Clss 07-8 MATHEMATICS FOR JEE MAIN & ADVANCED SECOND EDITION Ehustiv Thory (Now Rvisd) Formul Sht 9000+ Problms bsd on ltst JEE pttrn 500 + 000 (Nw) Problms of prvious 35 yrs of AIEEE (JEE Min) nd IIT-JEE (JEE Adv) 5000+Illustrtions nd Solvd Empls Dtild Solutions of ll problms vilbl Topi Covrd Indfinit Intgrtion Mstrj Conpts Tips & Triks, Fts, Nots, Misonptions, Ky Tk Awys, Problm Solving Ttis MstrjEssntil Qustions rommndd for rvision
. INDEFINITE INTEGRATION. INTRODUCTION Intgrtion is rvrs pross of diffrntition. Th intgrl or primitiv of funtion f() with rspt to is diffrntil funtion φ() suh tht th drivtiv of φ() with rspt to is th givn funtion f(). It is prssd symbolilly s f()d φ() d Thus. f()d φ () () f() d φ. Th pross of finding th intgrl of funtion is lld Intgrtion nd th givn funtion is lld Intgrnd. Now, it is obvious tht th oprtion of intgrtion is th invrs oprtion of diffrntition. Hn th intgrl of funtion is lso nmd s th nti-drivtiv of tht funtion. Furthr w obsrv tht ( ) ( ) d d d d + k d d d + + onstnt So w lwys dd onstnt to th intgrl of funtion, whih is lld th onstnt of Intgrtion. It is gnrlly dnotd by. Du to th prsn of this rbitrry onstnt suh n intgrl is lld n Indfinit Intgrl.. ELEMENTERY INTEGRATION Th following intgrls r dirtly obtind from th drivtivs of stndrd funtions. () 0.d (b).d + () k.d k + (k R) (d) () (f) n+ n d + (n ) n+ d log + d +
. Indfinit Intgrtion (g) d + log + log (h) sin d os + (i) os d sin + 3. BASIC THEOREMS OF INTEGRATION If f(), g() r two funtions of vribl nd k is onstnt, thn () k f()d k f()d (b) f() ± g() d f()d ± g()d d () ( f()d) (d) d f() d f() d f() d + MASTERJEE CONCEPTS Th rsults of intgrtion r vry diffrnt from diffrntition. Thr is no stndrd formul for intgrtion. Alwys mk sur to writ th onstnt of intgrtion. NEVER ssum it s zro from your sid. Vibhv Gupt (JEE 009, AIR54) Illustrtion : Evlut: sin d os thrfor w n split Sol: As w know, f() ± g() d f()d ± g()d sin d os s sin d os sin d d nd thn by solving w n gt rsult. os os sin d d s d tns d tn s + os os Illustrtion : Evlut: + sin d Sol: Hr rsult. sin + os nd sin sin os, thrfor by using ths formul nd solving w will gt th + sin d sin + os + sin os d (sin + os ) d (sin+ os)d sin d + os d os + sin +
Mthmtis. 3 4 Illustrtion 3: Evlut: sin d Sol: Hr s w know, os sin, Now by putting this in th bov intgrtion nd solving w will gt th trm ( os os + )d 4, Aftr tht by using th formul + os 4 os w n solv th problm givn bov. sin 4 d os d ( os + os )d 4 + os 4 os + d (3 4 os + os 4)d 4 8 3 sin + + C 8 4 sin4 3 3 Illustrtion 4: If f'() 4 suh tht f()0,thn, find f() (JEE ADVANCED) 4 3 3 Sol: Hr f'() 4 3 3 thrfor f() 4 d 4 4 hn by splitting this intgrtion nd solving w will gt th rsult. W hv, d 3 3 f() 4 d 4 3 3 3 3 3 4 f() 4 d 4 d d 4 d 3 d 4 4 3+ 4+ 4 4 3 + C + + C 3 + 4 + 3 (i) 4 Givn f() + + C 0 3 Putting th vlu of C in (i), w gt 9 0 6 + + C C 8 8 4. METHODS OF INTEGRATION 4 9 f() + 3 8 Whn th intgrnd n t b rdud into som stndrd form thn intgrtion is prformd using following mthods 4. Intgrtion by Substitution 4.. Intgrnd is Funtion of Anothr Funtion ' If th intgrl is of th form f φ() φ ()d, thn w put φ () t so tht φ ' () ddt. Now intgrl is rdud f(t) dt. MASTERJEE CONCEPTS In this mthod th funtion is brokn into two ftors so tht on ftor n b prssd in trms of th funtion whos diffrntil offiint is th sond ftor. In s of objtiv qustions in whih dirt indfinit intgrtion is skd, funtion bing vry omplitd to intgrt, thn try diffrntiting th options.
.4 Indfinit Intgrtion If d I, thn I is Equl to sin( )os( b) () sin( ) log sin( b) os( b) + C (b) sin( ) log os( b) os( b) + C () sin( ) log sin( + b) os( b) + C (d) sin( ) log os( + b) os( b) + C Vibhv KrishnnJEE 009, AIR Illustrtion 5: Evlut: tn s d Sol:This problm is bsd on intgrtion using substitution mthod. In this w n put t nd thrfor ddt nd thn solving w will gt th rsult. Lt t ddt d dt tn s d tntstdt st s + + 4.. Intgrnd is th Produt of Funtion nd its Drivtiv If th intgrl is of th form I f ' () f() d w put f() t nd onvrt it into stndrd intgrl. Illustrtion 6: Evlut: tns d Sol: Hr s is drivtivs of tn hn w n put tn t nd givn problm. Lt tn t s.d dt t I tns d tdt tn + + 4..3 Intgrnd is Funtion of th Form f(+b) Hr w put +b t nd onvrt it into stndrd intgrl. Now if, f()d φ(), thn f( + b)d φ ( + b) s.d dt thrftr w n solv th Illustrtion 7: Evlut: os3os5d Sol: By multiplying nd dividing by in th givn intgrtion nd using th formul osa.osb os( A + B) + os( A B) w n solv it. I os3os5d os8 + os d sin8 + sin + 8
Mthmtis.5 Illustrtion 8: Evlut: I d (JEE ADVANCED) 4 + + Sol: Hr by putting 3 putting t + tnθ, w n solv it. Lt t dt d w will gt th trm dt dt t dt d I t t + + t + (/) + 3/ f + d t + tn + 3 3 f tn + f' + tn + 3 3 dt dt t + t+ t + I tn + 3 3 nd thn by ( t + (/)) + ( 3/) Now put 3 t + tnθ 3 dt s θdθ ( 3 / )s θdθ t + d tn tn + θ θ+ + + (3 / 4) tn θ+ 3 3 3 3 3 3 Stndrd intgrtion rsults () (b) () ' f () d log f() + f() n+ n ' f() f() f ()d + (providd n -) n+ ' f () d f() + f() s Illustrtion 9: Evlut: d tn Sol: Hr simply substituting ttn dt s d w n solv it. Lt ttn dt s d I dt t tn t + + 4..4 Intgrl of th Form d b b thn substitut r os θ nd br sin θ, tnθ θ tn,w gt sin + bos I d +θ os( +θ )d logtn + r sin( +θ) r r ( + ) logtn ( / ) ( / )tn (b / ) + b +
. 6 Indfinit Intgrtion MASTERJEE CONCEPTS m n sin os d, whr m, n ϵ N If m is odd put os t If n is odd put sin t If both m nd n r odd, put sin t if m n nd os t othrwis. If both m nd n r vn, us powr rduing formul os + os sin or os If m+n is ngtiv vn intgr, put tn t Shriknt Ngori (JEE 009, AIR 30) Illustrtion 0: Evlut: d (JEE ADVANCED) sin + os d Sol: As w know, if intgrtion is in th form of sin + bos thn w n put r os θ nd br sin θ hn th intgrtion will b log tn + θ +. r Hr & b π So d logtn + tn + logtn + + sin + os + 8 4..5 Stndrd Substitutions Th following stndrd substitutions will b usful Intgrnd form or Substitutions sin θ or os θ + or + tn θ or ot θ or sinh θ or s θ or os θ + or + or ( + ) or ( + ) tn θ or or ( ) or ( ) sin θ or os θ
Mthmtis. 7 or or ( ) or ( ) s θ or os θ + or + os θ α β or ( α)( β ) ( β > α) αos θ + βsin θ Som Stndrd Intgrls () tnd logs + logos + (b) ot d logsin + logos + () s d log(s + tn) + (d) osd log(os + ot ) + () s tnd s + (f) os ot d os + (g) (h) s d tn + os d ot + logd log + log + + (i) π log(s tn) + logtn + + 4 log(os ot) + log tn + MASTERJEE CONCEPTS p q If th intgrl is of th form R,, r... d, whr R is rtionl funtion thn, Lt lm of (p,q,r,.) nd put t Nitish Jhwr (JEE 009, AIR 7) Illustrtion : Prov tht: d logn( + ) + C (JEE ADVANCED) Sol: By putting s θ d sθtnθdθ, w n solv th problm givn bov. Lt s θ d sθtnθdθ d + log ( + ) + C log ( + + ) + C' s θ tn θ d θ s θ d θ tnθ log (s θ + tn θ ) + C
.8 Indfinit Intgrtion Illustrtion : Evlut: os.os 4.os6d Sol: By multiplying nd dividing by in th givn intgrtion nd thn by using osa.osb os( A + B) + os( A B) w n solv it. Lt I os.os 4.os6d (os.os 4)os6d (os6 + os)os6d os Aos B os ( A + B) + os( A B) ( os 6 + os 6 os) d ( os 6 + os 6 os) d 4 4 + + + ( os ) ( os8 os 4) d + + + sin f os f d + C f' sin sin8 sin4 I + + + + C 4 8 4 d osd os8d os 4d 4 4 4 4 Illustrtion 3: Evlut: os osα d (JEE ADVANCED) os osα Sol: Hr in this problm by using th formul C+ D C D osc osd sin.sin, sin A sin A os A nd os C os D os (C+D)+os (C-D) W n solv th problm bov stp by stp. W hv, os os α sinsin d +α α d os osα sin ( +α) / sin ( α) / ( +α ) ( +α ) ( α ) ( α ) sin (( +α) / ) sin (( α) / ) sin( +α)sin( α) d sin / sin / ( +α ) ( α ) sin / os /.sin / os / d (sin A sin A os A) +α α 4 os os d [ os C os D os (C+D)+os (C-D)] (os + os α )d osd + osα d sin + osα+ C Illustrtion 4: Evlut: 8 8 sin os d (JEE ADVANCED) sin.os Sol: Hr by using th formul b ( b)( b) + nd putting (sin + os ) in pl of in th dnomintor, w n rdu th bov intgrtion nd thn using os os sin w n solv it. W hv, 8 8 sin os d sin.os 4 4 4 4 (sin + os )(sin os ) d (sin + os ) sin.os
Mthmtis. 9 4 4 (sin + os )(sin + os )(sin os ) d 4 4 (sin + os ).(sin os )d osd os os sin sin + C 4. Intgrtion by Prts If u nd v r two funtions of, thn du (u.v)d u v d v d d d This is lso known s uv rul of intgrtion. This mthod of intgrting is lld intgrtion by prts. MASTERJEE CONCEPTS From th first lttr of th words invrs irulr, logrithmi, Algbri, Trigonomtri, Eponntil funtions, w gt word ILATE. Thrfor th prfrn of slting th u funtion will b ording to th ordr ILATE. In som problms w hv to giv prfrn to logrithmi funtion ovr invrs trigonomtri funtions. Hn somtims th word LIATE is usd for rfrn. For th intgrtion of Logrithmi or Invrs trigonomtri funtions lon, tk unity () s th v funtion. Shivm Agrwl (JEE 009, AIR 7) Illustrtion 5: Evlut: ( + )logd Sol: Hr w n intgrt th givn problm by using Intgrtion by prts i.. du (u.v)d u v d v d d d Hr u log nd v ( + ). Lt I ( + )logd Intgrting by prts, tking log s st funtion, (by LIATE rul) w gt d d I log ( + ) d (log). ( + )d d log +. + d + log + d + log + + C 4 Illustrtion 6: Evlut: 3 s d (JEE ADVANCED) Sol: Hr w n solv by intgrting by prts, tking s s th first funtion. I 3 s d s. s d Lt I s tn ( s tn ).tn d u s & v s s tn s tn d s tn s (s )d
.0 Indfinit Intgrtion 3 I s tn I s d I s tn s d + s d I s.tn + log ( s + tn ) + C Illustrtion 7: Evlut : + + I s tn log(s tn) C + + (sin ) d (JEE ADVANCED) nd thn tking Sol: W n writ th givn intgrtion s (sin ).d u sin & v solving by intgrtion by prts. d I (sin ). (sin ). d d ( ) (sin ). sin.. d Now, putting sin t sin t so tht d dt I (sin ) t. sintdt (sin ) tost + ostdt (gin Intgrting by prts) { } { } (sin ) tost + sint + C (sin ) + tost sint + C (sin ) + sin. + C Illustrtion 8: Evlut: sin os d (JEE ADVANCED) sin + os π Sol: By using th formul sin + os, w n solv th bov problm. Lt I sin os d sin + os sin ( π/ ) sin d ( π / ) π sin + os π 4 sin d sin d d π π 4 sin d π (i) Putting sin θ sin θ so tht d sin θ. os θ d θ sin θ d θ. sin d θ. sinθdθlt u & V sin θ θ, thn intging by prts w gt θ sin θ + sin θ.os θ 4 osθ θ θ. + osθdθ osθ + sinθ θ( sin θ+ ) sinθ sin θ 4 (sin )( ) +. + C (ii) From (i) nd (ii), w gt 4 I ( )sin + + C π { } π ( )sin + C 4.. Intgrtion by Cnlltion Illustrtion 9: Evlut : 3 tn s d
Mthmtis. Sol: Lt 3 tn s d 3 tn d s d du i.. (u.v)d u v d v d d d nd thn by using th intgrtion by prts formul w n solv th problm bov. 3 tn s d 4.. Intgrtion of th Form: 3 tn d s d If th intgrl is of th form f() + f ' () d, thn us th formul; f() + f'() d f() + 3 3 tn s d s d 3 tn + Illustrtion 0: Evlut: (log + / )d Sol: Solution of this problm is bsd on th mthod mntiond bov, hr f() log nd f () /. log + d I log + d log + Hr, f() log ' &f () / If th intgrl is of th form ' f () f() + d thn us th formul; ' f () + f() d f() + Illustrtion : Evlut : (s + tn)d Sol: Similr to th problm bov. Hr I (s + tn)d f'() + f() d 4..3 Spil Intgrls sinbd (sinb bosb) + + b osbd (bsinb + osb) + + b osbd + b Illustrtion : Evlut : [whr f() tn ].tn + (b sin b+ os b)+ sin d Sol: By putting sin t sin t d os t dt nd thn intgrting by prts w n solv th givn problm. I Lt sin I sin d t sin t d os t dt t sin t ostdt (sint + ost) + ( + ) +
. Indfinit Intgrtion 4.3 Intgrtion of Rtionl Funtions 4.3. Whn th Dnomintor n b Ftorizd (Using Prtil Frtion) Lt th intgrnd b of th form f(), whr both f() nd g() r polynomils. If dgr of f() is grtr thn g() dgr of g(), thn first divid f() by g() till th dgr of th rmindr boms lss thn th dgr of g(). Lt Q() b th quotint nd R(), th rmindr thn f() R() Q() + g() g() Now in R()/g(), ftoriz g() nd thn writ prtil frtions in th following mnnr: () For vry non-rptd linr ftor in th dnomintor. Writ A B + ( )( b) b (b) For vry rptd linr ftor in th dnomintor. Writ A B C D + + + 3 3 ( ) ( b) ( ) ( ) ( b) () For vry non-rptd qudrti ftor in th dnomintor. Writ A + B C + ( + b + )( d) + b + d (d) For vry rptd qudrti ftor in th dnomintor. Writ A + B C + D E + + ( + b + ) ( d) ( + b + ) + b + d MASTERJEE CONCEPTS Considr f() s th funtion w nd to ftoriz. For non- rptd linr ftor in th dnomintor. A B Lt f() + ( )( b) ( ) ( b) To obtin th vlu of A rmov ( ) from f() nd find f(). Similrly, to obtin vlu of B, rmov (-b) from f() nd find f(b).. For rptd linr ftor in th dnomintor. Lt f() A B C D + + + 3 3 ( ) ( b) ( ) ( ) ( b) To obtin vlu of D rmov ( b) from f() nd find f(b). To obtin vlu of rmov ( ) 3 from f() nd find f(). Now tht w hv rdud th numbr of unknowns from 4 to, w n find A nd B sily by quting.
Mthmtis.3 Now lt s try this mthod for Prtil frtion will b of th form 4 3 4 3 + + + 4 3 ( + )( + ) + + + 4 A B+ C D+ E F+ G H+ I + + + + ( + )( + ) ( + ) ( + ) ( + ) ( + ) 3 3 Now rmov nd put 0, w gt A Now rmov 3 ( + ) nd put W gt Hi+I 3i. Hn H 3 nd I. - i.. i (you n lso substitut i ). Now rmov ( + ) nd put i. W gt B ( i ) +C i +3. Hn B nd C 3 Now th numbr of unknowns hv rdud from 9 to 4 nd th rmining unknowns n b solvd sily. This mthod vry usful instd of solving for ll th unknowns t th sm tim. Also rmmbr tht substituting n imginry numbr for is not disussd nywhr in NCERT. So, us this mthod only for omptitiv ms. Rvi Vood (JEE 009, AIR 7) Illustrtion 3: Evlut : d Sol: Hr th givn intgrtion is in th form of, hn by using prtil frtions w n split it s ( )( b) A B + nd thn by solving w will gt th rquird rsult. ( ) ( b) Hr I d d ( )( + ) + 3 + log log log 3 + + + + + 3 Illustrtion 4: Evlut : d 4 3 8 + (JEE ADVANCED) Sol: Hr simply by putting t dt d nd thn by using prtil frtions w n solv th givn problm. I d d 4 3 8 + dt 3t 8t + (Put t dt d) dt dt dt 6 t 6t ( / 3) 6 (t 3) (6 / 3) 6 + (t 3) 4 / 3 (t 3) (4 / 3) 3 3t 3 3 4 log + C log + C 6 (4 / 3) (t 3) + (4 / 3) 48 3t 3 3 + 4 3 3 3 3 4 log + C 48 3 3 3 + 4
. 4 Indfinit Intgrtion 4.3. Whn th Dnomintor nnot b Ftorizd In this s th intgrl my b in th form (i) d (ii) (p q) + b + Mthod: (i) Hr tking th offiint of + d + b + ommon from th dnomintor, writ b 4 + (b / )+ / (+ b/ ) 4 Now th intgrnd obtind n b vlutd sily by using stndrd formul. (ii) Hr suppos tht p + q d A ( + b + ) + B d A(+b)+B..(i) Now ompring offiint of nd onstnt trms. W gt Ap/, Bq-(pb/) p + b pb d I d q + b + + b + Now w n intgrt it sily. 4.3.3 Intgrnd Contining Only Evn Powrs of To find intgrl of suh funtions, first w divid numrtor nd dnomintor by, thn prss th numrtor s d(±/) nd th dnomintor s funtion of (±/). Th following mpls illustrt it. MASTERJEE CONCEPTS R(sin,os )d whr R is rtionl funtion (univrsl substitution tn(/)t) Spil ss: () If R ( sin, os ) R(sin, os ) Put os t (b) If R (sin, os ) R(sin, os ) Put sin t () If R ( sin, os ) R(sin, os ) Put tn t Aksht Khry (JEE 009,AIR 35) Illustrtion 5: Evlut: + d 4 (JEE ADVANCED) + Sol: Hr dividing th numrtor nd dnomintor by thn by putting -/t [+/ ]d dt, w n solv it., w gt + ( / ) + ( / ) d d nd + ( / ) { ( / ) } +
Mthmtis.5 I + ( / ) ( / ) d d ( / ) + + { ( /) } + Now tking -/t [+/ ]d dt, w gt I dt t tn + t + tn + 4.4 Intgrtion of Irrtionl Funtions If ny on trm in numrtor or dnomintor is irrtionl thn it is md rtionl by suitbl substitution. Also if th intgrl is of th form d or + b + + b + Thn w intgrt it by prssing Also for intgrls of th form d + b + ( +α ) +β p + q d or + b + (p + q) b + d First w prss p+q in th form d p+q A [ ( b + ) ]+B nd thn prod s usul with stndrd form. d Illustrtion 6: Evlut : Sol: Simply by putting Put t, thn d 5 4 d dt d 5 4 t, thn dt 5 4t t d dt, w n solv th givn problm. dt 5 (t + 4t) dt t+ sin + C sin + + C (3) (t + ) 3 3 Illustrtion 7: Evlut : dt 5 (t + 4t + 4) + 4 dt 9 t+ d ( )( b) (JEE ADVANCED) + b Sol: Hr first pnd ( )( b) nd thn dding nd subtrting by, w n rdu th bov + b intgrtion. Aftr tht by putting u, w n solv th givn problm. Lt, I d d d ( )( b) ( + b)+ b ( + b) + ( + b) / ( + b) / + b d d ( (( b) / )) + (( + b + b) / 4) b ( ( + b) / ) ( b ) / 4 d (i) ( + b) / ( b) / ( )
.6 Indfinit Intgrtion + b On putting u so tht d du in (i), w gt du I d log + b log u + u + u ( b) / Putting u + b, w gt + b + b b I log + + + b + + log ( )( b) 5. STANDARD INTEGRALS () (b) d tn + + d log + + + () d log + (d) () (f) d sin + os + d sinh log + + + + + d osh log + + + (g) d + sin + (Substitut osθ or sinθ nd prod) (h) + d + + log + + + (Substitut tnθ or otθ nd prod) (i) d logn + + (Substitut sθ or osθ nd prod) (j) d s + (Vlid for > > 0) (k) sinbd (sinb bosb) + + b + b b sin b tn +
Mthmtis.7 (l) osbd (osb + bsinb) + + b + b b os b tn + Intgrtion of irrtionl lgbri funtions: Typ : () d (Put : ( α) ( α)( β ) os q + bsin q ) (b) d (Put : ( α) ( β) s q b tn q ) Typ : d (Put: p + q t ) ( + b) p+ q Typ 3: d (Put: ( + b) p + q+ r + b ) t Typ 4: Typ 5: d ( b + ) p + q (Put:. p + q t ) d ( + b + ) p + q + r Cs I: Whn I Put ( b + ) brks up into two linr ftors,.g. d d thn + + Put t Cs II: If + t + b + is prft squr sy Cs III: If b, q 0 A B + d A + + + (l m) +, thn put l + m t.g. d thn, put or trigonomtri substitutions r lso hlpful. ( + b) p + r t d Intgrl of th form, whr P, Q r linr or qudrti funtions of. P Q d d + B ( ) + + ( + ) + + Intgrl d + b + d d + b + p + q d p + q + b + d + b + d Substitutions + d z p + q z p + q z z
.8 Indfinit Intgrtion d + b + b + + b / t m Illustrtion 8: Evlut : d ( + ) Sol: Simply by putting d ( + ) t, d t dt w n solv th givn problm by using th pproprit formul. Put t d t dt I t dt (t + 3)t dt t + ( 3) t tn 3 3 + tn + 3 3 ( t ( ) ) Illustrtion 9: Evlut : d ( 4) Sol: Hr first put t thrfor d t dt nd thn using prtil frtions w rdu th givn intgrtion in stndrd form. Aftr tht by solving w will gt th rsult. d Lt I ( 4) t dt Put t d t dt thn I dt 4 (t 4)t (t + )(t ) Put t z (t )(t ) A B + + (z+ )(z ) z+ z A nd B 4 4 + (t + )(t ) 4(t + ) 4(t ) I dt dt (t + )(t ) t + t + t t tn + log + 4 t+ tn + log + 4 + ( t ) 6. SPECIAL TRIGONOMETRIC FUNCTIONS Hr w shll study th mthods for vlution of th following typs of intgrls. Typ (i) d (ii) + bsin d + bos (iii) d (iv) os + bsinos + sin d (sin + bos) Mthod: Divid th numrtor nd dnomintor by os in ll suh typs of intgrls nd thn put tn t
Mthmtis.9 Illustrtion 30: Evlut : d + 3sin Sol: Hr dividing th numrtor nd dnomintor by os w n solv it. I s d s + 3tn s d tn (tn) + + 4 tn Typ (i) d (ii) + bos d + bsin (iii) d (iv) os + bsin d sin + bos + Mthod: In suh typs of intgrls w us th following substitutions tn / t sin + tn / + t tn / t dt, os ; d + tn / + t + t nd intgrt nothr mthod for th vlution of th intgrl. d Illustrtion 3: Evlut: d 5 + 4os tn ( / ) Sol: Hr by putting os tn ( / ) d I 5 + 4 ( tn ( / )) / ( + tn ( / )) + nd thn by tking tn (/) t w n solv th givn problm s ( / ) d 9 + tn ( / ) dt 3 + t whr tn (/) t t tn + C 3 3 tn / tn + C 3 3 Typ 3 (i) psin + qos d (ii) sin + bos psin d sin + bos (iii) qos d sin + bos For thir intgrtion, w first prss numrtor s follows- Numrtor A (dnomintor) + B (drivtiv of dnomintor) Thn intgrl A + B log (dnomintor) + C Illustrtion 3: Evlut : 6 + 3sin + 4os d (JEE ADVANCED) 3 + 4sin+ 5os Sol: By using prtil frtions, w n rdu th givn intgrtion to th stndrd form. 6 + 3sin + 4os d 3 + 4sin+ 5os 6 + 3sin + 4os A ( 3 + 4sin+ 5os ) + B ( 4os 5sin ) + Solving R.H.S. & ompring both sids, w gt 4A 5B 3 5A + 4B 4 Also, 3A+C6 A(3 + 4sin + 5os) + B(4os 5sin) + 3 + 4sin+ 5os
.0 Indfinit Intgrtion C d A + logn(3 + 4sin + 5os) + 3 + 4sin+ 5os this is of typ sinφ osφ Illustrtion 33: Evlut : dφ (JEE ADVANCED) 6 os φ 4sinφ Sol: Hr w n writ th givn intgrtion s th drivtiv of 6 os φ 4sinφ hn by putting (sinφ 4os φ ) + 7osφ dφ 6 os φ 4sinφ nd s w know (sinφ 4os ) 6 os φ 4sinφ t, w n solv th givn problm. φ is I sinφ osφ dφ 6 os φ 4sinφ d (6 os φ 4sin φ ) dφ osφsinφ 4osφ sinφ 4osφ sinφ osφ (sinφ 4os φ ) + 7osφ (sinφ 4os φ ) + 7osφ I dφ 6 os φ 4sinφ (sinφ 4os φ)dφ 7osφdφ + + 6 os φ 4sinφ 6 os φ 4sinφ t 6 ( sin ) 4sin dt 7osφdφ φ φ logt + C + 7osφdφ 5 + sin φ 4sinφ 7d log(6 os φ 4sin φ ) + C + ( sinφ ) 4 + 5 7d log(6 os φ 4sin φ ) + C + log(6 os φ 4sin φ ) + C + 7tn + C ( ) + log(6 os φ 4sin φ ) + 7tn (sinφ ) + C 7. SPECIAL EXPONENTIAL FUNCTIONS () (b) d [put b + t] d [Multiply nd divid by + nd t] () (d) d [Multiply nd divid by d [Multiply nd divid by ] nd t] () (f) (g) (h) + d ' f () form f() + d [Multiply nd divid by d [Multiply nd divid by ( + )( ) / ] nd put d [Multiply nd divid by / ] t]
Mthmtis. (i) (j) d [Multiply nd divid by / + ] d [Multiply nd divid by / ] (k) d [Multiply nd divid by / ] (l) d [Intgrnd ( )/ ] (m) + d [Intgrnd (+ )/ + ] (n) d [Intgrnd ( )/ ] (o) + d [Intgrnd ( + ) / ] Illustrtion 34: Evlut : d Sol: Hr by multiplying nd dividing by vlut th givn intgrtion. Hr I Lt d t, thn d d t dt in th givn intgrtion nd thn by putting t w n d d I dt dt t + t tn (t) + tn + Illustrtion 35: Evlut : d 5 4 Sol: W hv, 5 4 d Put t, thn d dt d 5 4 dt 5 4t t d dt 5 (t + 4t) dt 5 (t + 4t + 4) + 4 dt 9 (t + ) dt (3) (t + ) t+ sin + C 3 sin + 3 + C
. Indfinit Intgrtion PROBLEM-SOLVING TACTICS Intgrtion by Prts () Intgrtion by prts is usful for dling with intgrls of th produts of th following funtions << (log) k u tn, sin,os Priority for hoosing u nd dv: ILATE sin, os >> dv (b) Intgrtion by prts is somtims usful for finding intgrls of funtions involving invrs funtions suh s n nd sin. () Somtims whn dling with intgrls, th intgrnd involvs invrs funtions (lik sin ), it is usful to substitut th invrs of tht invrs funtion (lik sin u), thn do intgrtion by prts. 3 (d) Somtims you will hv to do intgrtion by prts mor thn on (for mpl, d nd sind. Somtims you nd to do it twi by prts, thn mnipult th qution (for mpl, sind ). () Try u substitution first bfor intgrtion by prts. Trigonomtri Intgrl () Intgrl Typ : m n sin os d Cs : On of m or n r vn, nd th othr odd Us u substitution by stting u sin or os tht with n vn powr. Us th idntity Cs : Both m nd n r odd Us u substitution by stting u sin or os tht with highr powr. Us th idntity Cs 3: Both m nd n r vn (hrd s) sin + os. sin + os. Do not us u substitution. Us th hlf doubl ngl formul to rdu th intgrnd into s o r: sinos sin ; sin ( os) ; os ( os) 3 3 o (Not: 0 is lso n vn numbr. For mpl, sin sin os, so it is in s ) Just rmmbr tht whn both r vn, you n t us u-substitution, but you n us th hlf doubl ngl formul. Whn it is not tht s, lt u sin or os, nd on will work (t th nd thr is no squr root trm ftr substitution). (b) Intgrl typ : () m n tn s d Cs : s is odd powr, tn is vn powr. Hrd to do, w omit (most likly won t pop out in th m). Cs : Els St u s or tn, nd us + tn s. On will work t th nd (thr is no squr root trm ftr substitution). Intgrl typ : sin(a)os(b)d, os(a)os(b)d, sin(a)sin(b)d Us th produt to sum formul: osθos φ (os θ φ ) + (os θ+φ )) ; sinθos φ (os θ φ) (os θ+φ)) sinθos φ (sin θ φ ) + (sin θ+φ )) Rdu produt into sum nd thn intgrt.
Mthmtis.3 Trigonomtri Substitution () Trigonomtri substitution is usful for qudrti form with squr root: : Lt sinθ + : Lt tnθ : Lt sθ (b) Gnrl produr for doing trig sub: Stp : Drw th right tringl, nd did wht trigonomtri funtion to substitut for. Stp : Find d, thn substitut th intgrnd using tringl, onvrt intgrl into trigonomtri intgrl. Stp 3: Solv th trigonomtri intgrl. Stp 4: Substitut bk using tringl. (i) If th qudrti form is not in th Pythgors form (for mpl, th squr mthod to trnsform it into Pythgors form). (ii) Try u substitution bfor trigonomtri substitution. + +, thn us th prfting (iii) Intgrls involving ( ) nd ( ) without squr roots n b solvd sily with prtil frtions. So don t us trigonomtri substitution. Rtionl Intgrl nd Prtil Frtion () Gnrl stp for solving rtionl intgrl: Stp : Do long division for th rtionl funtion if th dgr of th numrtor is highr thn th dnomintor. Stp : Do prtil frtion domposition. Stp 3: Evlut th intgrl of h simpl frtion. (b) Gnrl stp for prtil frtion: Stp : Ftoriz th dnomintor. Stp : St th prtil frtion ording to rul. Stp 3: Solv th unknown of th numrtor of th prtil frtion. Impropr Intgrl () Gnrl stps for vluting impropr intgrl: Stp : Chng th impropr intgrl into th pproprit limit. [Chng ± or singulr point (whr) to pproprit limit.] Stp : Evlut th intgrl. Stp 3: Find th limit. (b) Th vry first stp to tst impropr intgrl involving is to hk its limit. If its limit is not zro, thn th intgrl divrgs. () Whnvr you s impropr intgrls involving th quotint of rtionl or irrtionl funtion, suh s 3 + 3 3 3 (8 + 7) d Us limit omprison tst. Th pproprit ompring funtion n b found by looking t th Intgrnd (quotint of rtionl irrtionl). Disrd th lowr dgr trms.
.4 Indfinit Intgrtion (d) Somtims, using u substitution bfor using ny tst will b sir. () Somtims, to dtrmin if n impropr intgrl onvrgs or divrgs, dirtly vluting th impropr intgrl is sir. (f) Whn doing omprison tst, bwr of th ompring funtion tht you hoos. It might not giv n pproprit onlusion if th ompring funtion is not orrt. (g) Try th limit omprison tst bfor th omprison tst. (h) Usful ompring funtion, whih is good to know thir onvrgn or divrgn k β d < For k 0, β> 0 d p d p < if p > ; if p ; < if p < ; if p ; FORMULAE SHEET Bsi thorms of Intgrtion:. f() ± g() d f()d ± g()d. k f()d k f()d d 3. ( f()d) f() d 4. d f() d f() d Elmntry Intgrtion:. 0.d..d + 3. k.d k + (k R) 4. n+ n d + (n ) n+ 5. d log + 6. d + 7. d + log + log 8. sind os + 9. osd sin + 0. n+ n ( + b) ( + b) d + (n + ). d log + b +. + b f ' () f() d f() + 3. logd log + 4. log d log + log
Mthmtis.5 Stndrd substitution:. or sin θ or os θ. + or + tn θ or ot θ 3. or s θ or os θ 4. +, +, ( + ) nd ( + ) tn θ 5. or ( ) nd ( ) sin θ or os 6. or or ( ) or ( ) s θ or os θ 7. + nd + os θ 8. α β or ( α)( β ) ( β>α) αos θ+βsin θ Som stndrd Intgrls:. tnd logs + logos +. ot d logsin + 3. s d log(s + tn) + π log(s tn) + logtn + + 4 4. osd log(os + ot ) + log(os ot) + log tn + 5. s tnd s + 6. os ot d os + 7. s d tn + 8. os d ot + logd log + log + + 0. 9. sin sin d + C ( sinos) + C
.6 Indfinit Intgrtion. sin os d + + C ( + sinos) + C. 3 s d s tn + log s + tn + 3. n sin n d n n sin os sin d + n n 4. n os n d n n os sin os d + n n Intgrtion by Prts: du.. (u.v)d u v d v d d d f() + f'() d f() + Stndrd Intgrls:.. 3. d tn + + d log + + + d log + 4. 5. 6. 7. d sin + os + d sinh log + + + + + d osh log + + + d + sin + 8. 9. + d + + log n + + + d log + +
Mthmtis.7 0. d s + (Vlid for > > 0). sinb d (sinb bosb) + + b + b b sin b tn +. osb d (osb + bsinb) + + b + b b os b tn + Solvd Empls JEE Min/Bords Empl : Evlut : + sin d + os Sol: Hr by using th formul sin sin os nd + os os w n solv th givn problm. + sin d + os + sin/ os/ d os / s / + tn d tn / + Empl : Evlut : + d Sol: By pplying intgrtion by prts nd tking + s th first funtion w n solv th givn problm. + d ( + ) + d + d + + + Put tn θ + d + d s θ d θ + log ( sθ+ tnθ ) logn + + I + + log + + + Empl 3: Evlut : sin tn d + sin Sol: Hr first writ os (( π/ ) ) t th pl of sin thn by using th formul os sin And + os os w n solv it. sin os ( π/ ) I tn d tn + sin + os ( π/ ) tn ( π ) ( π ) sin ( / 4) ( / ) d os ( / 4) ( / ) π tn tn d 4 Empl 4: Evlut : π d 4 log( + )d d π + C 4 4 Sol: Hr intgrting by prts by tking log( + ) s th first funtion w n solv th givn problm.