Tài liệu Giải nhanh bài toán nguyên hàm và tích phân dành cho học sinh khối 11 và 12 part 1

515.076 GI-103N TRAN TUAN ANH Gliil NHANH BAI TOAN HGIIYJNHAH &TiCH PHAN D A N H C H O H Q C S I H H L0P11-12 TRAN TUAN ANH m r m m m • 1 » - ^ NGUYEN HAM ^ T I C H PHAN T!U/ VIEN TiNHBlNH THUAI>2 NHA XUAT BAN DAI HQC QUOC GIA THANH PHO HO CHI MINH GIAI NHANH BAI TOAN NGUYEN HAM VA TICH PHAN La>i noi dau Nha xua't ban DHQG-HCM va tdc gii/doi tdc lien ke't gifl ban quy^n® Copyright © by VNU-HCM Publishing House and author/co-partnership All rights reserved Viec giai mot bai toan noi chung la mot qua trinh tu duy cao do, dua tren hilu biet cua nguai giai toan. Viec tinh mot bai toan nguyen ham hay mot bai toan tich phan cung vay. Co nguai tham chi khong giai dugc, c6 nguai giai dugc nhimg can qua trinh may mo rat lau, thu het each nay den each khac mai giai xong, trong khi c6 nguai lai tim dugc each giai rat nhanh. Vay dau la bi quyet de giai nhanh dugc mot bai toan nguyen ham, mot bai toan tich phan noi rieng? Cach ren luyen de c6 each giai nhanh? Cuon sach nay viet ra nhSm dem lai cho ban doe nhimg each hieu, nhijng huang di, thu thuat de tilp can nhanh tai lai giai thoa dang cho mot bai toan nguyen ham, mot bai toan tich phan. Cac cong thuc dua tai nguai doc khong CONG TY TNHH M^T THANH VIEN SACH VIET . 391/15A Hajnh T i n Phat, P.T§n Ttw^n Dong, QuSn 7, TP.HCM, BT: {06} Jf.720.837 • F a x P) 38.726,052 • MST: 03114307135 Email: W i f i a c l R f i e t c o x o m - Website: «»w.sachvie!co.«ni mang tinh ap dat ma theo huang de hieu, de nha de nguai doe c6 thien cam han ve cac cong thuc do, phuc vu cho viec van dung tinh toan sau nay. Cuon sach viet theo loi dien giang nen kho tranh khoi khiem khuyet, rat mong nhan dugc nhung gop y thiet thuc ciia ban doe gan xa. Xin chan thanh cam an nhung gop y, chi dan cua quy thay: - TS. Nguyen Viet Dong, Truang Bo mon Giao due Toan hoc, DHKHTN, DHQG TP. H6 Chi Minh. - Thhy Nguyen Dinh Do, Pho Hieu truong Truang THPT Thanh Nhan TP. Ho Chi Minh. - ThSy Le Hoanh Sir, Giang vien DHQG TP. HCM Te Luat. i ' - Truang DH Kinh ' - Thhy Nguyen Tat Thu, Giao vien Truang chuyen Luang Th6 Vinh Bien Hoa - Dong Nai. Tran Tuan Anh Xufi't ban nam 2013 G I A I NHANH B A I TOAN T I C H PHAN T R O N G D E T H I T U Y E N S I N H D A I H O C NAM 2 0 1 3 Cau 1; Tinh tich phan / = | — ~ 1" • (E>H kh6i A, A i - 2013) Cdch gidi thong thir&ng w - I 2 Cdch 1: / = ——.lnxdx= + A = In 1 \nxdx + In xdx. Ta xet: xdx. Dat u = lnx=> du = —dx ; dv = dx=> v = x. X In xdx = x\nx -1 + h = 2 ' 1 -dx = 2\n2-\. X In xdx Dat u = \nx^ , du = — dx; dv = X -1 1 rfX = > V = — . J X \X f 1\ In xdx = 2 ^f 1 1, dx = — Inx 1 X X = —Inx 2 1 X I + X 1 = -ln2—. 2 2 Vay / = / , + / 2 = 2 1 n 2 - l + - l n 2 - i = - ( 5 1 n 2 - 3 ) . V-1 Dat /• = In X = > X . In xdx - 1- 1 = e' va dx = e' dt. Inxc/x Doi can: x = l = > / = 0; x = 2=i>/ = ln2. In 2 / 1= , \n 2 Cau 2; Tinh tich phan / = ^x^l-x'dx In 2 ^\\—L-\e'dt= \[e'-e")dt= \td(e'+6-) Cdch giai thong thu&ng 0 V = t(e'+e-) In 2 0 In 2 • " nen ta chon an phu x = V2 sin?. Cdch 1: Do dau hieu '(e'+e-')dt Dat X = y/l sin t => dx^-Jl Doi c a n : x = 0 = ln2. V 2, Cdch 3: cdc ban di y quan he giita X X X ^x'-\^ Xet tich phan J = 272 |sin / cos^ tdt. Vdy ta CO the giai dx = d x + - dx = ..2 x'-\ 4 ; quan he — va \ : -^dx = d nhanh bdi todn tren nhu sau : 1= / = 0;x = 1 => / = —. = 27^ sin ^ cos/. cos/L// = 2V2 sin/cos tdt. giiia X vd 1 la : \dx = dx. 2 1- . In xdx = Dat u = cos/ ^du = -s'mtdt. X. x+ - 2 - V 1 \^2 = Inx. x + - 1^ In xdx = In xd x + * 2 O T a c o : J = -l4l x+ - X rv 2 1+ 72 K Doi can : / = 0 =o w = 1;/= — => w = — . 4 2 1 = ln n n Taco: / = JV2 sin / V2 - 2 sin" /.72 cos /Jr = 2 V2 Jsin cos Vl - sin^ / J/ Cac/i ^w/ nhanh CO : cos tdt, te 2 - i .l(51n2-3). 2 Vay / = | ( 5 1 n 2 - 3 ) . Do do, ta ( D H kh6i B - 2013 1^ \u^du = 2^ 1 3 \u^du = 2^.— 7^ = 272-1 2 r 02— ( dx = In X . x + — = |(51n2-3). 1 X 1' — Vay / = 272-1 Cdch 2: Theo kinh nghiem thi thay can thuc ta dat can thuc la an phu ! Dat / = 72-x^ ^t^ =2-x^ =>tdt = -xdx. I,^/ giai that nhanh ggn so v&i hai cdch tren ! D6i can:' x = 0 => / = 72;x = 1 =i> / = 1. Taco: I = -\t^dt= \t^^t= — V2 2V2-1 Cdch 2: Ta c6 : /= i Cdch gidi nhanh Cdch 3: Cdc ban de y quan he giua x vd x^ la: xdx = ~^d{^x~^ = -^d{2-x^y ;—'—dx= X +\ 1 2x Xet tich phan J = Nen viec ta chon an phu cdch 2) la hodn todn tu nhien ! khong mang tinh dp dat cua kinh nghiem trong 2x_ \dx+ \— dx^\ J 0r - -4-1 0 x'+\ J V 2x x'+l -dx •dx. x^+1 (0 t ; dx^ x ^ + 1J Dat x = t a n / = > ( i x = —^—dt = {\-^ian-t]dt, cos"/ ^ ' ti 2 '2 suy nghi Id : "thay cd can thiic thi dat can thitc la an phu". Chung ta c6 the gidi nhanh nhie sau: Doican: x = 0 ^^^-p!- / = 0;x = 1 = ^ / = — . 4 3 x42^dx = ^\2-xjd(2-x') = 1 _ 2V2-I 0" 3 Tadugc / = f ^ ^ ( l tan / + 1 sin/ + t a n ^ / ) ^ / = 2 f ^ c J/ = 2xdx. D6ican: x = 0 (x^ + 1 ) . Nen viec ta chon dn phu t = x' -\-\(o cdch 1) Id hodn todn tu nhien ! x'+\ Dat / = 2xdx = ci(x^ j = -dx / = 2. = X + -flX = I , dx = \dx + f ^"^ dx 0 0 + ln x ^ + l i/(x'+l) =x x^+1 1 = l + ln2. 0 = ln2. LM gidi that nhanh gon ! Vay / = l + l n 2 . D6 CO each nhin "tudng minh" vh each giai nhanh Nguyen ham va Tich phan, mai ban doc t i m hieu nhirng kien giai trong cuon sach nay ! ChLPcng 1 . N G U Y E N H A M (2) Cong thii-c 2 : \dx=l Ta suy nghi : ham so nao c6 dao ham bac nhat bang 1? De dang nhan thay do la X vi x' — 1. Vay ta c6 cong thuc thu haii Bai 1 .NGUYEN H A M \dx = x + C 1. Dinh nghIa (3) Cong thii-c 3 : Cho ham so f(x) xac dinh tren K (K la khoang ho^c doan hoac nua khoang cua M ) . Ham s6 F(x) dugc goi la nguyen ham ciia ham s6 f(x) tren K nSu x"dx =? Ta suy nghi: ham s6 nao c6 dao ham bac nhat bang jc"? Chung ta lien tuong ngay toi cong thuc dao ham {x")' = nx"'^ hay F'(x) = f(x) vai mgi x thugc K. = x" . Ta thay n-\^a Mgi ham s6 f(x) lien tuc tren K d^u c6 nguyen ham tren K. hay « = a + 1 , thu dugc cong thuc = X « +l Sau nay, yeu chu tim nguyen ham cua mot ham s6 dugc hieu la tim nguyen = x" . Vay la ham so hay ham tren tung khoang xac dinh cua no. F(x) la mot nguyen ham ciia ham f(x) thi F(x) + C (C la hang s6) la ho nguyen ham cua ham f(x) hay tich phan hk dinh cua ham f(x). Ki hieu : fix)dx c6 dao ham bac nhat bang x". Suy ra a +\ ..a+\ cong thuc thu ba : x"dx=-— + C (a^-1). ar + 1 = F{x) + C (4) Cong thuc 4 : f—c/x =? Ta suy nghi: ham so nao c6 dao ham bac nhat Vi du 1 bang — Ta lien tuong toi cong thuc ( i n x ) =— thi thu duac cong thuc a) J2xdx = x^+C vi ( x ' + C ) ' = 2x. b) X 9 X cosxdx = smx + C vi (sinx + C)' = cosx. * Luu y: di hiiu nhanh nhung noi dung kien thuc trong cuon sdch nay, ban doc nen ren luyen thdnh thgo viec tinh dgo ham ! . Chiing ta lay dau gia tri tuyet doi vi dieu kien ciia ham Logarit! V (5) Cong thij-c 5 : 2. Tinh chat thii- nhat f'{x)dx=fix) \-dx^\nx+C J a''dx=7 Ta suy nghi : ham so nao c6 dao ham bac nhk bang a''? Tu cong thuc tinh dao ham quen thugc (^a"^ = a ' ' l n a +C Tinh chat thu nhSt dugc suy true tilp tir dinh nghia nguyen ham. Trong thuc hanh, tinh chk nay giup ta tim ra nguyen ham cua mot ham so don gian, cung vlna; = a", tiic la ham so In a hay c6 dao ham bac nh^t bang a". Vay ta d l dang nhu viec xac dinh lai nguyen ham tim ra c6 dung khong theo each nghi: ''muon tim nguyen ham ciia ham so f(x), chiing ta tim ham so md dgo ham bgc nhat thu dugc cong thiic cm no phdi chinh la f(x)'\i each hieu do, chung ta c6 the thanh lap Bang a +C a'dx = — (a>0,fl^l). \na (6) Cong thuc 6 : cong thuc nguyen ham co ban nhu sau : e''dx=? Ta suy nghi: ham so nao c6 dao ham bac nhSt J (1) Cong thirc 1 : Qdx =? Ta suy nghi : ham so nao c6 dao ham bac nhat bang 0? Hien nhien do la hang so ! Vay ta c6 cong thuc thii nhSt: Qdx = C bang e"') De dang ta nhan thay do la ham e' vi (e'') = € ' , suy ra cong thiic thu sau : e^dx = 6" + C . Cong thuc thii sau la truofng hgp rieng ciia cong thiic thii nam khi thay "a" bang "e" ! (7) Cong thu-c 7 : jcosxdx =? Ta suy nghi : ham so nao c6 dao ham b$c nhk bang cosx? Tir cong thuc quen thuoc (sinx) thuc thu bay la : =cosx, ta c6 ngay cong — sin^ X — cos^ X sin^ X cosxi/x =sinx + C (8) Cong thii'c 8 : sin xdx =? Ta suy nghT: ham so nao c6 dao ham bac nhat sin^ X sm^ X smx ^ . Vay ham so c6 dao ham bac nhat bang ^ la sin^ X sin^ X ham so - cos X hay ( - cotx). Suy ra cong thuc thu m u a i : sinx bang sinx? Tu cong thuc quen thupc (cosx) = - s i n x hay (-cosx) = s i n x , dx = - cot X + C ta CO ham so ma dao ham bac nhat cua no bang "sinx" la " - cosx", suy ra cong thuc thu tarn la : | sin xdx = - cos x + C 1 (9) Cong thuc 9 : hop tren ! chung ta dir doan ham so can tim c6 dang A. cosx + sinx. A "cos^x"). Ta c6: cos a; ' ^ cos cos 1 ' -. Vay ham so c6 dao ham bac cos X , ro rang neu chon A = sinx thi X cos^x + sin^o; X X —Y~ ^ cos x +C •dx = ? Ta suy nghi : ham so nao c6 dao ham <-' s i n ' sm X ? Tuang tu cong thuc 9 ! Minh du doan ham so can tim (chu y do mau thuc "sin^ x "). Ta c6: sinx X y +C / x"dx = - f J — dx = \n + C(a ^ - 1 ) X + C{cy > 0; Q ^ 1) cos xdx — sin x + C sin xdx = — cos x + C f — ^ - — dx = tanx + C ^ cos X +C X e'dx = e' J In a +C f—— dx = — cot X + C sin^ X Hieu va thuoc bang nguyen ham ca ban la dieu kien thiet yeu de chung ta su dung thanh thao cac cong thuc trong bang nguyen ham ca ban. (10) Congthiic 10 CO dang dx = adx = tinh dugc nguyen ham cung nhu tich phan sau nay. Chinh vi vay, chung ta can — X bac nhat bang J Sill X nhat bang — - — la ham so hay t a n x. Suy ra cong thuc thu chin : cos a; cos X r f =C (chu y do mau thuc cos a; cos^ Jodx X 1 nhat bang — r — ? Truang hop nay khong de tim nguyen ham hon cac truoTig cos X A. COS X + sinx.A X Vay ta c6 Bang nguyen ham ca ban sau : -dx=1 Ta suy nghT : ham so nao c6 dao ham bac cos sin^ 3. Tinh chat thu- hai J kf{x)dx = kj f{x)dx Trong cong thuc nay, dieu ma chung ta can chu y la he s6 "k" (he so k c6 the "ra", "vao" qua dau nguyen ham!), tat nhien k phai la hang so, con bien so khong dua ra ngoai dSu nguyen ham dugc. Vi du 2. Ap dung tinh chSt thu hai va Bang nguyen ham ca ban, ta c6 : a) J 6xdx = GJ xdx (dp dung tinh chat thu hai) = 4 •2 = 2a;^ + 3 sin X + C = 6 — + C (dp dung bang nguyen ham ca ban) 2 b) / = r (5e" ' thif ba) = 3a;' + C . , . r cos X , I f h 3 sin a; + C (dp dung bang nguyen ham ca ban) 1 cos xdx = — s inx + C • 3 = r Se'dx- f dx (dp dung tinh chdt cos' x *^ ^)dx cos' x = 5 r e^'dx - 7 r — - — dx (dp dung tinh chdt thu hai) . "J ^ cos' X = 56^^-7 tan x + C . (dp dung bang nguyen ham ca ban) c) j e^^'dx = J e.e'dx = e J e'dx = e.e' + C. d) \mx\lx = \Q\x'dx = \0.^— + C = 6x' +C. - +1 3 Noi chung, khi tinh nguyen ham cung nhu tich phan sau nay, chung ta c6 c) I, =: j 3 - + ax + a + 6 = x, Vx e R a-l = 0 |'a = l , Ci>< . a + b-0 b = -l thi F ( x ) la mot nguyen ham cua f(x). Vay vai a =l va Vi du 6, Chung minh rang F(x) - sin xe"" la mot nguyen ham cua ham so / ( x ) = (sin X + cos x y . Gidi + 1)^ dx X 2 1 + — + -I CO : J 7^ = X X -I 1 - dx -I 1 x = ^ + - e^x^ 1 dx X- + Tap xac djnh cua F(x) va / ( x ) la IR . Taco: F ' ( x ) = ( s i n x ) ' ^ ' ' + sinx(e'')' 4^2!^ + In = cos xe"" + sin xe' = (cos x + sin x)e'' = / ( x ) . + C. X Vay F(x) = sin xe' la mot nguyen ham cua ham so / ( x ) = (sin x + cos x •dx e'x^ 3 3 ,3 JC JC 1 2 X X dx = — - + In X X c) T a c o ': 73 = JT.2,'dx 1 + dx x = J(2.3)'dx X ~e'' dx BAITAP 1. Tinh : +C. 25a.'' + 1 2 2 ' + 1991 = jG'dx d) T a c o : re = \—;<^ 42 a) Ta thi /^'(^) = fix) Ta CO : ^ c) Tap xac djnh cua F(x) . 'J tan a; -dx', sin 2x SI b) 7^ = J " dx cos X sm X 3. Tinh : Cong t/iii'c nguyen ham cff ban a) / = / c) • dx e'dx, X = JZ'e^dx; — xe dx. X 4. Tinh : Jodx Jdx = C = x +C f x"dx = ^ ^ r(VJ+4^ + b^)dx; b) / = + + - 1 =^)dx 5. Cho ham s6 / ( x ) = ( x ' + x)e" va F(JC) = (ax' +bx + c)e'. Vai gia tri nao ciia a, b va c thi F{x) la mot nguyen ham cua / { x ) ? Cong thuc nguyen ham m& rpng J a + 1 + C{a * - 1 ) f{ax+bfdc='^.^'^^^^'^\c{a^ f -dx = hi 1 X 1 + C f —-— dx = — An \ + b \ ^ ax + b a J X J e'dx = 1) a+l a +C f e'"'-'dx = -e'^^'' a +C 'J Bai 2. BANG NGUYEN HAM M Q RONG Sail day chung ta se ma rong cac cong thuc nguyen ham ca ban de dugc f adx = — + C{a>Q]a^ In a 1) ^ ra'^^'dx = ± . ^ + C{a > 0;a ^ 1) a \na Bang nguyen ham ma rong. Bang nguyen ham ma rong la cong cu giup chung ta tinh nhanh nguyen ham va tich phdn. Truac tien ta xet dinh li sau : j cos xdx = sin x -\- C / cos (ax + b)dx = — sin (ax + b) + C 1. Dinh li Nh J f(u)du J f{u(x)).u\x)dx =F{u) + C vau = u(x) la ham s6 c6 dgo ham lien tuc thi: =F{u{x)) f sin(ax + b)dx = — — cos(ax +b) + C a +C 2. Cong thirc nguyen ham mo" rOng Ap dung dinh li tren trong truong hop u = ax + b{a^0),t^c6 jfiax J sin xdx — — cos x + C + b)dx= jf{ax + b).{ax + f —\ dx — tanx ^ cos X : by.-dx f —\ dx = — cot x + C ^ sin X r + C I 1 1 — dx = — ta n{ax + b) + C ^ cos (ax + b) a r 1 1 / — dx = cot(ax + 6)-1-C sin (ox+ 6) a Trong thuc hanh tinh nguyen ham cung nhu tich phan sau nay, a nhieu f(ax + b).{ax + bydx =-F{ax + b) + C a Tom lai, ta c6 cong thuc dk mo rong bang nguyen ham ca ban: '/{ax + b)dx = -F(ax + b) + C a Cong thuc nguyen ham ca ban va cong thuc nguyen ham ma rpng dugc cho tuang ung duai bang sau : truang hop viec ap dung bang nguyen ham ma rong cho ta lai giai bai toan nhanh va " sang " han ! Chang han vai bai toan sau : Tinh nguyen ham : I — J (2x + l^dx. Neu khong ap dung cong thuc nguyen ham ma rong thi ta khai trien bieu thuc {2x + 1)^, sau do mai ap dung cong thuc nguyen ham ca ban de tinh : I = J{2x + lydx = J(2x Giai + l)(2x + Ifdx = J {2x + l)i8x''+12x^+6x Ap dung cac cong thuc trong Bang nguyen ham ma rong ta c6 : + l)dx a) = J (16a;' + 24a;' + 12a;' +2x + 8x'' + 12a;' + 6x + l)dx = J (16x' 16x' 5 24x' ^ 2a; + 3 J \ 8x' ^ 2 -1 + C(a ^ - 1 ) ) d^ tinh/, ta c6 : 4+1 nhau mot hang so xde dinh!) Neu bai toan tren ta thay s6 mu 4 bang so mu 2013 chang han thi lai giai nhu each dSu tien se phuc tap nhu the nao? Con neu chung ta ap dung cong thuc nguyen ham ma rong ta c6 ngay lai giai ngan gon cho bai toan do la / = / (2a; + If^^dx = ^ '— + C . Ke ca chung ta dung phuang phap J 4028 d6i biln s6 (se hoc a bai sau) thi cung c6 lofi giai khong gon bang each nay ! R6 rang cong thuc nguyen ham ma rong to ra uu diem han cong thuc nguyen dx = -(3-2a;)' L + C. = __V -7 + C- 18(3a;-l) d) /, = {• , J^A: = 2 dx= 3' -5 {3x-2)idx +1 * Nil an xet: - Chung ta c6 the trinh bay nhanh nhu sau : a ) / , = / 23;+ 3 12 dx = 3 — 2xI dx = 2a;+ 3) vl3 26 -(3-2.)' 8 1; i = 0 thi ta thu dugc cong thuc nguyen ham ca ban. 1 dx Vi du 1. Tinh : J vl3 2a; + 3 = ^ - L 26 (3-2a;f .-^ L + ham ca ban ! Cac cong thuc nguyen ham ma rong, neu chung ta cho he so a = = / 10 (C/zw >• rang, each nay va each tren deu cho kit qua dung, no chi sai khdc a) \13 r I ci \-7 1 (3a;-l] c) / = / -(^a; = / ( 3 a ; - l dx = -> '—^ C ' ^ (3:.-lf ^ ^ ^ 3 - 6 + Sa;" + 8a;' + 4a;^ + a; + C . thuc r{ax + bydx = - . ^"•'^ ^ ^ o a +1 1 v3 b) / = / 3 - 2 a ; Bay gia chung ta ap dung cong thuc nguyen ham ma rong (cong «^ ^ / = r / 1 2a;+ 3 dx = ^.^ 2 13 xi2 r, + 32a;' + 24a;' + 8a; + l)dx 32x* 16x-' ' c, 18(3a;-l)' (2a; + s)'' dx; b) = J'(3 - 2a;)' dx dx • (3i-l) ^(33;-2) 2 -5 .,,.2_(3x-2)3 -1 #x-2)^ 3 - D6i vai Cdu c) vd cdu d), neu sir dung cong thitc + C. •dx = —.a (n-l){ax -1 C{n^l;a^O) se — + + b) cho ta lai gidi nhanh han nua (vi gidm duac mot buac biin doi) ! -1 = C. -1 rdx = - A p dung cac cong thuc trong Bang nguyen ham m a rong ta c6 : 1 a) 7^= f - ^ d x = —In 2 2-5a; -5 b) 7 = 18(3a;-l) (3x-l) d)L + Gidi ' • + C. c) day chung ta dan cu cdu a) duac gidi bdng phuang phdp doi bien so de so sdnh hai cdch \e ^''dx = ^ e l3= Ta M = CO 5 2a; + 3 12 2a: + 3 => (iti = 2dx :/ = 1 \v}^-du J ^''+C = -3e r, rfx 1 = - du 2 = - . — + 2cy 13- • d) 7 = ^ b) 7 = ' — + C 26 fe-''dx —e-'+C = ^ r3'^'-'dx - J c)/3= Ap dung cong thuc nguyen ham ma rong ta c6 lai gidi gon vd nhanh han: \13 2x + 3 12 2a; + 3 dx = + C 26 +C. -e-^+C. = J _1 + C f 3.3'^rfa; = - . — - + C 5 In 3 _1 \e 3^c/x = -3e ^\c. V i du 3. Tinh : b) 7^ = J a)7j = J s i n ( 3 - 4 a ; y a ; ; :) I = Do vdy, viec nha vd van dung tot cong thuc nguyen ham ma rong Id can f V i du 2. Tinh : 5cos{—^)dx •dxsin ( 2 z - l ) - L - •dx' cos^(3;r) thiet de chung ta tinh nhanh duac nguyen hdm vd tich phdn sau nay. Gidi A p dung cac cong thuc trong Bang nguyen ham m a rong ta c6 : f—^dx; 1 —X e ^ dx; C• + C. 26 Rd rang cdch nay to ra khd phuc tap so vai mot bdi todn dan gidn nhu vdy ! c) h= + 5 ln3 3 a) 7 = / — - — dx = —^ In 2-5x ' J 2-5a; 5 dx Thay u = 2x + 3 , ta duac: /, = a) /, = C = - Chung ta c6 the trinh bdy nhanh nhu sau : gidi: / = Dat C = — I n 2 - 5 a ' + C. 5 f S'^'^dx = 3 r3'^ia; = S.-.—+ J 5 ln3 J - Neu khong dp dung cong thuc nguyen ham ma rong, chung ta gidi bdi todn bang phuang phdp doi bien so md chung ta se xet trong bdi hoc sau. O — 5a;+ b ) / , = / 3 d) 7^ =. J dx e-'dx a) /j = J s i n ( 3 - Ax)dx cos(3 - 4a;) 4 = ^ . ( - c o s ( 3 - 4a;)) + C ^ + C--15sin dx = 5.—.sin b) h = JScos 3. Tinh : \ / \ / = f c) I +C dx = f cos\3x) 1 ^ dx = -temSx 3 cos'(3.x) r ( s i n ( 3 - - ) + cos53;)t/x; b) / 2 + C- 1 .a: — h sm — sin 3a; 2 ; f ^ ^ cos (4 — x) d) ^4 = / 2cos(3--)rfa;. 1 -dx = 3.(--cot(2a; - 1)) + C sin^(22;-l) 2 3 ^) I . = Bai 3. PHl/QNG PHAP D O I B I E N S O = - | c o t ( 2 a ; - l ) + C- Bai nay chiing ta se xet hai truong hop Ichi tinh nguyen ham - Chung ta c6 thi trinh bay nhanh nhu sau : /{x)dx bang phuong phap doi bien so : a) / = fsin{3-4x)dx=:^^^^^^-^C. 4 J = J 5cos{^)dx b) = -15sin(^) + C - Truong hop 1 : Dat u Id mot ham so cua x. — Truong hop 2 : Dat x la mot ham so cua u. A. Phep dat u la mot ham so cua x : u = u(x) Gia su can tinh / = /{x)dx, ta thuc hien nhu sau : c) / ' = f J cos\3x) dx^-tanSx 3 + C- Buoc 1 : Chon an phu thich hop u = u(x). Buoc 2 : Xdc dinh viphdn du = du(x) hoac du^ - du^(x) ... ^ ^ da; = - | c o t ( 2 x - l ) + C. sa ii ln ' ( 2 x - l ) f (x)dx = g{u)du. Chu V : chon an phu u = w(x) sao cho viec tinh / = g(u)du phai de hon la 1. Tinh : = theo u va du. Gidsurang Buoc 4 : Tinh I = g(u)du. Sau do thay u = w(x) de dime ket qua can tim. BAITAP a) Buoc 3 : Bieuthi f(x)dx J (4x + 2)' dx; tinh / = jf(x)dx b) ^. = / - ^ ^ - ; ^ (3-2x) ! Khi nhin vao mot bai giai cho bai toan tinh nguyen ham hay tich phan bang phuong phap dat an phu (hay phuong phap doi bien so), ban doc thuong c6 cau hoi : tai sao lai chon dat an phu nhu vay? Lam sao chon an phu thich hop?... ^(41 + 5) Nhirng Icien thuc duai day se giiip cae ban dinh huong dugc phep dat an phu cho minh mot each nhanh chong ma Ichong phai may mo lam giam toe do tinh 2. Tinh: b) = J nguyen ham, tich phan cua cac ban. Truoc tien cac ban c^n luu y hai ket qua ma chiing ta thuong dung sau day : (1) c)/2= |e 3 Jx; 4 J df{x) = f\x)dx . (2) Niu J f(u)du =F{u) + C va u = u(x) la ham sS c6 dqo ham lien tuc x^dx = -d(2x^ +1), nen ta c6 x'{2x^ + ifdx = -(2a;' + lfd{2x^ + 1). 6 thl: J f{u{x)).u\x)du = J i\u{x))du{x) =F{u{x)) +C 6 Do do, viec chung ta chon an phu la u-2x^ mang tinh dp dat. Vi du L&i giai cua bai todn a) J cos(2x^ + 3a; + l)d{2x^ + 3a; + 1) = sin(2a;^ + 3a,' + 1) + C. (ta hieu trong suy nghi " 2x^ + 3x + 1 " J x'(2x' lau) b) f ^- d(x' + 1) = l n a;' + 1 + C = ln(a-' + 1) + C v/ ^ (a;' + 1 ) hieu trong suy nghi " -\-\" la u ) + Ifdx Dat u = 2x^ +1 + 1 > 0 (ta = J-(2x' + lfd{2x' + !)• du = d(2x^ + 1 ) . Taco: T = - / u M u = - . + C = + C' 6 - ^ 6 10 60 Sau day chiing ta tim hi6u cac moi quan he quan trpng giup chung ta tim Thay u = 2x^+1 ta duac: / nhanh phep dat in phu va dinh huong nhanh each giai cho bai toan nguyen + ' ham, tich phan bang phuang phap doi bien so. l.Quan hegiua x" va + \ hodn todn ty nhien, khong 60 b) Phan tich bai todn: Cac ban de y quan he giica x vd x : x"*\n^-\) d{ax"^'+b), Ta CO : dx'"' = (« + \)x"dx o x"dx = —dx"*' ' n + \ + \) do a^O con b tuy y tren R . Vay ta c6 quan he giira x" va x"^\n^-1) xdx = - d(x^ +1) nen ta c6 x^x^ +\dx = - Vx^ +1 d(x^ +1). Do vdy, ta c6 thi 2 2 trong chon an phu Id w = nhu u= 1 d{ax"''+b) sau : x"dx = a{n + \) de bieu thuc dual ddu nguyen ham khong con can thuc.. (ta hieu cong thicc tren mot each don gidn nhu sau : dua x" vdo trong vi phan thl thdnh {ax"*^ -^b), voi a ^ +1 hodc u = Vx" + 1 . Trong truang hap nay ta nen chon L&i giai cHa bai todn Q vd h tuy fxyjx'+ldx = J^Vx'+l d(x' + l ) . ytren Dat u = V x ' + l ^ u ' = x ' + l : ^ d u ' = : d ( x ' + l ) . Vidu l . T i n h : 1 a) Jx^{2x^ +lfdx\ = Jx4^+ Taco: Idx Giai bieu thuc (2x + 1) , sau do nhdn vai X de dua ve nguyen ham de tinh han. la khong dan gidn? Do vdy, 1 d u ' = J - u 2udu = JuMu ' Dat u = x^+l=:>du = d ( x ' + l ) . 3 each nay da to ra khong hieu qua ! Niu giai bai todn nay bang phuang phap doi biin so, ta chon an phu la u-lx^ + \. Tgi sao Igi chon duac an phu nhu vay? Bay gia cac ban de y quan he giica x^ va x^ nhu sau : ^ = • — + 0- /"I—T t A (Vx^ + 1 ) Thay u ^ Vx +1 ta dugc: i = ' 3 Cach khac : a) Phan tich hai todn: Theo I6i giai thong thuang, cac ban se khai trien The nhung viec khai trien bieu thicc (2x^ + if = Jiu Ta c6: I2 = j-Judu 2 = — ju^du = 2 2 + C = ^^^^^ + C 3 2 (Vx^ +1 f 1+- Thay u x +1 ta duac: I = 1- C • ' 3 * Nhgn xet: Neu da thanh thgo trong viec sic dung phuang phdp nay, cdc ban CO thi trinh bay loi gidi nhanh hon nhu sau : + C • (ta hieu trong suy nghi "2x^ + 1" la "u") a) == J x'(2x' 60 + l)Mx = J -{2x' + l)M(2x' + 1) = -4x^+\d{x^ b)l2= \x^x^+ldx +1) = - f(x^ +1)2d{x^ +1) Ta CO giua '1' a) Ii = ^ \—r-dx; X X = — reMu- —e" +C. X X L&i gidi cua bdi todn X 1 1 b) / = I — sin — cos — dx. X X 1 ta CO the chon an phu Id u = — . r I ^ va — 1 , dx — - .sin1 — cos1 —, d nen. ,ta .co1 —. sm1 — cos — . Do do, X X X X X X l4 r 3^ e 1 + ^d 1+ - X 1 dx = —d phdn thi thanh — + b, voi a ^ 0 vdbtuyy tren R ) re -dx = f-l b) Phdn tick bdi todn : Cdc ban de y quan he giua X (ta hieu cong thuc tren mot each don gidn nhu sau : dua \ trong vi Vi du 2. Tinh : 1.1 3 -1 Thay u = 1 + — ta duac: Ii = — e ^ + C. X • 3 nen quan he can xet giila — va ^ la: X X L&i gidi cua bdi todn Ta CO : I va — X 3^ ^d 1 + - Dat u = ! + - = > d u = d(l + - ) . . + C • (ta hieu trong suy nghi " x + 1 " la "u") 2. Quan . , —1 va , 1— 1; —, dx = -—1 a, giua , e — d\-—e-1 X nen ta co X X 3 Do do, ta CO thi chon an phu la w = 1 + —. X X Dat u = Gidi a) Phdn tick bdi todn : Niu chua dugc biet din quan he giua \ — thi X Taco: i X that khong de de chung ta tim ra ngay phep dgt an phu! Cdc ban de y quan he = 1 r • 2J 1 / sin —d — X 2^ X i =^ du = d X 2 _ i f sin(2u) du = - i ( - - c o s 2 u ) + C - - c o s 2 u + C9.J ^ 2 2 4 Thay u = — ta duac: I = — cos X 2 4 + C- * Nhan xet: Niu dd thanh thao trong viec sic dung phuang phdp nay, cdc ban CO thi trinh bay lai gidi nhanh han nhu sau : 1+- -dx = e \ b) 1 = / — X 2 Thay u = 21nx + 3 tadugc: 1+- -1 ^of X X X — 10 / ' 3 / s m — cos — ax = — i s m — c o s —a X "J 1+-- =ifuMu=1.^:^+0=^+0. 6:1 CO 1 3 a)Ii = Ta = f sin 2^ X 20 ^ ( 2 I n x + 3)^" ^ 20 ^ b) Phdn tich bai todn: Cdc ban dSy quan he giua — va \nx : —dx = d{\n x) x X —a X 2 l n ^ x + 51n^x , 21n^ x + 51n^ x \ , , ^ , dx = d[\n x). Do vay, ta chon an phu Inx xlnx yi^n ta CO = — cos 4 /fl w = In X. 1 3. Q u a n he g i u a — v a line LM gidi cda bai todn X — dx X = — d{a a \nx + 2 In^ r 2211nn-''xx + 5 l n ' x dx In X Ta CO (In x ) = — nen quan he can xet giiia — va I n x la : X + 5 In^ x Inx d(lnx). Dat u = Inx => du = d(lnx). b) •2u' + 5 u ' Taco: = j : 2u' du u (ta hieu cong thuc tren mot each dan gidn nhu sau : dua — vdo trong vi + 5u' — du = u J (2u^ + 5u) du X phdn thi thanh {a I n x -\-h), vai a ^Ovdb tiiyy tren R) 3 V i d u 3. Tinh : 2 In X + 3 -dx; b ) I , = / 21n'x + 5 l n ' x xlnx X 2 Thay u = Inx ta duac : i ='^i^RJ^ +^A\}12^+ c • ' 3 2 * Nhan xet: Niu da thanh thao trong viec su dung phuang phdp nay, cdc ban CO the trinh bay lai gidi nhanh han nhu sau: dx. Gidi: a) 1 , = / i 2 In X + 3 p ^dx= \ K9 /•-(21nx + 3 d(21nx + 3) a) Phan tich bai todn: Cdc ban di y quan he giita — va I n x ; X _(21nx + 3r ^dx = ^d(21na, + 3) nen ta c6 i l l ! l f l 2 L a ! x = l(21njc + 3)V(21nx + 3). vay, ta chon an phu Id u (2 In 2lnx + 3. Ij = J Dat u = 2 p i , X + ^dx In X + ., + 3). d(21nx + r 2 In^ x + 5 In^ X , / xlnx dx = r 2 In^ x + 5 In^ x , , / ; d(lnx) Inx ^ "-/, ^2 ^, 1,., X 2(lrix)^ 5(lnx)^ „ 2 ( l n x ) +51nx d(\nx) = -A L . 4. _A L . 4- Q . -•^ ^ J 3 2 >9 = J ^ ( 2 1 n x + 3) 3 => du = d ( 2 In X 20 b) I Ldi gidi cda bai todn n Do ^ ^ 3). 4. Q u a n he giira e^ v a ae^ + b Ta CO {^ae' = ae' nen quan he can xet giira va ae' + b la: Cdc ban de y quan he giiea (a^O) a (ta hiiu cong thuc tren mot each dan gicin nhu sau : dua 6 1 • .e'^^dx + 1 • dx — + 1 vao trong vi -d{e' + 1 +!)• LM gidi cua bdi todn V i d u 4. Tinh : ' 1 e'^dx = d(e* + 1) nen ta c6: Do do, ta chon dn phu la u = e" + \. phdn thi thanh {ae"" + h), voi a ^Ovdb tiiyytren M) 1 [—1—dx. b) va e'^ + 1 l + e- X -J 2e' +1 -dx = —-dx = -dx = -.e'dx e^+1 e'+\ 1+- Gidi a)Phan tich bdi todn : Cdc ban de y quan he giua ^ va 2e^ + 1 .• + 1 Dat u = e ' ' + l ^ d u = d ( e ' ' + l ) . e'^dx = ^ d(2e^ + 1) nen ta c6 dx 2e" + 1 3 1 .e'dx = 2e" + 1 Taco: I = f-du u . - d(2e' + 1) 2e^ + l 2 (i(2e^ +1). Do vgy ta chon dnphu la u = 2e' +\. 2 2e' +1 Ld'i gidi cua bdi todn I = r-^-Hl-dx= r - ^ . e M x = .id(2e''+l) 1 J 2e^ + 1 ^ 2e^ + 1 ^ 2e'' + 1 2 ' = ln|u|+C. ; Thay u = e" + 1 ta dugc: * Nhan xet: Neu da thanh thao trong viec su dung phuang phdp nay, cdc ban CO the trinh bay lai gidi nhanh han nhu sau: a) I = r — d ' -'20^+1 b) I2 = x = r — - — . - d ( 2 e ' ' + l ) = -ln(2e^+1) + C. = r-.^^d(2e^+l) = - f—^d(2e^+l). Ta CO 3 r l , ' 2 ^ 1 1 = 3 In u + C 2 tuyet d6i vi 2e'' +1 > 0) l + e-X -dx = ^ -dx e^+1 Ta CO (sinx) =cosx va (cosx) = - s i n x nen quan he can xet giua sin a; X la: s inxdx = — — d(a cos x + b) a (Ta hieu cong thuc tren mot cdch dan gidn nhu sau: dim cos x vao trong 1 1+ ^ f — ^ d(e'' + 1) = ln(e^ + 1) + C • c'' + 1 cos xdx = — d{a s i n x + b ) a b) Phdn tich bdi todn : Ta Men doi h = 1+ 2 5. Quan he giua sinx va cosx va cos 1_ -dx = ^ 26" + 1 Thay u = 26" + 1 ta dugc : 1 = - ln(2e"' + 1) + C • (ta khong lay dau gia tri r 2e^ + 1 2 ^ -dx = l +e 2 20" + 1 2 Dat u = 26" +1 =^ du = d ( 2 e ' + 1 ) . = ln(e" + 1) + C 1 -dx = -dx viphdn thanh (asinx+h); dua sinx vao trong vi phdn thanh -(acosx + b), vai * Nhan x4t: V i du 5.-Tinh : b) = Jcos^xsin'xdx; a) Jcosxe-'"""+'da:. Neu da thanh thgo trong viec sir dung phucmg phdp nay, cdc bgn CO thi trinh bay lai gidi nhanh han nhu sau : a) I j = J cos^ x sin^ x d x = J cos x ( l — sin^ x) sin^ xdx Gidi — J (I- a) Phan tich bai todn : Ta bien doi : cos^ X sin'^ x = cos x cos^ x sin^ x = cos x{\ sin^ x ) sin^ x . Cdc ban de y quan he giua sinx va cosx; cosxdx sin^ X = d(sinx) 5 b)/, = LM gidi cua bai todn f{l- / cos xe - 3 1 = J cos^ x sin^ x d x = J cos x cos^ x siii^ x d x cos x ( l - sin^ x ) sin^ x d x = sin^ x nen ta c6 cosx(l -sin^x)sin^xdx = (1 -sin^x)sin^xd(siwc). Do vdy, ta chon an phu la u = sinx. =J sin^ x ) sin^ x d ( s i n x) = J* (sin^ x - sin^ x ) d ( s i n x ) - 33sinx+2 si 3 ^ sm x ) sin^ xd(sin x ) . sin + c. - 1 1+2 dx^j -38mx+2 +c. 6. Quan he gifra sin^x, cos^x va sin2x Ta CO (sin^ x ) = 2 s i n x c o s x = sin2x Dat u = sinx => du = d(sinx). c / ( - 3 s i n x + 2) va (cos^ x) = - 2 c o s x s i n x = - s i n 2 x nen quan he can xet giua sin^x, cos'^x va sin2x la : Taco: = J (1 - u ' ) u M u = J ( u ' - u ' ) d u = y ™ • . i T (sinx)^ Thay u =^ sinjc ta duac: I =-^^ ^ ' 3 - y+0- (sinx)^ „ ^ -^ —+C5 b) Phan tich bai todn : Cdc ban de y quan he giua sinx va cosx; cos xdx = d{-3 sin 2xdx = — d{a sin^ x -\-b) a (ta hieu cdng thuc tren mot cdch don gidn nhu sau: dua sin2x vao trong vi phan thanh (a sin^ x + b) hogc —[a cos^ x + b), voi a ^0 va b tuyy tren M) V i du 6. Tinh : sin x + 2) nen ta c6 a) cos = — e-'^'^'^^di-?, 3 Do vdy, ta chon an phu Id u = -3sinx sin 2xdx = — — d{a cos^ x + b) a = j(3siu^x+l)sin2x(ix; b) = i n22 xz r _ ^ _ s is n ^ sin x + 2)- V2sin'rz;+• 3 cos rdX • X Gidi + 2. a) Phan tich bai todn : Cdc ban diyquan he giCta sin^x va sin2x; LM gidi cua bai todn = Jcosxe-'''"''^'dx = J ^ e - " ' " ' ^ + ' ( i ( - 3 s i n x + 2). Thay u = - 3 sin x + 2 ta dugc: ta c6 (3sin^ x+l)sin2xda; L&i gidi cua bai todn : +C. = — nen = •^(3sin^ x + l ) d ( 3 s i n ^ x + 1 ) . Do vdy, ta chon dnphu la u = 3sin^x + I . o Dat u = - 3 sin x + 2 =^ du = d ( - 3 sin x + 2 ) . Taco: I = — f e M u - — e " ' 3 ^ 3 sin2xrfa; = - d ( 3 s i n ^ x + 1 ) 3 = J(3sin'x+l)sin2xdx e-3»""'+2 ^ = J^(3sin^ x+l)d(3sin^ x+1)- Q . E>at u = 3 s i n ^ x + 1 ^ du = d(3sin^ x + 1 ) . Taco : I = - / udu = - . hC= i hC- Thay u = 3 s i n ' x + l tadugc : I ^ ( 3 s i n x + 1 ) ' 6 • * Nhan xet: - A'ew trinh bay lai gidi nhanh han nhu sau : a) b) Phan tick bai todn : Ta bien doi: sin 2x sin 2x sin 2x V2sin^ X + 3cos^ x -^2(sin^ x + cos^ re) + cos^ x Vs + cos^x thanh thgo trong viec sir dung phuang phdp nay, cdc ban c6 the = J(Ssin^ s i n 2a: _ V 2 + cos^a; =d{2 + cos^ .x)- 6 V 2 s i n ^ X + 3cos^ x = - '(2 + cos^x) 2 t/(2 + cos^x) = -2V(2 + cos^x) + C V2 + cos^x u = V2 + cos" X cfe Z>/ew //zii-c i/j/OT i/aw nguyen ham khong con can thuc. - A'ew chung ta de y den quan he giua = s\n2x , r —1 2 \ -/ , dx = J , c^(2 + cos^ x) • ^ V 2 s i n ^ x + 3cos^x ^ V 2 + cos^ x r X => = s i n a; va c o s x thi chung ta c6 them each gidi theo huang khdc nhu sau : a) / j = J(3sin^ L&i gidi cua bdi todn : Dat u = V2 + cos^ yJ2 + cos^ x do, ta c6 the chon an /fl w = 2 + cos^ X /zoac w = V2 + cos^x. r/-o«g truang hap nay ta nen chon h=J x+l)c/(3sin^ x+1) _(3sin^x+l)^ Cdc ban di y quan he giua cos^x va sin2x.- sin2xda; = -d{2 + cos^ x) nen ta CO x + l ) s i n 2 x c ? x = j^(3sin^ X x4-l)(ix x + l ) r f ( s i n x ) = y (6sin^ x + 2 s i n x ) ( i ( s i n x ) J2sinx(3sin^ „ sin^ = 6. 4 2 + cos^ x => du^ = d(2 + cos^ x ) . x + l ) s i n 2 x ( i x = j2sinxcosx(3sin^ „ sin^ x „ 3 . 4 . , ^ h2. + C = -sin^ x+sm^ x + C • 2 2 Chii y rang each nay va each tren deu cho ket qua dung, no chi sai khdc Tac6:l^=r j d i d u ^ = - J ^ d u = - 2 j d u = - 2 u + C- nhau mot hang so xdc dinh ! b) Thay u = y/l + cos' x ta dugc : I2 = - 2 \ / 2 + cos^ x + C . * Cach khac : , /2 = r / Ta CO : ^"^^^ dx = flpi^dx ^ V2sin^x + 3cos^x ^ V 2 + cos^ x (sau do dua sinx vdo trong vi phan) , sm2x , dx = COS" X - 1 r i ^ V2sin^x + 3cos^x Dat u = 2 + =f ^l2 + , ^ =rf(2 + cos''x)- (chu y r / -2cosx r cos^x thuc dao ham 2 - ^ 1 = - 2 (2 + cos^x)2 +C = - 2 V ( 2 + cos^x) + C . 2Vx Thay u = 2 + cos^x ta dugc: T = -2V2 + cos^ x 4 - C• x) (2±£^ij0i_ ^ ^ 1 (V^)'=4=)- 2x ^ V 2 + cos^x -1 = - |(2 + cos^x) 2 J ( 2 + cos^x) = _ cong 1/ 1 ==rd(cosx) = — I — = = = = = r d ( c o s ^ V2 + cos^x =^ du = d(2 + cos^ x ) . \ f - r = d u = —2Vu+C =