giải nhanh bài tóan nguyên hàm và tích phân

Vay dau la bi quyet de giai nhanh dugc mot bai toan nguyen ham, mot bai toan tich phan noi rieng?. Cuon sach nay viet ra nhSm dem lai cho ban doe nhimg each hieu, nhijng huang di, thu th

TRAN TUAN ANH

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

N G U Y E N H A M V A T I C H P H A N

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

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 • Fax P) 38.726,052 • MST: 03114307135

Email: WifiaclRfietcoxom- Website: «»w.sachvie!co.«ni

Xufi't ban nam 2013

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 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,

Thhy Nguyen Tat Thu, Giao vien Truang chuyen Luang Th6 Vinh

-Bien Hoa - Dong Nai

Tran Tuan Anh

G I A I NHANH B A I TOAN T I C H PHAN T R O N G

Doi can: x = l = > / = 0; x = 2=i>/ = ln2

dx = dx = d x + - Vdy ta CO the giai

nhanh bdi todn tren nhu sau :

I,^/ giai that nhanh ggn so v&i hai cdch tren !

Cau 2; Tinh tich phan / = ^x^l-x'dx (DH kh6i B - 2013

Cdch giai thong thu&ng Cdch 1: Do dau hieu " nen ta chon an phu x = V2 sin?

Dat X = y/l sin t => dx^-Jl cos tdt, te n n

Doi c a n : x = 0 / = 0;x = 1 => / = —

4

Taco: / = JV2 sin / V2 - 2 sin" /.72 cos / Jr = 2 V2 Jsin cos Vl - sin^ / J/

= 27^ sin ^ cos/ cos/L// = 2V2 sin/cos tdt

Xet tich phan J = 272 |sin / cos^ tdt

Dat u = cos/ ^du = -s'mtdt

V2 2V2-1

T a c o : I = -\t^dt= \t^^t= —

Cdch gidi nhanh

Cdch 3: Cdc ban de y quan he giua x vd x^ la:

xdx = ~^d{^x~^ = -^d{2-x^y Nen viec ta chon an phu t (0

cdch 2) la hodn todn tu nhien ! khong mang tinh dp dat cua kinh nghiem trong

suy nghi Id : "thay cd can thiic thi dat can thitc la an phu" Chung ta c6 the

gidi nhanh nhie sau:

3

0 " 3

L&i gidi that nhanh gon !

Cau 3: Tinh tich phan / = ( x + 1)^ , -<lx ( D H k h 6 i D - 2 0 1 3 )

LM gidi that nhanh gon !

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 !

C h L P c n g 1 N G U Y E N H A M

B a i 1 N G U Y E N H A M

1 Dinh nghIa

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

F'(x) = f(x) vai mgi x thugc K

Mgi ham s6 f(x) lien tuc tren K d^u c6 nguyen ham tren K

Sau nay, yeu chu tim nguyen ham cua mot ham s6 dugc hieu la tim nguyen

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 = F{x) + C

Vi du 1

a) J2xdx = x^+C vi ( x ' + C ) ' = 2x

b) 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 !

2 Tinh chat thii- nhat

f'{x)dx=fix) + 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

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

cm no phdi chinh la f(x)'\i each hieu do, chung ta c6 the thanh lap Bang

cong thuc nguyen ham co ban nhu sau :

(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

(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

\dx = = x + C

(3) Cong thii-c 3 : 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

= x" Ta thay n-\^a hay « = a + 1 , thu dugc cong thuc

(4) Cong thuc 4 : f—c/x =? Ta suy nghi: ham so nao c6 dao ham bac nhat

bang — Ta lien tuong toi cong thuc ( i n x ) =— thi thu duac cong thuc

X 9 X

\-dx^\nx+C

J V

Chiing ta lay dau gia tri tuyet doi vi dieu kien ciia ham Logarit!

(5) Cong thij-c 5 : 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 hay

= a", tiic la ham so c6 dao ham bac nh^t bang a" Vay ta d l dang

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) =cosx, ta c6 ngay cong

thuc thu bay la : cosxi/x =sinx + C

(8) Cong thii'c 8 : sin xdx =? Ta suy nghT: ham so nao c6 dao ham bac nhat

bang sinx? Tu cong thuc quen thupc (cosx) = - s i n x hay ( - c o s x ) = s i n x ,

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

(9) Cong thuc 9 :

1

1 cos X

-dx=1 Ta suy nghT : ham so nao c6 dao ham bac

nhat bang — r — ? Truang hop nay khong de tim nguyen ham hon cac truoTig

A COS X + sinx.A cos^x + sin^o;

cos^ X cos X cos X

- Vay ham so c6 dao ham bac

bac nhat bang ? Tuang tu cong thuc 9 ! Minh du doan ham so can tim

CO dang (chu y do mau thuc "sin^ x ") Ta c6:

sin^ X

dx = - cot X + C

Vay ta c6 Bang nguyen ham ca ban sau :

su dung thanh thao cac cong thuc trong bang nguyen ham ca ban

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)

= 6 — + C (dp dung bang nguyen ham ca ban)

2

= 3a;' + C , r cos X , I f 1

gang bidn d6i ham s6 dual dau nguyen ham hay dual dau tich phan xuat hien

nhung ham s6 c6 trong bang nguyen ham ca ban Do vay, viec nam dugc Bang

nguyen ham co ban la di^u kien rSt quan trong de chung ta tinh dugc nguyen

ham, tich phan

4 Tinh chat thir ba

J" {J{x) ± g{x))dx = J f{x)dx ± J g{x)dx

Chung ta c6 the hieu mot each dan gian cong thiic tren nhu sau: nguyen

ham ciia tong (hieu) cua hai ham so, bang tong (hieu) cac nguyen ham cua hai

ham so do

Cong thuc CO the ma rgng nhu sau :

/ U^{x)±f^(x)± ±l{x))dx=^ f^{x)dx± f f^{x)dx± ± J l{x)dx

Bay gia chiing ta di xet cac vi du minh hoa :

V i du 3 Ap dung cac tinh chat va Bang nguyen ham ca ban, ta c6 :

a) = J {4x + 3cosa;)(ia; = J Axdx + J 3cosa;c?x- (dp dung tinh chat

thu ba)

= 4 J" xdx +3 J cos xdx (dp dung tinh chdi thu hai)

= 4 h 3 sin a; + C (dp dung bang nguyen ham ca ban)

• 2

= 2a;^ + 3 sin X + C b) / = r (5e" ^)dx = r Se'dx- f dx (dp dung tinh chdt

Tiiy theo kha nang cua nguai lam todn met ta c6 the lucre bo di nhirng buac

gidi khong can thiet

Ta bien doi ham so dual ddu nguyen ham ve dang ham cd chira cdc ham

trong bdng nguyen ham co bdn de tinh

Vi du 5 Cho ham s6 f{x) = xe' va F{x) = {ax + b)e'' V a i gia tri nao cua a va

b thi la mot nguyen ham ciia f(x) 9

Gidi

Tap xac djnh cua F(x) va / ( x ) la R Ham s6 F(x) la mpt nguyen ham cua f(x) thi /^'(^) = fix) vai Vx e R

Ta CO : F'(x) = ae" + (ax + b)e''(ax + a + b)e" nen F'(x)=-f(x) vai

Vx e R thi (ax + a + b)e' = ' vai Vx e R <=> ax + a + 6 = x, Vx e R

a - l = 0 |'a = l

< ^ ( o - l ) x + a + Z ) - 0 , V x e R <=><^ , Ci>< Vay vai a = l va

a + b-0 b = -l b l thi F(x) la mot nguyen ham cua f(x)

Vi du 6, Chung minh rang F(x) - sin xe"" la mot nguyen ham cua ham so

/ ( x ) = (sin X + cos x y

Gidi Tap xac djnh cua F(x) va / ( x ) la IR

Taco: F'(x) = (sinx)'^''+ sinx(e'')'

= cos xe"" + sin xe' = (cos x + sin x)e'' = / ( x ) Vay F(x) = sin xe' la mot nguyen ham cua ham so / ( x ) = (sin x + cos x)e''

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 ) ?

B a i 2 B A N G N G U Y E N H A M M Q R O N G

Sail day chung ta se ma rong cac cong thuc nguyen ham ca ban de dugc

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 :

1 Dinh li

Nh J f(u)du =F{u) + C vau = u(x) la ham s6 c6 dgo ham lien tuc thi:

J f{u(x)).u\x)dx =F{u{x)) + C

2 Cong thirc nguyen ham mo" rOng

Ap dung dinh li tren trong truong hop u = ax + b{a^0),t^c6 :

jfiax + b)dx= jf{ax + b).{ax + by.-dx

a f(ax + b).{ax + bydx =-F{ax + b) + C

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 :

Cong t/iii'c nguyen ham cff ban Cong thuc nguyen ham m& rpng

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 + l)(2x + Ifdx

= J {2x + l)i8x''+12x^+6x + l)dx

= J (16a;' + 24a;' + 12a;' +2x + 8x'' + 12a;' + 6x + l)dx

= J (16x' + 32a;' + 24a;' + 8a; + l)dx

16x' 32x* 24x' 8x'

+ Sa;" + 8a;' + 4a;^ + a; + C

16x-'

5

Bay gia chung ta ap dung cong thuc nguyen ham ma rong (cong

thuc r{ax + bydx = - ^"•'^ ^ + C(a ^ - 1 ) ) d^ tinh/, ta c6 :

^ o a + 1

«^ ^ ^ 2 4 + 1 10

(C/zw >• rang, each nay va each tren deu cho kit qua dung, no chi sai khdc

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

ham ca ban ! Cac cong thuc nguyen ham ma rong, neu chung ta cho he so a =

1; i = 0 thi ta thu dugc cong thuc nguyen ham ca ban

•dx = —.- - 1

a (n-l){ax + b) — + C{n^l;a^O) se cho

ta lai gidi nhanh han nua (vi gidm duac mot buac biin doi) !

- 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

day chung ta dan cu cdu a) duac gidi bdng phuang phdp doi bien so de so

sdnh hai cdch gidi:

/ = 2a; + 3 dx 12

1 Dat M = 2a: + 3 => (iti = 2dx rfx = - du

Rd rang cdch nay to ra khd phuc tap so vai mot bdi todn dan gidn nhu vdy !

Ap dung cong thuc nguyen ham ma rong ta c6 lai gidi gon vd nhanh han:

\13

12

2a; + 3 dx = 2x + 3

26 + C

Do vdy, viec nha vd van dung tot cong thuc nguyen ham ma rong Id can

thiet de chung ta tinh nhanh duac nguyen hdm vd tich phdn sau nay

d) 7 = fe-''dx = —e-'+C = -e-^+C

- Chung ta c6 the trinh bdy nhanh nhu sau :

; d) ^ 4 = / 2cos(3 )rfa;

Bai 3 PHl/QNG PHAP D O I BIEN SO

Bai nay chiing ta se xet hai truong hop Ichi tinh nguyen ham /{x)dx bang

phuong phap doi bien so :

- 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 :

Buoc 1 : Chon an phu thich hop u = u(x)

Buoc 2 : Xdc dinh viphdn du = du(x) hoac du^ - du^(x)

Buoc 3 : Bieuthi f(x)dx theo u va du Gidsurang f (x)dx = g{u)du Buoc 4 : Tinh I = g(u)du Sau do thay u = w(x) de dime ket qua can tim

Chu V : chon an phu u = w(x) sao cho viec tinh / = g(u)du phai de hon la

tinh / = jf(x)dx !

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? 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 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) 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

thl: J f{u{x)).u\x)du = J i\u{x))du{x) =F{u{x)) + C

Vi du

a) J cos(2x^ + 3a; + l)d{2x^ + 3a; + 1) = sin(2a;^ + 3a,' + 1) + C

(ta hieu trong suy nghi " 2x^ + 3x + 1 " lau)

b) f ^- d(x' + 1) = l n a;' + 1 + C = ln(a-' + 1) + C v/ + 1 > 0 (ta

^ (a;' + 1 )

hieu trong suy nghi " -\-\" la u )

Sau day chiing ta tim hi6u cac moi quan he quan trpng giup chung ta tim

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 x"*\n^-\)

Ta CO : dx'"' = (« + \)x"dx o x"dx = —dx"*' '

n + \ + \)

d{ax"^'+b), trong

do a^O con b tuy y tren R Vay ta c6 quan he giira x" va x"^\n^-1) nhu

(ta hieu cong thicc tren mot each don gidn

sau : x"dx =

-1

a{n + \) d{ax"''+b) nhu sau : dua x" vdo trong vi phan thl thdnh {ax"*^ -^b), voi a ^ Q vd h tuy

ytren

V i d u l T i n h :

a) Jx^{2x^ +lfdx\ = Jx4^+ Idx

Giai

a) Phan tich hai todn: Theo I6i giai thong thuang, cac ban se khai trien

bieu thuc (2x + 1) , sau do nhdn vai X de dua ve nguyen ham de tinh han

The nhung viec khai trien bieu thicc (2x^ + if la khong dan gidn? Do vdy,

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 :

x^dx = -d(2x^ +1), nen ta c6 x'{2x^ + ifdx = -(2a;' + lfd{2x^ + 1)

' 60

b) Phan tich bai todn: Cac ban de y quan he giica x vd x :

xdx = - d(x^ +1) nen ta c6 x^x^ +\dx = - Vx^ +1 d(x^ +1) Do vdy, ta c6 thi

2 2

chon an phu Id w = +1 hodc u = Vx" + 1 Trong truang hap nay ta nen chon

u = de bieu thuc dual ddu nguyen ham khong con can thuc

L&i giai cHa bai todn

(Vx^ +1 f

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 :

a) = J x'(2x' + l)Mx = J -{2x' + l)M(2x' + 1) = + C • (ta hieu trong suy nghi "2x^ + 1" la "u") 60

(ta hieu cong thuc tren mot each don gidn nhu sau : dua \ trong vi

phdn thi thanh — + b, voi a ^ 0 vdbtuyy tren R)

Thay u = 1 + — ta duac: Ii = —e ^ + C X • 3

b) Phdn tick bdi todn : Cdc ban de y quan he giua va —

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 :

vay, ta chon an phu Id u 2lnx + 3

Ldi gidi cda bai todn

Inx

LM gidi cda bai todn

21n'x + 5 l n ' x r21n-'x

In X dx 2 In^ X + 5 In^ x Dat u = Inx => du = d(lnx)

a

( a ^ O )

(ta hiiu cong thuc tren mot each dan gicin nhu sau : dua 6 vao trong vi

phdn thi thanh {ae"" + h), voi a ^Ovdb tiiyytren M)

V i du 4 Tinh :

' -J 2e' +1

b) [—1—dx

Gidi

a)Phan tich bdi todn : Cdc ban de y quan he giua va 2e^ + 1 •

e'^dx = ^ d(2e^ + 1) nen ta c6

(i(2e^ +1) Do vgy ta chon dnphu la u = 2e' +\

Ld'i gidi cua bdi todn

I = r - ^ - H l - d x = r - ^ e M x = i d ( 2 e ' ' + l )

1 J 2e^ + 1 ^ 2e^ + 1 ^ 2e'' + 1 2 '

= r - ^ ^ d ( 2 e ^ + l ) = - f — ^ d ( 2 e ^ + l )

2 20" + 1 2 26" + 1 Dat u = 26" + 1 =^ du = d ( 2 e ' + 1 )

3 r l , 3

Ta CO

' 2 ^ 1 1 2 In u + C

-Thay u = 26" + 1 ta dugc : 1 = - ln(2e"' + 1) + C • (ta khong lay dau gia tri

tuyet d6i vi 2e'' +1 > 0)

b) Phdn tich bdi todn : Ta Men doi

Do do, ta chon dn phu la u = e" + \

LM gidi cua bdi todn

^ + 1 Dat u = e ' ' + l ^ d u = d ( e ' ' + l ) Taco: I = f-du = l n | u | + C

u

; Thay u = e" + 1 ta dugc: = ln(e" + 1) + C

* 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 x = r—-—.-d(2e''+ l ) = - l n ( 2 e ^ + 1 ) + C ' - ' 2 0 ^ + 1 2e^ + 1 2 ^ ^ 2 ^ ^

5 Quan he giua sinx va cosx

Ta CO (sinx) =cosx va (cosx) = - s i n x nen quan he can xet giua sin a;

va cos X la:

cos xdx = — d{a s i n x + b )

a s inxdx = — — d(a cos x + b) a (Ta hieu cong thuc tren mot cdch dan gidn nhu sau: dim cos x vao trong viphdn thanh (asinx+h); dua sinx vao trong vi phdn thanh -(acosx + b), vai

Vi du 5.-Tinh :

a) = Jcos^xsin'xdx; b) Jcosxe-'"""+'da:

Gidi

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 = d(sinx) nen ta c6

cosx(l -sin^x)sin^xdx = (1 -sin^x)sin^xd(siwc) Do vdy, ta chon an phu la u = sinx

LM gidi cua bai todn

= J cos^ x sin^ x d x = J cos x cos^ x siii^ x d x

=J cos x ( l - sin^ x ) sin^ x d x = f{l- sm x ) sin^ xd(sin x )

Dat u = sinx => du = d(sinx)

Taco: = J (1 u ' ) u M u = J ( u ' u ' ) d u = y y + 0

-™ • i T ( s i n x ) ^ ( s i n x ) ^ „

Thay u =^ sinjc ta duac: I =-^^ ^ -^^

—+C-' 3 5

b) Phan tich bai todn : Cdc ban de y quan he giua sinx va cosx;

cos xdx = d{-3 sin x + 2) nen ta c6

cos = — e-'^'^'^^di-?, sin x +

2)-3

Do vdy, ta chon an phu Id u = -3sinx + 2

LM gidi cua bai todn

= Jcosxe-'''"''^'dx = J ^ e - " ' " ' ^ + ' ( i ( - 3 s i n x + 2)

Dat u = - 3 sin x + 2 =^ du = d ( - 3 sin x + 2 )

Taco: I = — f e M u - — e " + C

' 3 ^ 3

Thay u = - 3 sin x + 2 ta dugc: = — e - 3 » " " ' + 2 ^ Q

* Nhan x4t: 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

— J (I- sin^ x) sin^ xd(sin x) = J* (sin^ x - sin^ x)d(sin x)

6 Quan he gifra sin^x, cos^x va sin2x

Ta CO (sin^ x) =2sinxcosx = sin2x 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 :

sin 2xdx = — d{a sin^ x -\-b)

a sin 2xdx = — — d{a cos^ 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)

a) Phan tich bai todn : Cdc ban diyquan he giCta sin^x va sin2x;

sin2xrfa; = - d ( 3 s i n ^ x + 1 ) nen ta c6 (3sin^ x+l)sin2xda;

-Taco : I = - / u d u = - h C =

hC-Thay u = 3 s i n ' x + l tadugc : I ^ ( 3 s i n x + 1 )

' 6

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

Cdc ban di y quan he giua cos^x va sin2x.- sin2xda; = -d{2 + cos^ x) nen

ta CO s i n 2a: _ =d{2 + cos^ x)- do, ta c6 the chon an

V 2 + cos^a; V 2 + c o s ^ x

/fl w = 2 + cos^ X /zoac w = V2 + cos^x r/-o«g truang hap nay ta nen chon

u = V2 + cos" X cfe Z>/ew //zii-c i/j/OT i/aw nguyen ham khong con can thuc

L&i gidi cua bdi todn :

- A'ew thanh thgo trong viec sir dung phuang phdp nay, cdc ban c6 the

trinh bay lai gidi nhanh han nhu sau :

a) = J(Ssin^ x + l ) s i n 2 x c ? x = j^(3sin^ x + l ) c / ( 3 s i n ^ x + 1 )

_ ( 3 s i n ^ x + l ) ^

6

V2sin^ X + 3cos^ x yJ2 + cos^ x

= - '(2 + cos^x) 2 t/(2 + cos^x) = -2V(2 + cos^x) + C

- A'ew chung ta de y den quan he giua s i n a; va c o s x thi chung ta c6

them each gidi theo huang khdc nhu sau :

1 1

7 Quan hf giira — va tanx , cos^x s i n ^ X va cotx

Ta CO ( t a n x ) = — ^ - j - va ( c o t x ) = ] nen quan he can xet giua —^

3tana; + 4 3tana; + 4 (3tana; + 4) 1

ta chon an phu la M = 3 tan x + 4

LM gidi cua bai todn

' 12

b) tick bai todn :

Ta biin doi cotx

s i n x sin'' x ~ cot^ X —\ • Cdc ban di y quan he

= - J cot' a-(i(cot a;)

Dat u = cot X => du = d(cot x )

Taco: = - J u ' d u = - y + C

Thay u = cotx ta dugc: I2 = - + C •

3

* Nhgn xet: Niu da thanh thgo trong viec sir dung phuang phdp nay, cdc

ban CO thi trinh bay lai gidi nhanh han nhu sau :

-Vay la, chung ta da nghien cihi xong 7 m6i quan he ca ban giup chung ta

dinh huang nhanh each giai cho mot bai toan nguyen ham, ciing nhu tich phan

sau nay Trong truoiig hop bai toan khong c6 xuat hien mot trong 7 moi quan

he tren, chung ta lam theo huang giai khac, c6 tinh chat tong quat hon nhu sau:

dat an phu u = u(x) de tit nguyen ham theo bien x chung ta bieu dien duac

nguyen ham do theo bien u! (tiic Id ta can biiu dien bien "x" theo bien u ,

a) Phan tich bdi toan : Niu khai triin (x +12)^'"^ rdi nhdn x vac di tinh thi

khong kha thi rdi ! O day chung ta cUng khong nhin thdy su xuat hien cua mot

trong 7 moi quan he de dinh huang phep dqt an phu, nhung theo huang giai

tong quat, chung ta chon an phu la u = x + 12 thi tie nguyen ham theo biin x

chung ta bieu dien duac nguyen ham do theo bien u rdi! Vi tit u = x + 12 ta

CO X = u — 12 va dx — d{u -12) — du (tuc la x duac biiu diin theo u vd

dx duac bieu dien theo du)

L&i giai cHa bdi toan

b) Plian tich bdi toan : Doi vai nguyen ham nay, vice su dimg moi quan he

giita xvdx khong dem Igi lai giai thoa dang! Nhung neu chon an phu la u = x

1 thi tit nguyen ham theo biin x chung ta bieu dien duac nguyen ham do theo

biin u rdi! Vi tit u = x + \ c6 x = u - \ dx = d(u - 1) = du (tiic Id x duac

•• biiu diin theo u vd dx duac bieu dien theo du)

\ giai cua bdi toan

a) Phan tich bdi toan : Trong bdi nay cUng vay, su dung moi quan he giita x

vd x^ khong dem Igi lai giai thoa dang! Neu chon an phu la u = x - 2 thi tit nguyen ham theo biin x chung ta bieu dien duac nguyen ham do theo bien u!

Vi tit u = X - 2 ta CO X = u + 2 vd dx = d(u + 2) = du (tuc la x duac bieu dien

theo u vd dx duac bieu dien theo du)

Lcfi giai cua bdi toan

Dat u = 2 + x^ du = d(2 + x^) va x ' = u - 2

3 u 3 u u 3 u u

= i ( l n | u | + ^ ) + C

3 u Thay w = 2 + xMac6 : I = - ( I n

phu thich hap la u = tanx + 1

LM gidi cua bai todn

C-b) Phan tick bai todn : Ta de y quan he giita sin x vd cos x de dinh

huangphep dgt an phu : sin xdx = —d(cos x ) Ta cd

sin^ xVcosxdx = sin x sin^ xVcosxdx

= sin x ( l — cos^ x)Vcosxdx

= (1 - cos^ x)Vcos xd(— cos x)

vgy ta chon an phu la u = cosx hoac u = Vcosx Trong truang hap nay

chung ta nen chon u = Vcosx de bieu thuc duai ddu nguyen hdm khong con

chica can thuc

Ldi giai cua bai todn

\ J sin^ xV cos xdx = J sin x sin^ xVcosxdx

= sin x(l - cos^ x) V cos xdx = (1 - cos" x)Vcosxc/(-cosx)

Dat u = VcosX =4> = cosx va d(—cosx) = d(—u^)

Tadugc: = J(1 - u^)ud(-u') = J(1 - u')u(-2u)du

= r ( - 2 u ^ + 2 u « ) d u = Z ^ + 2 u ^ + C

^ 3 7

Thay u - ta c6 : = -2{sf^f ^ 2 ( V ^ ) ^

* Cach khac : Neu dat M = cos x ta c6 each giai khac nhu sau :

\—j'^ii^^ xVcosxdx — J'sinXsin^ xVcosxdx

= j sin x ( l - cos^ x)Vcosxdx = J{1- cos^ x)Vcosxd(- cos x)

Cdc ban nhdn thdy, bdng cdch dat u = Vcosx chung ta c6 lai gidi gon han

vd khong phuc tap nhu cdch dat u = cosx

Vidu 11 T i n h :

a) I, = r d X ' b) = J xVl + xdx

Gidi a) Phan tick bai todn : Bai todn ndy sit dung 7 quan he de dinh huang phep

dat dn phu Id khong khd thi Neu chon an phu Id u = V l — x ihi tit nguyen hdm theo biin x chung ta bieu diin duac nguyen hdm do theo bien u vi tit

u = Vl - X ta suy ra x = 1 - vd dx = d(l - u^) = - 2 u d u (tuc Id X

duac bieu dien theo u; dx duac bieu dien theo u vd du )

L&i gidi cHa bai todn

Dat Li = V l - X =^ = 1 - X hay X = 1 - dx = d(l - u^)

=

-2udu-Ta duac gc : = -Ji-iil2udu = 2J(u' - l)du = 2 u u + c

(Vr^)^ + c

-Thay u = Vl - x ta c6: = 2

b) Tuang tu cdu a), ta chon dn phu la u ^ Vl + x Lai gidi cua bdi todn Id

Dat u = V l + x = 1 - f X hay X = - 1 dx = d(u^ - 1 )

= 2udu

e — e

d x

Gidi

a) P/tan tick bai todn : Neu chon an phu la M = + 1 thi tir nguyen ham

theo bien x chung ta bieu dien duac nguyen ham do theo bien u vi

du — rf(e^ + 1 ) = e'^da: ma =-u-\

LM gidi cua bai todn

qua e"" Lai de y d e " = e^dx hay = d x nen ta chon dn phu la u = e\

L&i gidi cua bai todn

* Cdch khdc : Cdc ban diyrdng 1 1 _ je-f vd

d{e"'') = -e~\ix hay ^—^ = dx thi ta chon duac dn phu la u — e

Ta CO lai gidi cho bai todn:

= r_ J _ ^ i x = f — ^ — d x = f i ^ dx = I - ^ ^ ^ x

d u Dat u = e " d u = de " = - e ''dx hay — = dx •

- u Tadugc: 1 = - f ' - ^ i d u = f —

Sau day, chung ta tiep tuc xet mot vai nguyen ham giai duQc bhng phuang

phdp ddi bien so nhung vai each dgt an phu dac biet!

a) Phan tick bai todn : Ddt u = Vx^ + 2 0 1 3 => = + 2013 va

u d u = x d x =^ dx = udu :, ro rang phep ddt nay c6 the ddi bien tit

V u ' - 2 0 1 3

bien "x " qua bien "u " duoc Nhung nguyen ham theo bien "u " Igi khd phuc

tap! Khong thoa man dieu kien: chon an phu u = u{x) sac cho vice tinh

/ = g{u)du phdi de hon la tinh I - f{x)dx Tuang tu trong truang hap ddt

u = x^ + 2013 Cling vdy Bdy gia ta xet cdch ddt giai duac bai todn

Ldi giai cHa bai todn

2013 u 2u 2

+c

= hi V x ' + 2013 + - h i 2 0 1 3 + C

- Ta chon phep ddt u^x = ^x^ +2013 hodc u - x = Vx^ +2013 di khi binh phuang hai vi ta triet tieu dugc vd khu can!

~ Nguyen ham dang I = f —— '"or nguyen ham ca ban nen can

thdnh thgo cdch tinh no

b) Phan tich bai todn : Ta biin dSi — = (chia ca tu va mau

cho x^) Tai day cdc ban lai de y rang 1

x 2 - x V 2 + l

+ C

Vx + 1 + V x + T (X -1)^ dx

(7/a/

a) Phan tich bai todn : Di y cdc can thuc Vx + 1 , + 1 //zeo tu la

can bac hai, can bdc ba Ma 6 la boi chung nho nhdt ciia 2 va 3 nen ta chon dn phu Id u = yjx + 1 hay x + 1 = u'^ de lam mdt can thicc

LM gidi cua bai todn

Dat u = Vx + l =^ = X + 1 va 6 u M u = d x

^ g j u V - l ) ( u ^ + l )

+ 1 ^ u

u + 1 _ g r u V - l ) ( u + l ) ( u ^ - u + l)^

= 6 j - l)(u^ - u + l)du = 6 J ( u « - u^)(u^ - u + l)du

= 6 j( u « - u ^ + u « - u ^ + u^-u^)du

= 6 u« u«

9 8 ^ 7 6 ^ 5 4 + C Thay u = Vx + 1 ta dugc :

I : = 6 if^i)' {4^if {4^iy {4^if

x + 1

x - 1

Phil Id u = •*

-dx <^ — d (x - 1 ) ^

, / i T i

x - 1 '

x + l

x - 1 (x -1)^ -dx- Do do, ta chon dn

LM gidi cua bai todn

X + 1 Dat u = Wx - 1 x - 1 =^ u " =

a) P/raw //c/i 6fl/ todn : Hai each ddt u = V ? + T va u = x^ + 1 ^ew Ja/i

din thdt bai! Cdc ban de y dx xdx «e/7 neu ddt u = 7? + 1

Vdy ta da dSi bien thanh cong (khir duac cdn thuc)

LM gidi cua bai todn

u ^ - 1 2u

Ti^p theo, chung ta su dung phucmg phap d6i bien so de tinh nguyen ham c6

dac dilm : biiu thuc duai ddu nguyen ham chua bieu thuc u(x) vd dao ham bdc

nhdt cua no u '(x) Chung ta thuong gap truong hop dac biet: tic thuc Id dao

ham cua mdu thuc Khi do ta chon an phu la u = u(x) va tat nhien an phu phai

dam bao nguyen tSc : nguyen ham moi theo bien u phai de tinh hon nguyen

ham ban dau theo bien x

Vi du 16 Tinh :

x(x-l)Mx (x + 1) 6 ' b) 1 = / X + 1

* Cdch khdc ; £)<5/ u = x + 1C Cdc ban tu lam, coi nhu mot bdi tap )

Cdch ndy se gap kho khan niu ta thay so mu 4 vd 6 bdi s6 mu cao han, chdng

x(x - l)Mx

han = J- (x + 1) 11

b) Phan tCch bai todn :

Cdc ban liruy (\ xe""^ =e'' + xe" =6" (l + x) - xe^ 'l + x^

V A /

Dieu nay cd nghia la: bieu thuc duai ddu nguyen ham chua u ( x ) = 1 + xe'' va dao ham

a) Phdn ticit bdi todn : Cdc ban di ;/ x^ + 2x^ - 3 = (x^ + 1 ) ^ - 4 vd

+ = 2x- f^'^'" i^dy CO nghia la: bieu thuc duai ddu nguyen ham chua

u(x) = x^ + 1 va dao ham bdc nhdt cua no u '(x) = 2 x Ta chgn dn phu Id

u = + 1

L&i gidi cua bdi todn

2xdx r 2xdx , ^ r 2xdx r

* Cdclt khdc : (cdc ban xem trongphdn phuangphdp bien doi)

b) Phdn tick bdi todn : Ta bien doi

x ' + 2 x ' + 3 x ' + 2x - 3 = [ ( x ' f + x'^ + 1' + 2 x l x + 2 x l + 2 x ' l ] - 4

= (x^ + X + 1)^ - 4 va (x^ + X + 1 ) ' = 2x + 1 Diiu nay cd nghia Id:

biiu thuc dirai ddu nguyen ham chua u(x) = x^ + x + 1 vd dao ham bdc nhdt

cua no u '(x) = 2x + 1 Ta chon dn phu /a u = x^ + x + 1

Loi gidi ciia bdi todn

x ' + 2 x ' +3x'^ + 2 x - 3

-dx 2x + 1

(x' + X + 1)' - 4 -dx Dat u = x'^ + X + 1 d u = (2x + l ) d x

x^ + X + 3 + c

* Luu y : Nguyen ham ca ban thuang duQc dp dung (cdc ban xem each tinh

trong bdi " phuang phdp hien doi") :

a) Phdn tich bai todn : Cdc ban de y (yjx

2 V I Diiu nay cd nghia la:

biiu thuc duoi ddu nguyen ham chua u(x) = 1 + Vx hodc u(x) = Vx va dgo

ham bdc nhdt cua no u '(x) = —4= • Ta chon dn phu /a u = 1 4- Vx

* Cdch khdc : Ddt u = (cdc ban tu trinh bay)

b) Phdn tich bdi todn : Ta c6 (sin x — cos x ) ' = sin x + cos x vd luu y

(sin x — cos x)^ = 1 — sin 2 x Diiu nay cd nghia la: bieu thuc duoi ddu nguyen ham chua u(x) = sin x — cos x vd dgo ham bdc nhdt cua no

u ' ( x ) = s i n x 4 - c o s x Ta chon dn phu Id u ^ s i n x —cosx hogc

u = Vsin X — cosX Ta nen chon u = Vsinx - cosx de nguyen ham theo

bien u khong con chua ddu can third

Loi gidi cda bdi todn

Dat u - Vsin x - cos x u^ = (sin x - cos x ) va

2udu = (sin x + cos x ) d x

Suy ra: T = r l l z _ ^ 2 u d u = 2 f (1 - u*)du = 2u - ^ + C

^ r 2(Vsinx-cosx)^

= i V s m x — C O S x — + C

5

7 Phep dat x la mot ham so cua u : x = ip{u)

Gia su can tinh / = f / { x ) d x , ta thuc hien nhir sau :

Birac 1: Chon an phu thich hap x — ip{u) (theo ddu hieu cho trong bang duai)

Buac 2 : Xdc dinh viphdn dx = d^f{u) — ^\u)du

Buac 3 : Bieu thi f{x)dx theo u vd du Gid sir rang f(x)dx = g{u)du

Buac 4 : Tinh I = g(u)du

J

* Cliiiy : chon an phu x = ip{u) sac cho viec tinh / = jg(u)du phai de honti

la tinh / = | / { x ) d x ! Cac dSu hieu dan tai viec lua chon i n phu theo kiku tren

dugc cho duai bang sau:

a) Phan tick bai todn : Ta nhln thdy ddu hieu ^J(l-xy nen chon dn phu

Id X = s i n u (— — < w < — ) Ta khong idy u — - — \u = — vi dieu kien

^ 2 2 2 2

x ^ ± 1

LM gidi cua bdi todn

Dat X = s i n u d x = c o s u d u ;

dx = , cos udu = ^ = cos udu

1

V(cos= u)^

1 cosu

Chii i Ta c6 ^(1 - x^)^ = ^(1 - sin^ u)^ ^ ^{cos' uf =

cos u = cos u va t a n u = = , v i : ( < < _ ) nen

cosu J ^ _ ^ 2 2 2

cos u > 0

V l - x Vcos^ u = cos u

•dx = + 1)' u + 1)' ' cos' u d u

a) Phan tick bai todn : De y cdc can thuc theo thic tu Id can bdc

hai, can bdc ba Ma 6 la boi chung nhd nhdt cua 2; 3 nen ta c6 the chqn dn phu de lam mdl can thuc Id x

L&i gidi cua bai todn :

Dat X = u ' =^ dx = 6 u ' d u : 6 : I , ^ r ^ _ 6 u M u ^ 6 r - ^ d u = 6 r i ^ i - t ^ - l

Lai co: tan - = = =

2 " o 2W 1 + cosw cos — 2cos —

2 2

Ma t a n u = x nen sin u = va cosu =

4 1 + x ' (vi ta luon c6 sin^ u + cos^ u = 1)

x + 1

dung quan he giua x^ va x ta chpn hn phu la u - y x ^ + 1 Vay lai giai khac

cho bai toan la :

- Nha-ng vi du minh hoa cho cdc phep dat tiep theo, cdc ban theo doi trong Phcln tich phdn !

^ sin X cos X ^ cos x — sin x cos a;

/ / £ ) ; Su dung quan hf giua —-— va t a n x, — - — va cot x

B a i 4 P H l / a N G P H A P B I E N D O I

Phin nay chung ta se kit hap cdc cdch trinh bay a nhung bai tren vai cdc

ky thugt phdn tich, biin ddi di tinh nhanh bai todn nguyen ham cUng nhu bai

todn tich phdn sau nay

Trong qud trinh lam todn chung ta thuang gap nguyen ham ca bdn duoi

ddy Do do cdc ban can thdnh thao trong viec tinh nguyen ham dang bay

Bai toan cff ban : T i n h l = f—-—dx (a > 0)

2015 2014 ( x ' + l ) - x ' 1 x"

= J tan^ xd(tan x) — (tan^ x + 1) — 1 dx

— J taii^ xd(tan ^) " J (tan^ x + l)dx + J'dx

P (1 — sin^ x) + sin^ x

sin^ x(l — sin^ x) d(sin x) = ^

1 1 + •

sin X 1 — sin x d(sin x)

sin X I s i n(sinx — l)(sinx + 1) Y — I 11 s n

* Nhan xet: ta su dung phep nhan lien hop de dua mdu thirc ve dang gon

hon, nhdm tdch nguyen ham da cho thdnh cdc nguyen ham de tinh hon

* Nhan xet: ta phdn tich vd su dung quan he giiia cos x va sin x de dinh

huong each gidi cho hdi nay

jnh huong each gidi cho hdi nay

d) Ta CO : = j cos^ xd x = j cos^ x cos xd x =J (1- s i n ' x ) ' cos xdx

= J {I - sin' x)'d(sin x) = J {1-2 sin' x + sin'* x)d(sin x)

= - cot' x.(l + cot' x ) ' d ( c o t x )

= - J cot' x ( l + 2 cot' x + cot^ x)d(cot x)

= - J (cot' X + 2 cot" x + cot*" x)d(cot x)

cot^ X 2 cot^ X cot'^ X „

3 5 7 b) Ta CO : = J ( t a n ' x + tan" x)dx = J t a n ' x ( l + t a n ' x)dx

= f t a n ' X — - — dx = r t a n ' xd(tan x) = ^ + C •