第4章不定积分习题课 本章主要内容 1.原函数的概念 2.不定积分的概念与性质 3.求不定积分的方法. 4.特殊函数的不定积分求法 一原函数概念 1.定义设f)在区间有定义,如果区间上可微函数Fe)满足F)=f)(任 I).则称F(x)是f(x)在I上的一个原函数. 注意定义中:F(z)是fz):(1)在I上的(2)一个原函数 2原函数的性质 (①)如果F(x)是f(x)的原函数,则对任意常数c,F()+c也是f(x)的原函数, ()f(工)的任意两个原函数之差为常数,所以 {F(x)+cIc∈R 就是f()的全体原函数组成的集合, ()f()在区间I上的原函数F(r)在I上一定是连续函数. 3原函数存在定理设fx)在区间1上连续,则在I上一定有原函数. 注:()区间上的连续函数一定有原函数,但原函数未必能用初等函数表示例 如 ,台e,-e产0<k<) sin r e (②)函数若在区间上有第一类间断点,则一定在上没有原函数。 三不定积分 1.定义设F(e)是f(e)的一个原函数,则称集合{F(x)+c|c∈R)为f(e)的不定 积分,记成∫fe)d=F()+C 2.性质 是∫reh=fe,a((e)=fe地 F(r)dz =F()+c. dF(z)=F(a)+c. 3.不定积分的几何意义
· 1 · 1 4 Ÿ ÿ½»©SKë ŸÃáSN 1. ºÍVg. 2. ÿ½»©VgÜ5ü 3. ¶ÿ½»©ê{. 4.AœºÍÿ½»©¶{ ò.ºÍVg 1. ½¬ f(x)3´mIk½¬,XJ´m˛åáºÍF(x)˜vF 0 (x) = f(x) (x ∈ I),K°F(x)¥f(x)3I˛òáºÍ. 5ø½¬•:F(x)¥f(x): (1)3I˛(2)òáºÍ. 2ºÍ5ü (i) XJF(x)¥f(x)ºÍ,KÈ?ø~Íc, F(x) + cè¥f(x)ºÍ. (ii) f(x)?ø¸áºÍÉè~Í,§± {F(x) + c | c ∈ R} “¥f(x)NºÍ|§8‹. (iii) f(x)3´mI˛ºÍF(x)3I˛ò½¥ÎYºÍ. 3ºÍ3½n f(x)3´mI˛ÎY,K3I˛ò½kºÍ. 5µ(1) ´mI˛ÎYºÍò½kºÍ,ºÍô7U^–ºÍL´.~ X sin x x , 1 ln x , e x x , e−x 2 , p 1 − k 2 sin2 x (0 < k < 1) (2) ºÍe3´mI˛k1òam‰:,Kò½3I˛vkºÍ. n.ÿ½»© 1.½¬ F(x)¥f(x)òáºÍ,K°8‹{F(x) +c | c ∈ R}èf(x)ÿ½ »©,P§ R f(x)dx = F(x) + c. 2.5ü d dx Z f(x)dx = f(x), d Z f(x)dx = f(x)dx. Z F 0 (x)dx = F(x) + c, Z dF(x) = F(x) + c. 3.ÿ½»©A¤ø¬
2 {F(E)+c|c∈是一族积分曲线,它们是y=F(x)沿Oy轴上下平移而得的在 这些曲线上,对应着那一点的切线都有相同的斜率F'(z)=f) 4.不定积分的运算法则 如果F(x)和G(r)分别是f(a)和g(c)的原函数,则aF(e)+bG(a)是af(r)+bg(z)的 原函数(其中a,b是常数).jaf)+bg(=afr+b∫g( 5.不定积分的公式 H+1 d= =-cscx+ci rd证 ∫r在-品+6∫平-mm+6 dr cos xdr sin +c chrdz shz+c sin rdr =-cos+c shrdr chr+c /n-a业=血k+V2-网+e 1 F+a业=+VP+@)+. 1 6.求不定积分的方法 ()直接利用公式及运算法则 (间)第一类换元法(凑微分法) ()第二类换元法 (w)分部积分法 7.常见凑微分
· 2 · {F(x) + c | c ∈ R}¥òx»©Ç,ßÇ¥y = F(x)˜Oy¶˛e²£ .3 ˘ Dz,ÈAXx @ò:ÉÇ—kÉ”«F 0 (x) = f(x). 4.ÿ½»©$é{K XJF(x)⁄G(x)©O¥f(x)⁄g(x)ºÍ, KaF(x)+bG(x)¥af(x)+bg(x) ºÍ(Ÿ•a, b¥~Í). R [af(x) + bg(x)]dx = a R f(x)dx + b R g(x)dx. 5.ÿ½»©˙™ Z 0dx = c; Z sec x tan xdx = sec x + c; Z x µ dx = x µ+1 µ + 1 + c, µ 6= −1; Z csc x cot xdx = − csc x + c; Z dx x = ln |x| + c; Z dx p 1 − x 2 = arcsin x + c; Z a x dx = a x ln a + c; Z dx 1 + x 2 = arctan x + c; Z cos xdx = sin x + c; Z chxdx = shx + c; Z sin xdx = − cos x + c; Z shxdx = chx + c; Z csc 2xdx = − cot x + c; Z sec2 xdx = tan x + c; Z 1 √ x 2 − a 2 dx = ln |x + p x 2 − a 2| + c. Z 1 √ x 2 + a 2 dx = ln(x + p x 2 + a 2) + c; . 6.¶ÿ½»©ê{ (i) Ü|^˙™9$é{K (ii) 1òaÜ{(ná©{) (iii) 1aÜ{ (iv) ©‹»©{ 7. ~Ñná©
-3 在=dar+外 Idz=d(nk)i edz=d(e"); cos rdz=d(sinz)片 sinx=-d(cosx片 sec2 rdr =d(tanz); cse2rdk=-d(cotx月 i+7=darctan水 1 chrdr dshr: shrdr=dchr; V-山=d(arcsin功 1 2=dv 8.常用换元法 (1.)三角代换(2)双曲代换(3.)倒代换 一,求不定积分 要求熟练掌握凑微分法、第二换元法、分部积分法、有理函数及三角有理函数 的积分法 1.(17)(6分)求不定积分max1,x2山 2(10(6分)求不定积分1=V+ 3m分)求不定积分1= 4.(16)(6分)求不定积分min{r2,x}dk(-o<x<+oo), 5.(15,16)(5分,6分)求不定积分x2 arctanrd 6.)6分)求不定积分∫0+可业 1 8.(14(6分)I血x
· 3 · dx = 1 a d(ax + b); x ndx = 1 n + 1 d(x n+1); 1 x dx = d(ln |x|); e xdx = d(e x ); cos xdx = d(sin x); sin x = −d(cos x); sec2 xdx = d(tan x); csc2 xdx = −d(cot x); 1 1 + x 2 dx = d(arctan x); chxdx = dshx; 1 1 + x 2 dx = −d(arccotx); shxdx = dchx; 1 2 √ x dx = d( √ x); 1 √ 1 − x 2 dx = d(arcsin x). 8.~^Ü{ (1.)nìÜ (2.) VìÜ (3.)ìÜ ò. ¶ÿ½»© ᶟˆ›ºná©{!1Ü{!©‹»©{!knºÍ9nknºÍ »©{ 1. (17) (6©) ¶ÿ½»© Z max{1, x2 }dx. 2. (17) (6©§¶ÿ½»©I = Z p a 2 + x 2dx. 3. (17) (6©§¶ÿ½»©I = Z dx 1 + x 3 . 4. (16) (6©) ¶ÿ½»© Z min{x 2 , x5 } dx (−∞ < x < +∞). 5. (15,16) (5©,6©)¶ÿ½»© Z x 2 arctan xdx. 6. (15) (5©)¶ÿ½»© Z 1 x(1 + x 4) dx. 7. (14) (10©)Æf 00(x)ÎY,f 0 (x) 6= 0,¶ Z f(x) f 0(x) − f 2 (x)f 00(x) (f 0(x))3 dx. 8. (14) (6©) Z | ln x|dx. 9. (14) (6©) Z x 3 − x 1 + x 4 dx.
·4 0a6分/在 11.(13)(6分)max{x2,x 12.(13)(10分)设fe)可微,且xfr)=r2csr-4rimr-6osr+C,求fa. 13.(13)(8分)设f()是(-0,+∞)上的可微函数且有反函数,己知F()是f(x)的一 个原函数,求f-1(x)dz 14.(12)5分)x-1)d(m>0) 15.(12)()sin(2r)cos2 rdr 16.(12)(6分)simV证 1.2)6分)/e+VP+ sms0∫二 19.a0)6分)ed 20a0)6分京 1 21.(09)(8分)221n2xb 2侧严 a四o9/+4 点四e/= 26.(03)(xdr
· 4 · 10. (13) (6©) Z 1 1 − x 4 dx. 11. (13) (6©) Z max{x 2 , x4 }dx. 12. (13) (10©) f(x)åá,Ö Z x 3 f 0 (x)dx = x 2 cos x−4x sin x−6 cos x+C,¶f(x). 13. (13) (8©) f(x)¥(−∞, +∞)˛åáºÍÖkáºÍ,ÆF(x)¥f(x)ò áºÍ,¶ Z f −1 (x)dx. 14. (12) (5©) Z x(x − 1)n dx (n > 0) 15. (12) (5©) Z sin(2x) cos2 xdx 16. (12) (5©) Z sin √ xdx 17. (12) (5©) Z ln(x + p x 2 + 1)dx 18. (11)(8©) Z xex √ e x − 1 dx 19. (10) (5©) Z x 2 e x dx 20. (10) (5©) Z 1 √ e x + 1 dx 21. (09) (8©) Z x 2 ln2 xdx 22. (08) (5©) Z arctan x x 3 dx 23. (06) (9©) Z cos x sin x(sin2 x + 1) dx 24. (05) (9©) Z ln x √ x − 1 dx 25. (04) (6©) Z sin 2x √ 1 + cos2 x dx 26. (03) (5©) Z x tan x sec2 xdx
5 27.02(8分ed在 28.(02)(8)分z-3 sin co+2ms27 d 例题: xtanzsec+rdr. 解 framrwe rdrfrd rdsefrdce) (rsc-sc'zdr)=irse(1+tan)d(tanz) =女a-tmr-立tam3z+e 2∫-“ 解 /二-二9aw- (以下自己完成) 解: ∫女=-小aamw- =-+∫是 -(redin-子+2zamx-2/2≥ =-(arcsin r)2v1-x2+2r arcsin +2v1-2+c 解: s∫
· 5 · 27. (02) (8©) Z e √ x dx, 28. (02) (8©) Z dx sin2 x − 3 sin x cos x + 2 cos2 x ~Kµ 1. Z x tan x sec4 xdx. )µ Z x tan x sec4 xdx = Z x sec3 xd sec x = 1 4 Z xd(sec4 x) = 1 4 (x sec4 x − Z sec4xdx) = 1 4 x sec4 x − 1 4 Z (1 + tan2 x)d(tan x) = 1 4 x sec4 x − 1 4 tan x − 1 12 tan3 x + c 2. Z xex √ e x − 2 dx. )µ Z xex √ e x − 2 dx = Z x d(e x − 2) √ e x − 2 = 2 Z xd√ e x − 2 = 2x √ e x − 2 − 2 Z √ e x − 2dx = (±egC§) 3. Z x(arcsin x) 2 √ 1 − x 2 dx. )µ Z x(arcsin x) 2 √ 1 − x 2 dx = − Z (arcsin x) 2 d p 1 − x 2 = −(arcsin x) 2 p 1 − x 2 + Z √ 1 − x 22 arcsin x √ 1 − x 2 dx = −(arcsin x) 2 p 1 − x 2 + 2x arcsin x − 2 Z x √ 1 − x 2 dx = −(arcsin x) 2 p 1 − x 2 + 2x arcsin x + 2p 1 − x 2 + c 4. Z sin 2x √ 3 − cos4 x dx. )µ Z sin 2x √ 3 − cos4 x dx = − Z d(cos2 x) √ 3 − cos4 x = − arcsin cos2 x √ 3 + c. 5. Z x ln x p (x 2 − 1)3 dx
·6 解: ∫=-+ 话nd ∫mt=j岁mt-∫mm-∫ x2 -dr-/i+e女-号aota球 rarctanz-In(+)-(arctanz)+c. 解 ∫a-∫女-a女 =∫g-mf血=年+a女-∫af血=年+ e ∫-信点-盖- =品+∫-∫=品+e 解法二: ∫ne∫
· 6 · )µ Z x ln x p (x 2 − 1)3 dx = − Z ln xd( 1 √ x 2 − 1 ) = − ln x √ x 2 − 1 + Z 1 x √ x 2 − 1 dx = − ln x √ x 2 − 1 + Z dx x 2 q 1 − 1 x2 = − ln x √ x 2 − 1 − Z d( 1 x ) q 1 − ( 1 x ) 2 = − ln x √ x 2 − 1 − arcsin 1 x + c. 6. Z x 2 1 + x 2 arctan xdx. )µ Z x 2 1 + x 2 arctan xdx = Z x 2 − 1 + 1 1 + x 2 arctan xdx = Z arctan xdx − Z arctan x 1 + x 2 dx = x arctan x − Z x 1 + x 2 dx − 1 2 Z d(arctan x) 2 = x arctan x − 1 2 ln(1 + x 2 ) − 1 2 (arctan x) 2 + c. 7. Z xex (1 + x) 2 dx. )µ Z xex (1 + x) 2 dx = Z (1 + x − 1)e x (1 + x) 2 dx = Z e x 1 + x dx − Z e x (1 + x) 2 dx = Z d(e x ) 1 + x − Z e x (1 + x) 2 dx = e x 1 + x + Z e x (1 + x) 2 dx − Z e x (1 + x) 2 dx = e x 1 + x + c. 8. Z ln x − 1 (ln x) 2 dx. ){òµ Z ln x − 1 (ln x) 2 dx = Z ( 1 ln x − 1 ln2 x )dx = Z dx ln x − Z 1 ln2 x dx = x ln x + Z x ln2 x · 1 x dx − Z 1 ln2 x dx = x ln x + c. ){µ Z ln x − 1 (ln x) 2 dx ln x=t === Z t − 1 t 2 · e t dt
7 90+血山 1+cosr 解。 j0a-j女+∫4 2o专血+etam业-edam+∫rm5-ctm号te 10.(1+:-erttdz. 解: (1+r-)etidr=fetidr+(-)e*+idr -+-e+1-2d+e-是e+证=4+e 解: 品兰岛-∫(中+) 2∫0+可 ∫可=苦=++边+∫* =(信-2+-点+中)血=京+京-京++m+e 8出 解: =0:8-/的≌号/ rx9_8 r98 =/品。-h牛+c-h巴+e 143+4osdh 解: 2r+In|2sin+cos+c
· 7 · 9. Z e x (1 + sin x) 1 + cos x dx. )µ Z e x (1 + sin x) 1 + cos x dx = Z e x 1 + cos x dx + Z e x sin x 1 + cos x dx = Z e x 2 cos2 x 2 dx + Z e x tan x 2 dx = Z e x d(tan x 2 ) + Z e x tan x 2 dx = e x tan x 2 + c. 10. Z (1 + x − 1 x )e x+ 1 x dx. )µ Z (1 + x − 1 x )e x+ 1 x dx = Z e x+ 1 x dx + Z (x − 1 x )e x+ 1 x dx = xex+ 1 x − Z xex+ 1 x (1 − 1 x 2 )dx + Z (x − 1 x )e x+ 1 x dx = xex+ 1 x + c. 11. Z x x 8 − 1 dx. )µ Z x x 8 − 1 dx x 2=t === 1 2 Z dt t 4 − 1 = 1 4 Z −1 2(t + 1) + 1 2(t − 1) − 1 t 2 + 1 dt. 12. Z dx x 8(1 + x 2) . )µ Z dx x 8(1 + x 2) = Z 1 − x 8 + x 8 x 8(1 + x 2) dx = Z (1 + x 4 )(1 + x 2 )(1 − x 2 ) x 8(1 + x 2) dx + Z dx 1 + x 2 dx = Z 1 x 8 − 1 x 6 + 1 x 4 − 1 x 2 + 1 1 + x 2 dx = − 1 7x 7 + 1 5x 5 − 1 3x 3 + 1 x + arctan x + c. 13. Z x 9 − 8 x 10 + 8x dx. )µ Z x 9 − 8 x 10 + 8x dx = Z x 8 (x 9 − 8) x 9(x 8 + 8)dx = 1 9 Z x 9 − 8 x 9(x 9 + 8)d(x 9 ) x 9=t === 1 9 Z t − 8 t(t + 8)dt = 1 9 Z ( 2 t + 8 − 1 t )dt = 1 9 ln (t + 8)2 t + c = 1 9 ln (x 9 + 8)2 x 9 + c. 14. Z 3 sin x + 4 cos x 2 sin x + cos x dx. )µ Z 3 sin x + 4 cos x 2 sin x + cos x dx = Z 4 sin x + 2 cos x + 2 cos x − sin x 2 sin x + cos x dx = 2 Z dx + Z d(2 sin x + cos x) 2 sin x + cos x = 2x + ln |2 sin x + cos x| + c
8 ∫+9 解 ∫k=-/+哈-m9-分产个 -2+jg-a地-9+a时-0++e 解: ∫ot=∫mi女=jt4m 1 Ztmitac 解: ∫血=兰//e+-+] 5 =+-++c=+2-++e /(+) 解 ∫e(r+a)在=ert+三)=+nm+e 19./g 解: ∫兰∫后-吉a=jre=re+ac+2+e =(白2e-+2te-+2e-+c retan 20.∫+平
· 8 · 15. Z ln(1 + x 2 ) x 3 dx. )µ Z ln(1 + x 2 ) x 3 dx = − 1 2 Z ln(1 + x 2 )d( 1 x 2 ) = − 1 2 ln(1 + x 2 ) x 2 − Z 1 x 2 · 2x 1 + x 2 dx = − ln(1 + x 2 ) 2x 2 + Z ( 1 x − x 1 + x 2 )dx = − ln(1 + x 2 ) 2x 2 + ln |x| − 1 2 ln(1 + x 2 ) + c. 16. Z 1 sin3 x cos5 x dx. )µ Z 1 sin3 x cos5 x dx = Z 1 tan3 x cos8 x dx = Z (1 + tan2 x) 3 tan3x d(tan x) = −1 2 tan x 2 + 3 2 tan2 x + 3 ln |tan x| + 1 4 tan4 x + c. 17. Z x 5 √4 x 3 + 1 dx. )µ Z x 5 √4 x 3 + 1 dx = 1 3 Z x 3 √4 x 3 + 1 d(x 3 ) t=x 3 === 1 3 Z tdt √4 t + 1 = 1 3 Z h (t + 1) 3 4 − (t + 1) −1 4 i dt = 4 21 (1 + t) 7 4 − 4 9 (1 + t) 3 4 + c = 4 21 (1 + x 3 ) 7 4 − 4 9 (1 + x 3 ) 3 4 + c 18. Z e x 3 x + e −x x ln x dx. )µ Z e x 3 x + e −x x ln x dx = Z (3e) x dx + Z 1 ln x d(ln x) = (3e) x ln(3e) + ln | ln x| + c. 19. Z e − 1 x x 4 dx. )µ Z e − 1 x x 4 dx 1 x =t === Z t 4 e t · (− 1 t 2 )dt = Z t 2 d(e −t ) = t 2 e −t + 2te−t + 2e −t + c = ( 1 x ) 2 e − 1 x + 2te− 1 x + 2e − 1 x + c 20. Z xearctan x (1 + x 2) 2 dx
解: ∫后示世∫等 e'sin2tdt =/sin atde=e'sin at-2e cos2tdt =e'sin 2t-2cos2tde =e'sin 2t-2e'cos2t-4e'sin 2tdt fe sinz(sin2t-2e cont). +r=e24-2caw2刘=es2-2as2ctam+e 21.设 到={10s ,且f0)=0,求f) (工,1<x<+o∞ 解:令血x=t,x=e r阳-{09 m-{ f0-0)=f0+)=f0),得a=1+c2 m-{e 因为f0)=0,可得c=-1所以 -0<f≤0 f-{2-10<t<+ 2已知fa倒=求”)d 解: ∫ft=∫回=f回-ra恤 =f回-a)=fa-fa+e=-三+e
· 9 · )µ Z xearctan x (1 + x 2) 2 dx arctan x=t === Z tan tet sec2 t (1 + tan2 t) 2 dt = 1 2 Z e t sin 2tdt Z e t sin 2tdt = Z sin 2tdet = e t sin 2t − 2 Z e t cos 2tdt = e t sin 2t − 2 Z cos 2tdet = e t sin 2t − 2e t cos 2t − 4 Z e t sin 2tdt Z e t sin 2tdt = 1 5 (e t sin 2t − 2e t cos 2t). Z xearctan x (1 + x 2) 2 dx = 1 10 (e t sin 2t − 2e t cos 2t) = 1 10 (e arctan x sin 2 arctan x − 2e arctan x cos 2 arctan x) + c. 21. f 0 (ln x) = ( 1, 0 < x ≤ 1 x, 1 < x < +∞ ,Öf(0) = 0,¶f(x). )µ-ln x = t, x = e t f 0 (t) = ( 1, 0 < et ≤ 1 ⇐⇒ −∞ < t ≤ 0 e t , 1 < et < +∞ ⇐⇒ 0 < t < +∞ f(t) = ( t + c1, −∞ < t ≤ 0 e t + c2, 0 < t < +∞ f(0 − 0) = f(0+) = f(0),c1 = 1 + c2 f(x) = ( x + 1 + c, −∞ < t ≤ 0 e x + c, 0 < t < +∞ œèf(0) = 0,åc = −1 §± f(x) = ( x, −∞ < t ≤ 0 e x − 1, 0 < t < +∞ 22. Æf(x) = e x x ,¶ Z xf00(x)dx. )µ Z xf00(x)dx = Z xdf0 (x) = xf0 (x) − Z f 0 (x)dx = xf0 (x) − Z df(x) = xf0 (x) − f(x) + c = ( e x x ) 0x − e x x + c
·10 练习题 + 2.sin2x+2in正 ∫ 4.(tanz+1)2dr 5∫+ 6+P 7.x2sin2 rdr 8.asin bcos g.设fe2-)=n二2且(ea》=lnz,求ed 10.设fm)-+卫,计算1g
· 10 · ˆSK 1. Z ln x (1 − x) 2 dx 2. Z dx sin 2x + 2 sin x 3. Z arctan x x 2(1 + x 2) dx 4. Z e 2x (tan x + 1)2 dx 5. Z dx x √ 1 + x 2 6. Z dx x + √ x 2 − 1 7. Z x 2 sin2 xdx 8. Z dx a sin2 x + b cos2 x 9. f(x 2 − 1) = ln x 2 x 2 − 2 ,Öf(ϕ(x)) = ln x,¶ Z ϕ(x)dx. 10. f(ln x) = ln(1 + x) x ,Oé Z f(x)dx