnaxn,1n,2ax ,,e,sinbx,sinbx [a,e,sinbx,b,(n,1)sinbx,cosbx,e]dx,bb
1aaxnnaxaxn,1 ,,e,sinbx,sinbx,edx,(n,1)e,sinbx ,cosbx dx,,bb
1aaxn,1axnnax移项并整理得:? e,sinbx ,cosbx dx,,e,sinbx,sinbx,edx ,,bnbn
n,1ax将?式代入?式的得: sinbx d(e,cosbx) ,
2aaaxnnaxnax ,,e,sinbx,sinbx,edx,bsinbx,edx,,bnbn
22aa,bnaxnnax? ,,e,sinbx,sinbx,edx ,bnbn
1axn,22axn,1将?式代入?式得: e,sinbx,cosbx dx,,e,cosbx,sinbx,b,(n,1)
22aa,bnaxnnax ,,e,sinbx,sinbx,edx ?22,b,n,(n,1)b,n,(n,1)
1axnaxn,2axn,1将式代入?式得: ? e,sinbx dx,e,sinbx dx,,e,cosbx,sinbx,,b,(n,1)
22aa,bnaxnnax ,,e,sinbx,sinbx,edx 22,b,n,(n,1)b,n,(n,1)
axn移项并整理得: e,sinbx dx,
2,,n,(n,1)b11axn,2axn,1axn ,e,sinbx dx,,e,cosbx,sinbx,,e,sinbx,,2222,b,(n,1)a,bnn,(n,1)b,,
2n,(n,1)bbnaaxn,2axn,1axn ,,e,sinbx dx,,e,cosbx,sinbx,,e,sinbx222222222,a,bna,bna,bn
1axn,1 , ,e,sinbx(a,sinbx,nb,cosbx)222a,bn
2n,(n,1)baxn,2 , e,sinbx dx222,a,bn
- 76 -
1 axnaxn,1131. e,cosbx dx,, e,cosbx(a,cosbx,nb,sinbx)222,a,bn
2n,(n,1)baxn,2 , e,cosbx dx222,a,bn
axnaxn,22axn,22证明: e,cosbx dx,e,cosbx,cosbx dx,e,cosbx,(1,sinbx) dx,,,
axn,2axn,22? ,e,cosbx dx,e,cosbx,sinbx dx ,,
1axn,22axn,1又 e,cosbx,sinbx dx,e,sinbx dcosbx,,b,(1,n)
11axn,1n,1ax? ,,e,sinbx,cosbx, cosbx d(e,sinbx) ,b,(1,n)b,(1,n)
n,1axn,1axax又 cosbx d(e,sinbx),cosbx (a,e,sinbx,b,cosbx,e)dx,,
axn,1nax? ,ae,cosbx ,sinbx dx,bcosbx,edx ,,
1axn,1axn,1又 e,cosbx ,sinbx dx,,e,cosbx dcosbx,,b
11axnaxn,1 ,,,e,cosbx,cosbx d(e,cosbx),bb
11axnaxn,1n,2ax ,,,e,cosbx,cosbx [a,e,cosbx,b,(n,1)cosbx,sinbx,e]dx,bb
1aaxnnaxaxn,1 ,,,e,cosbx,cosbx,edx,(n,1)e,cosbx ,sinbx dx,,bb
1aaxn,1axnnax移项并整理得:? e,cosbx ,sinbx dx,,,e,cosbx,cosbx,edx ,,bnbn
n,1ax将?式代入?式的得: cosbx d(e,sinbx) ,
2aaaxnnaxnax ,,,e,cosbx,cosbx,edx,bcosbx,edx,,bnbn
22aa,bnaxnnax? ,,,e,cosbx,cosbx,edx ,bnbn
1axn,22axn,1将?式代入?式得: e,cosbx,sinbx dx,,e,sinbx,cosbx,b,(1,n)
22aa,bnaxnnax ,,e,cosbx,cosbx,edx ?22,b,n,(1,n)b,n,(1,n)
1axnaxn,2axn,1将式代入?式得: ? e,cosbx dx,e,cosbx dx,,e,sinbx,cosbx,,b,(1,n)
22aa,bnaxnnax ,,e,cosbx,cosbx,edx 22,b,n,(n,1)b,n,(n,1)
axn移项并整理得: e,cosbx dx,
2,,n,(1,n)b1aaxn,2axn,1axn ,e,cosbx dx,,e,sinbx,cosbx,,e,cosbx,,2222,b,(1,n),a,bnn,(1,n)b,,
2n,(n,1)bbnaaxn,2axn,1axn ,,e,cosbx dx,,e,sinbx,cosbx,,e,cosbx222222222,a,bna,bna,bn
21n,(n,1)baxn,1axn,2 , ,e,cosbx(a,cosbx,nb,sinbx) , e,cosbx dx222222,a,bna,bn
- 77 -
(十四)含有对数函数的积分(132~136)
132. lnxdx, x,lnx,x,C ,
证明: lnxdx, x,lnx,x dlnx,,
1 ,x,lnx,x, dx,x
,x,lnx,dx ,
, x,lnx,x,C
dx 133. dx, ln lnx ,C,x,lnx
dx11 证明: dx,dlnx,提示: (lnx),,,x,lnxlnxx
,ln lnx ,C
11nn,1 134. x,lnx dx, ,x(lnx,),C,n,1n,1
lnxnn 证明: x,lnx dx,,(n,1),xdx,,n,1
lnxn,! , dx,n,1
lnx1n,1n,1 ,,x,xdlnx,n,1n,1
lnx1n,1n ,,x,xdx,n,1n,1
lnx1n,12n,1 ,,x,(),x,Cn,1n,1
11n,1 ,,x(lnx,),Cn,1n,1
- 78 -
,nnn1135. (lnx)dx,x,(lnx),n(lnx)dx ,,
,nk nn~k ,x (,1),,(lnx) ,k~,k0 nnn (lnx)dx,x,(lnx),xd(lnx)证明: ,,
1nn,1 ,x,(lnx),x,n,(lnx),dx ,x ,nn1 ,x,(lnx),n(lnx)dx,
,,nnn11 ,x,(lnx),n,x,(lnx),nxd(lnx), ,,nnn12 ,x,(lnx),n,x,(lnx),n,(n,1)(lnx)dx ,
,,,nnnn123 ,x,(lnx),n,x,(lnx),n,(n,1),x,(lnx),n,(n,1),(n,2)(lnx)dx,
.......
nn,n,n,123 ,x,(lnx),n,x,(lnx),n,(n,1),x,(lnx),n,(n,1),(n,2)(lnx)
,,nknk ,??,(,1),n,(n,1),(n,2)??(n,k,1),(lnx),?? ,231 ,(,1),n,(n,1),(n,2)5,4,3,(lnx),x??
,121 ,(,1),n,(n,1),(n,2)4,3,2,(lnx),x??
,011 ,(,1),n,(n,1),(n,2)3,2,1,(lnx),x??
,nknn~k ,x (,1),,(lnx),k~k, 0
1nmnm,1nmn,1 136. x,(lnx) dx, ,x,(lnx),x,(lnx) dx,,m,1m,1
1mnnm,1 证明: x,(lnx) dx, (lnx)dx,,m,1
11m,1nm,1n , ,x,(lnx),x d(lnx),m,1m,1
1n1m,1nm,1n,1 ,,x,(lnx),x,(lnx),dx,m,1m,1x
1nm,1nmn,1 , ,x,(lnx),x,(lnx) dx,m,1m,1
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(十五)含有双曲函数的积分(137~141)
.shxdx,chx,C137 , ,chx,shxchxshx 证明: ? () ,即为的原函数
?shxdx,dchx ,,
,chx,C
138. chx dx, shx,C ,
, 证明: ? (shx),chx ,即shx为chx的原函数
? chx dx, d shx,,
, shx,C
139. thx dx, lnchx,C, shx 证明: thx dx, dx ,,chx 1 , d chx,chx ,lnchx,C
x1 2140. shx dx,,,sh 2x,C, 24
2x,x x,xe,e,,e,e2提示 : (双曲余弦)chx,,,证明: shx dx,dx ,,,,22,, x,xe,e12x,2x shx, (双曲余弦) ,(e,e,2)dx,24 2x,2xeex ,,,,C882
2x,2xx1e,e ,,,,,C 242
x1 ,,,,sh 2x,C 24
x12141. chx dx,,,sh 2x,C ,24
2x,xx,x,,e,ee,e2,, chx dx,dx证明:提示 : (双曲余弦)chx,,,,,22,,
x,xe,e12x,2x shx, (双曲余弦) ,(e,e,2)dx,24
2x,2xeex ,,,,C882
2x,2xx1e,e ,,,,C242
x1 ,,,sh 2x,C24
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(十六)定积分(142~147)
ππ142. cosnx dx,sinnx dx,0,,,π,π
1ππ 证明?: cosnx dx,cosnx dnx,,,π,πn
1π ,,(sinnx) ,πn
11 ,,sin (nπ),,sin (,nπ)nn
2 ,,sin (nπ)n
,0
ππ1 证明?: sinnx dx,sinnx dnx,,,π,πn
1π ,,,(cosnx),πn
11 ,,,cos (nπ),,cos (,nπ)nn
,0
ππ 综合证明??得: cosnx dx,sinnx dx,0,,,π,π
π143. cosmx,sinnx dx,011,,π公式100 : sinax,cosbx dx,,,cos(a,b)x ,,cos(a,b)x,C,2(a,b)2(a,b) 证明: 1. 当m,n时
πππ11 cosmx,sinnx dx,,,cos(m,n)x,cos(n,m)x,,π2(m,n)2(n,m),π,π
11 ,,[cos(m,n)π,cos(m,n)π],[cos(n,m)π,cos(n,m)(,π)]2(m,n)2(n,m)
,0,0,0
2. 当m,n时
ππ cosmx,sinnx dx,cosmx,sinmx dx提示: sin2x,2,sinx,cosx,,,π,π
π1 ,sin2mx d mx,,π2m
π1 ,sin2mx d 2mx,,π4m
π1 ,, ,cos2mx4m,π
1 ,,,[cos2mπ,cos(,2mπ)]4m
,0
ππ 综合讨论 1 , 2 得: cosnx dx,cosmx,sinnx dx,0,,,π,π
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0 , m,n,π144. cosmx,cosnx dx,,,,π,π, mn,
证明: 1. 当m,n时 πππ11,,,,,, cosmxcosnx dxsin (mn)xsin (mn)x,,π,,2(mn)2(mn),,ππ
11,,,,,,,,,, [sin (mn)πsin (mn)(π)][sin (mn)πsin (mn)(π)] ,,2(mn)2(mn)
11,,, 000公式 102 : cosax,cosbx dx,,sin (a,b)x ,,sin (a,b)x,C,2(a,b)2(a,b) 2. 当m,n时
ππ ,,, cosmxcosnx dxcosmxcosmx dx,,,,ππx12π1公式94 : cosx dx,,,sin2x,C2,, cosmx d mx24,,πm
ππ 11,,,, sin2mxmx4m2m,,ππ
1ππ,,,,,, [sin2mπsin (2mπ)]4m22 , π
π0 , m,n, 综合讨论 1 , 2 得: cosmx,cosnx dx,,,,ππ, mn,,
0 , m,nπ,145. sinmx,sinnx dx,,,π,π, m,n,
证明: 1. 当m,n时 πππ11 sinmx,sinnx dx,, ,sin (m,n)x,sin (m,n)x,,π2(m,n)2(m,n),π,π
11 ,,[sin (m,n)π,sin (m,n)(,π)],[sin (m,n)π,sin (m,n)(,π)] 2(m,n)2(m,n)
11 ,0,0,0公式101 : sinax,sinbx dx,,,sin (a,b)x ,,sin (a,b)x,C,2(a,b)2(a,b), 2. 当mn时
ππ2x1 sinmx,sinnx dx ,sinmx dx2,,公式93 : sinx dx,,,sin2x,Cππ,,,24π12 ,sinmx d mx,π,m
ππ11 ,,mx,,sin2mx2m4mππ,,
1ππ ,,,[sin2mπ,sin (,2mπ)],,4m22
,π
π0 , m,n, 综合讨论 1 , 2 得: sinmx,sinnx dx,,,π,π, mn,,
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0 , m,n,,ππ 146. sinmx,sinnx dx,cosmx,cosnx dx,π,,, m,n00 ,2,
证明: 1. 当m,n时
ππ11π sinmx,sinnx dx,, ,sin (m,n)x,sin (m,n)x,2(m,n)2(m,n)000
11 ,,[sin (m,n)π,sin 0],[sin (m,n)π,sin 0]2(m,n)2(m,n)
,0,0,0
ππ11π cosmx,cosnx dx,,sin (m,n)x,sin (m,n)x,2(m,n)2(m,n)000
11 ,[sin (m,n)π,sin 0],[sin (m,n)π,sin 0]2(m,n)2(m,n)
,0,0,0
2. 当m,n时
ππ2 sinmx,sinnx dx ,sinmx dx,,00
1π2 ,sinmx d mx,0m
ππ11 ,,mx,,sin2mx2m4m00
1π ,,,[sin2mπ,sin 0],,04m2
π ,2
ππ cosmx,cosnx dx,cosmx,cosmx dx,,00
π12 ,cosmx d mx,0m
ππ11 ,,sin2mx,,mx4m2m00
1π ,,[sin2mπ,sin 0],,04m2
π ,2
0 , m,n,ππ, 综合讨论 1 , 2 得: sinmx,sinnx dx,cosmx,cosnx dx,π,,00, mn,,2,以上所用公式:
11公式101 : sinax,sinbx dx,,,sin (a,b)x ,,sin (a,b)x,C,2(a,b)2(a,b)
11公式 102 : cosax,cosbx dx,,sin (a,b)x ,,sin (a,b)x,C,2(a,b)2(a,b)
x12公式93 : sinx dx,,,sin2x,C,24
x12公式94 : cosx dx,,,sin2x,C,24
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ππnn147. I,sinx dx,cosx dx22,,n00
n,1 I,I,nn2n
n,1n,342,,,?,, (n为大于1的正奇数) , I,1,1,nn,253 ,,n,1n,331ππ,,,?,,, (n为正偶数) , I,,0nn,24222,
πππ1n,12,,nnn1222 证明?: I,sinx dx,,,sinx,cosx,sinx dx,,nnn000
π1ππn,1,,,nnn1122 ,,(sin,cos,sin0,cos0),sinx dx,n22n0
πn,1n,1,n22 ,sinx dx,I,,n20nn
当n为正奇数时
πn,1n,3422 I,,,?,,,sinx dx,n0nn,253
πn,1n,342 ,,,?,,,(,cosx)20nn,253
n,1n,342 ,,,?,,,1nn,253
ππ22 特别的,当n,1时,I,sinx dx,(,cosx),1n,00
当n为正偶数时
πn,1n,33102 I,,,?,,,sinx dxn,0nn,242
πn,1n,3312 ,,,?,,,(x)0nn,242
n,1n,331π ,,,?,,,nn,2422
πππ022 特别的,当n,0时,I,sinx dx,(x),n,002
πn2 证明?:I,cosx dx??亦同理可证n,0
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附录:常数和基本初等函数导数公式
,1. ( C ),0 (C为常数)
,,μμ12. ( x),μ,x (x?0),3. (sinx),cosx
,4. (cosx),,sinx
2,5. (tanx),secx
2,6. (cotx),,cscx
,7. (secx),secx,tanx
,8. (cscx),,cscx,cotx
xx,9. ( a),a,lna (a为常数)xx,10 . ( e),e
1,11. (logx), (a,0)ax,lna
1,12. (lnx),x
1,13. (arcsinx),21,x
1,14. (arccosx),2,1,x
1,15. (arctanx),21,x
1,16. (arccotx),,21,x
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1.感谢本团队诸成员的高数老师的谆谆教导,感谢本团队诸成员间的合作,感
谢所有支持本讲义编辑的支持者
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明过程的原则
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2013年5月
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