INTEGRALS 87 hpter 7 INTEGRALS Just s mountineer clims mountin ecuse it is there, so goo mthemtics stuent stuies new mteril ecuse it is there. JAMES B. BRISTOL 7. Introuction Differentil lculus is centre on the concept of the erivtive. The originl motivtion for the erivtive ws the prolem of efining tngent lines to the grphs of functions n clculting the slope of such lines. Integrl lculus is motivte y the prolem of efining n clculting the re of the region oune y the grph of the functions. If function f is ifferentile in n intervl I, i.e., its erivtive f eists t ech point of I, then nturl question rises tht given f t ech point of I, cn we etermine the function? The functions tht coul possily hve given function s erivtive re clle nti erivtives (or primitive) of the function. Further, the formul tht gives G.W. Leinitz (66-76) ll these nti erivtives is clle the inefinite integrl of the function n such process of fining nti erivtives is clle integrtion. Such type of prolems rise in mny prcticl situtions. For instnce, if we know the instntneous velocity of n oject t ny instnt, then there rises nturl question, i.e., cn we etermine the position of the oject t ny instnt? There re severl such prcticl n theoreticl situtions where the process of integrtion is involve. The evelopment of integrl clculus rises out of the efforts of solving the prolems of the following types: () the prolem of fining function whenever its erivtive is given, () the prolem of fining the re oune y the grph of function uner certin conitions. These two prolems le to the two forms of the integrls, e.g., inefinite n efinite integrls, which together constitute the Integrl lculus.
88 MATHEMATIS There is connection, known s the Funmentl Theorem of lculus, etween inefinite integrl n efinite integrl which mkes the efinite integrl s prcticl tool for science n engineering. The efinite integrl is lso use to solve mny interesting prolems from vrious isciplines like economics, finnce n proility. In this hpter, we shll confine ourselves to the stuy of inefinite n efinite integrls n their elementry properties incluing some techniques of integrtion. 7. Integrtion s n Inverse Process of Differentition Integrtion is the inverse process of ifferentition. Inste of ifferentiting function, we re given the erivtive of function n ske to fin its primitive, i.e., the originl function. Such process is clle integrtion or nti ifferentition. Let us consier the following emples: We know tht (sin ) cos... () ( )... () n ( e ) e... () We oserve tht in (), the function cos is the erive function of sin. We sy tht sin is n nti erivtive (or n integrl) of cos. Similrly, in () n (), n e re the nti erivtives (or integrls) of n e, respectively. Agin, we note tht for ny rel numer, trete s constnt function, its erivtive is zero n hence, we cn write (), () n () s follows : (sin + ) cos, ( + ) n ( e + ) e Thus, nti erivtives (or integrls) of the ove cite functions re not unique. Actully, there eist infinitely mny nti erivtives of ech of these functions which cn e otine y choosing ritrrily from the set of rel numers. For this reson is customrily referre to s ritrry constnt. In fct, is the prmeter y vrying which one gets ifferent nti erivtives (or integrls) of the given function. More generlly, if there is function F such tht F() f (), I (intervl), then for ny ritrry rel numer, (lso clle constnt of integrtion) [ F( )+] f (), I
INTEGRALS 89 Thus, {F +, R} enotes fmily of nti erivtives of f. Remrk Functions with sme erivtives iffer y constnt. To show this, let g n h e two functions hving the sme erivtives on n intervl I. onsier the function f g h efine y f () g() h(), I f Then f g h giving f () g () h () I or f (), I y hypothesis, i.e., the rte of chnge of f with respect to is zero on I n hence f is constnt. In view of the ove remrk, it is justifie to infer tht the fmily {F +, R} provies ll possile nti erivtives of f. We introuce new symol, nmely, f () which will represent the entire clss of nti erivtives re s the inefinite integrl of f with respect to. Symoliclly, we write f ( ) F( ) +. y Nottion Given tht f (), we write y f (). For the ske of convenience, we mention elow the following symols/terms/phrses with their menings s given in the Tle (7.). Symols/Terms/Phrses f () f () in f () in f () Integrte An integrl of f Integrtion onstnt of Integrtion Tle 7. Mening Integrl of f with respect to Integrn Vrile of integrtion Fin the integrl A function F such tht F () f () The process of fining the integrl Any rel numer, consiere s constnt function
9 MATHEMATIS We lrey know the formule for the erivtives of mny importnt functions. From these formule, we cn write own immeitely the corresponing formule (referre to s stnr formule) for the integrls of these functions, s liste elow which will e use to fin integrls of other functions. Derivtives (i) n+ Integrls (Anti erivtives) n+ n n ; n+ +, n n+ Prticulrly, we note tht () ; + sin cos ; cos sin + cos sin ; sin cos + tn sec ; sec tn + cot cosec ; cosec cot + sec sec tn ; sec tn sec + cosec cosec cot ; cosec cot cosec + sin ; sin + (ii) ( ) (iii) ( ) (iv) ( ) (v) ( ) (vi) ( ) (vii) ( ) (viii) ( ) (i) ( cos ) () ( tn ) (i) ( cot ) ; + + ; ; cos + tn + + cot + +
INTEGRALS 9 (ii) ( sec ) (iii) ( cosec ) ; ; sec + cosec + (iv) ( e ) e ; e e + (v) ( log ) ; log + (vi) ; log + log Note In prctice, we normlly o not mention the intervl over which the vrious functions re efine. However, in ny specific prolem one hs to keep it in min. 7.. Geometricl interprettion of inefinite integrl Let f (). Then f () +. For ifferent vlues of, we get ifferent integrls. But these integrls re very similr geometriclly. Thus, y +, where is ritrry constnt, represents fmily of integrls. By ssigning ifferent vlues to, we get ifferent memers of the fmily. These together constitute the inefinite integrl. In this cse, ech integrl represents prol with its is long y-is. lerly, for, we otin y, prol with its verte on the origin. The curve y + for is otine y shifting the prol y one unit long y-is in positive irection. For, y is otine y shifting the prol y one unit long y-is in the negtive irection. Thus, for ech positive vlue of, ech prol of the fmily hs its verte on the positive sie of the y-is n for negtive vlues of, ech hs its verte long the negtive sie of the y-is. Some of these hve een shown in the Fig 7.. Let us consier the intersection of ll these prols y line. In the Fig 7., we hve tken >. The sme is true when <. If the line intersects the prols y, y +, y +, y, y t P, P, P, P, P etc., respectively, then y t these points equls. This inictes tht the tngents to the curves t these points re prllel. Thus, + F () (sy), implies tht
9 MATHEMATIS Fig 7. the tngents to ll the curves y F (), R, t the points of intersection of the curves y the line, ( R), re prllel. Further, the following eqution (sttement) f ( ) F ( ) + y (sy), represents fmily of curves. The ifferent vlues of will correspon to ifferent memers of this fmily n these memers cn e otine y shifting ny one of the curves prllel to itself. This is the geometricl interprettion of inefinite integrl. 7.. Some properties of inefinite integrl In this su section, we shll erive some properties of inefinite integrls. (I) The process of ifferentition n integrtion re inverses of ech other in the sense of the following results : f () f () n f () f () +, where is ny ritrry constnt.
INTEGRALS 9 Proof Let F e ny nti erivtive of f, i.e., F( ) f () Then f () F() + Therefore f () ( F( )+) Similrly, we note tht F( ) f( ) f () f () (II) n hence f () f () + where is ritrry constnt clle constnt of integrtion. Two inefinite integrls with the sme erivtive le to the sme fmily of curves n so they re equivlent. Proof Let f n g e two functions such tht f () () g or f () g (), where is ny rel numer (Why?) Hence f () g () or f () g () + So the fmilies of curves { f ( ) +, R} n { g ( ) +, R} re ienticl. re equivlent. Hence, in this sense, f() n g()
9 MATHEMATIS Note The equivlence of the fmilies { f ( ) + }, { g( ) +, } f R n R is customrily epresse y writing () g() without mentioning the prmeter. (III) [ ] f ()+ g() f () + g() Proof By Property (I), we hve, [ f ()+ g()] f () + g ()... () On the otherhn, we fin tht f () + () f () + g() g f () + g ()... () Thus, in view of Property (II), it follows y () n () tht ( () + ()) f g f () + g(). (IV) For ny rel numer k, k f () k f () (V) Proof By the Property (I), k f() k f(). Also k f () k f() k f() Therefore, using the Property (II), we hve k f () k f (). Properties (III) n (IV) cn e generlise to finite numer of functions f, f,..., f n n the rel numers, k, k,..., k n giving [ kf() + kf() ++... knfn() ] k f() + k f () ++... kn fn(). To fin n nti erivtive of given function, we serch intuitively for function whose erivtive is the given function. The serch for the requisite function for fining n nti erivtive is known s integrtion y the metho of inspection. We illustrte it through some emples.
INTEGRALS 95 Emple Write n nti erivtive for ech of the following functions using the metho of inspection: (i) cos (ii) + (iii) Solution (i), We look for function whose erivtive is cos. Recll tht sin cos or cos (sin ) sin (ii) (iii) Therefore, n nti erivtive of cos is sin. We look for function whose erivtive is +. Note tht ( ) + +. Therefore, n nti erivtive of + is +. We know tht (log ), > n [log ( )] ( ),< omining ove, we get ( ) log, Therefore, log is one of the nti erivtives of. Emple Fin the following integrls: (i) (ii) ( + ) (iii) ( + e ) Solution (i) We hve (y Property V)
96 MATHEMATIS + + + + + + ;, re constnts of integrtion + + + + +, where is nother constnt of integrtion. Note From now onwrs, we shll write only one constnt of integrtion in the finl nswer. (ii) We hve ( + ) + (iii) We hve + 5 ++ + 5 ++ ( + e ) + e + + e log + + + e log + 5 Emple Fin the following integrls: (i) (sin + cos ) (ii) cosec (cosec + cot ) (iii) sin cos Solution (i) We hve (sin + cos ) sin + cos cos + sin + 5
INTEGRALS 97 (ii) (iii) We hve + (cosec (cosec + cot ) cosec cosec cot We hve sin sin cos cos cos cot cosec + sec tn sec tn sec + Emple Fin the nti erivtive F of f efine y f () 6, where F () Solution One nti erivtive of f () is 6 since ( 6 ) 6 Therefore, the nti erivtive F is given y F() 6 +, where is constnt. Given tht F(), which gives, 6 + or Hence, the require nti erivtive is the unique function F efine y Remrks (i) (ii) F() 6 +. We see tht if F is n nti erivtive of f, then so is F +, where is ny constnt. Thus, if we know one nti erivtive F of function f, we cn write own n infinite numer of nti erivtives of f y ing ny constnt to F epresse y F() +, R. In pplictions, it is often necessry to stisfy n itionl conition which then etermines specific vlue of giving unique nti erivtive of the given function. Sometimes, F is not epressile in terms of elementry functions viz., polynomil, logrithmic, eponentil, trigonometric functions n their inverses etc. We re therefore locke for fining f (). For emple, it is not possile to fin e y inspection since we cn not fin function whose erivtive is e
98 MATHEMATIS (iii) When the vrile of integrtion is enote y vrile other thn, the integrl formule re moifie ccoringly. For instnce + y 5 y y + y + + 5 7.. omprison etween ifferentition n integrtion. Both re opertions on functions.. Both stisfy the property of linerity, i.e., k f () + k f () k f () + k f () (i) [ ] (ii) [ + ] + k f () k f () k f () k f () Here k n k re constnts.. We hve lrey seen tht ll functions re not ifferentile. Similrly, ll functions re not integrle. We will lern more out nonifferentile functions n nonintegrle functions in higher clsses.. The erivtive of function, when it eists, is unique function. The integrl of function is not so. However, they re unique upto n itive constnt, i.e., ny two integrls of function iffer y constnt. 5. When polynomil function P is ifferentite, the result is polynomil whose egree is less thn the egree of P. When polynomil function P is integrte, the result is polynomil whose egree is more thn tht of P. 6. We cn spek of the erivtive t point. We never spek of the integrl t point, we spek of the integrl of function over n intervl on which the integrl is efine s will e seen in Section 7.7. 7. The erivtive of function hs geometricl mening, nmely, the slope of the tngent to the corresponing curve t point. Similrly, the inefinite integrl of function represents geometriclly, fmily of curves plce prllel to ech other hving prllel tngents t the points of intersection of the curves of the fmily with the lines orthogonl (perpeniculr) to the is representing the vrile of integrtion. 8. The erivtive is use for fining some physicl quntities like the velocity of moving prticle, when the istnce trverse t ny time t is known. Similrly, the integrl is use in clculting the istnce trverse when the velocity t time t is known. 9. Differentition is process involving limits. So is integrtion, s will e seen in Section 7.7.
INTEGRALS 99. The process of ifferentition n integrtion re inverses of ech other s iscusse in Section 7.. (i). EXERISE 7. Fin n nti erivtive (or integrl) of the following functions y the metho of inspection.. sin. cos. e. ( + ) 5. sin e Fin the following integrls in Eercises 6 to : 6. ( e + ) 7. ( ) 8. ( + + c) 9.. 5. 7. ( + e ).. + 5 + + +.. ( ) ( + + ) 6. ( cos + e ) ( sin + 5 ) 8. sec (sec + tn ) sec. sin 9. cosec. cos hoose the correct nswer in Eercises n.. The nti erivtive of + equls (A) + + (B) + + () + + (D) + +. If f ( ) such tht f (). Then f () is 9 (A) + 8 (B) 9 + + 8 9 () + + 8 (D) 9 + 8
MATHEMATIS 7. Methos of Integrtion In previous section, we iscusse integrls of those functions which were reily otinle from erivtives of some functions. It ws se on inspection, i.e., on the serch of function F whose erivtive is f which le us to the integrl of f. However, this metho, which epens on inspection, is not very suitle for mny functions. Hence, we nee to evelop itionl techniques or methos for fining the integrls y reucing them into stnr forms. Prominent mong them re methos se on:. Integrtion y Sustitution. Integrtion using Prtil Frctions. Integrtion y Prts 7.. Integrtion y sustitution In this section, we consier the metho of integrtion y sustitution. The given integrl f () cn e trnsforme into nother form y chnging the inepenent vrile to t y sustituting g (t). onsier I f () Put g(t) so tht g (t). t We write g (t) t Thus I f ( ) f ( g( t)) g ( t) t This chnge of vrile formul is one of the importnt tools ville to us in the nme of integrtion y sustitution. It is often importnt to guess wht will e the useful sustitution. Usully, we mke sustitution for function whose erivtive lso occurs in the integrn s illustrte in the following emples. Emple 5 Integrte the following functions w.r.t. : (i) sin m (ii) sin ( + ) (iii) tn sec (iv) sin (tn ) + Solution (i) We know tht erivtive of m is m. Thus, we mke the sustitution m t so tht m t. Therefore, sin m sin t t m m cos t + cos m + m
(ii) INTEGRALS Derivtive of + is. Thus, we use the sustitution + t so tht t. Therefore, sin ( + ) sin t t cos t + cos ( + ) + (iii) Derivtive of is. Thus, we use the sustitution t so tht t giving t t. tn sec ttn t sec t t Thus, tn t sec t t t Agin, we mke nother sustitution tn t u so tht sec t t u 5 u Therefore, tn t sec t t u u + 5 tn 5 5 t+ (since u tn t) (iv) Hence, tn sec tn 5 + (since t ) 5 tn 5 + 5 Alterntively, mke the sustitution tn t Derivtive of tn. Thus, we use the sustitution + tn t so tht t. + sin t t sin (tn ) Therefore, + cos t + cos(tn ) + Now, we iscuss some importnt integrls involving trigonometric functions n their stnr integrls using sustitution technique. These will e use lter without reference. (i) tn log sec + We hve sin tn cos
MATHEMATIS (ii) Put cos t so tht sin t Then tn t log t + log cos + t or tn log sec + cot log sin + cos We hve cot sin Put sin t so tht cos t t Then cot log t + log sin + t (iii) sec log sec + tn + We hve sec (sec + tn ) sec sec + tn Put sec + tn t so tht sec (tn + sec ) t Therefore, sec t log t + log sec + tn + t (iv) cosec log cosec cot + We hve cosec (cosec + cot ) cosec (cosec + cot ) Put cosec + cot t so tht cosec (cosec + cot ) t So cosec t log t log cosec + cot + t Emple 6 Fin the following integrls: (i) sin cos cosec cot log + cosec cot log cosec cot + (ii) sin (iii) sin ( + ) + tn
INTEGRALS Solution (i) We hve sin cos sin cos (sin ) ( cos ) cos (sin ) Put t cos so tht t sin Therefore, sin cos (sin ) ( t ) t t (ii) Put + t. Then t. Therefore sin sin ( t ) t sin ( + ) sin t 5 t t ( t t ) t + 5 5 cos + cos + 5 sin tcos cos tsin t sin t cos t sin cot t t (cos ) t (sin ) log sin t+ (cos ) ( + ) (sin ) log sin ( + ) + + + cos cos (sin ) log sin ( ) sin (iii) sin Hence, sin ( + ) cos sin log sin ( + ) +, where, sin + cos, is nother ritrry constnt. cos + tn cos + sin (cos + sin + cos sin ) cos + sin
MATHEMATIS cos sin + cos + sin cos sin + +... () cos + sin Now, consier I cos sin cos + sin Put cos + sin t so tht (cos sin ) t Therefore I t log t + log cos + sin + t Putting it in (), we get + + log cos sin + tn + + + log cos + sin + + + log cos + sin +, + EXERISE 7. Integrte the functions in Eercises to 7: ( log )... +. sin sin (cos ) 5. sin( + )cos( + ) + log 6. + 7. + 8. + 9. (+ ) ++.. 5 ( ). 5. 9 (+ ).. e + 7. e 6. +, > (log ) m, >, m
INTEGRALS 5 8. e tn + 9. e e +. e e e + e. tn ( ). sec (7 ). sin. cos sin 6cos + sin 7. sin cos 8. 5. cos ( tn ) cos + sin cos 6. 9. cot log sin. sin + cos. sin ( + cos ). + cot. tn ( ). tn sin cos ( ) 5. ( + log ) 6. ( + ) + log sin tn 7. + 8 hoose the correct nswer in Eercises 8 n 9. 9 + loge 8. equls + (A) + (B) + + () ( ) + (D) log ( + ) + 9. equls sin cos (A) tn + cot + (B) tn cot + () tn cot + (D) tn cot + 7.. Integrtion using trigonometric ientities When the integrn involves some trigonometric functions, we use some known ientities to fin the integrl s illustrte through the following emple. Emple 7 Fin (i) cos (ii) sin cos (iii) sin
6 MATHEMATIS Solution (i) Recll the ientity cos cos, which gives cos + cos Therefore, cos ( + cos ) cos + + sin + (ii) Recll the ientity sin cos y [sin ( + y) + sin ( y)] (Why?) Then sin cos sin 5 sin (iii) cos 5 cos + + 5 cos 5+ cos + From the ientity sin sin sin, we fin tht Therefore, Alterntively, sin sin sin sin sin sin cos + cos + sin sin sin Put cos t so tht sin t Therefore, sin ( t ) t ( cos ) sin t t+ t t t+ + cos + cos + Remrk It cn e shown using trigonometric ientities tht oth nswers re equivlent.
INTEGRALS 7 EXERISE 7. Fin the integrls of the functions in Eercises to :. sin ( + 5). sin cos. cos cos cos 6. sin ( + ) 5. sin cos 6. sin sin sin 7. sin sin 8 cos cos 8. 9. + cos + cos. sin. cos.. cos cos α cos cosα. 6. tn 7. 9. sin cos. cos sin + sin sin + cos sin cos cos ( cos + sin ). cos ( ) cos ( ) hoose the correct nswer in Eercises n.. sin cos is equl to sin cos (A) tn + cot + () tn + cot + e ( + ). equls cos ( e) (A) cot (e ) + () tn (e ) + sin + cos 5. tn sec 8. cos + sin cos. sin (cos ) (B) tn + cosec + (D) tn + sec + (B) tn (e ) + (D) cot (e ) + 7. Integrls of Some Prticulr Functions In this section, we mention elow some importnt formule of integrls n pply them for integrting mny other relte stnr integrls: () log + +
8 MATHEMATIS + () log + () + tn + () log + + (5) sin + (6) + log + + + We now prove the ove results: () We hve ( )( + ) ( + ) ( ) ( )( + ) + Therefore, + () In view of () ove, we hve log ( ) log ( + ) + [ ] log + + ( + ) + ( ) ( + )( ) + +
INTEGRALS 9 Therefore, + + [ log log ] + + + + log + Note The technique use in () will e epline in Section 7.5. () Put tn θ. Then sec θ θ. sec θθ Therefore, + tn θ+ tn θ + + θ () Let secθ. Then secθ tnθ θ. Therefore, secθ tnθθ secθ secθθ log sec θ + tn θ + log + + log + log + log (5) Let sinθ. Then cosθ θ. + +, where log Therefore, cosθθ sin θ θθ +sin + (6) Let tnθ. Then sec θ θ. Therefore, + sec θθ tn θ+ sec θθ log (secθ+ tn θ) +
MATHEMATIS log + + + log + + log + log + + +, where log Applying these stnr formule, we now otin some more formule which re useful from pplictions point of view n cn e pplie irectly to evlute other integrls. (7) To fin the integrl, we write + + c + + c c c + + + + Now, put + tso tht t n writing c k ±. We fin the t c integrl reuce to the form epening upon the sign of t ± k n hence cn e evlute. (8) To fin the integrl of the type, proceeing s in (7), we + + c otin the integrl using the stnr formule. p+ q (9) To fin the integrl of the type, where p, q,,, c re + + c constnts, we re to fin rel numers A, B such tht p + q A ( + + c)+ba( + )+B To etermine A n B, we equte from oth sies the coefficients of n the constnt terms. A n B re thus otine n hence the integrl is reuce to one of the known forms.
INTEGRALS ( p+ q) () For the evlution of the integrl of the type + + c s in (9) n trnsform the integrl into known stnr forms. Let us illustrte the ove methos y some emples. Emple 8 Fin the following integrls: (i) (ii) 6 Solution, we procee (i) We hve 6 log 8 + + [y 7. ()] (ii) ( ) Put t. Then t. Therefore, t t sin ( t ) + [y 7. (5)] sin ( ) + Emple 9 Fin the following integrls : (i) 6 + (ii) + (iii) 5 Solution (i) We hve 6 + 6 + + ( ) + So, Let Therefore, 6 + + ( ) 6 + t. Then t t tn t + [y 7. ()] t + tn +
MATHEMATIS (ii) The given integrl is of the form 7. (7). We write the enomintor of the integrn, + + 7 + 66 (completing the squre) Thus + 7 + 66 Put + t. Then t. 6 Therefore, + t 7 t 6 7 t log 6 + 7 7 t+ 66 [y 7. (i)] 7 + log 66+ 7 7 + + 66 6 log 7 6 + + log + + log 7 + 5 7 log +, where + log 7 + 5 7
INTEGRALS (iii) We hve 5 5 5 Put t. Then t. 5 Therefore, 5 (completing the squre) 5 5 5 t 5 5 t t log t+ + 5 5 log + + 5 5 5 [y 7. ()] Emple Fin the following integrls: (i) + + (ii) + 6 + 5 5 Solution (i) Using the formul 7. (9), we epress + ( ) A + 6+ 5+ B A (+ 6) + B Equting the coefficients of n the constnt terms from oth sies, we get A n 6A + B or A n B. Therefore, + + 6 + 6 + 5 + + 6 + 5 + 6 + 5 I+ I (sy)... ()
MATHEMATIS In I, put + 6 + 5 t, so tht ( + 6) t t Therefore, I log t + t log +++ 65... () n I + 6 + 5 + + 5 + + Put + t, so tht t, we get t I t + tn t+ [y 7. ()] tn + + Using () n () in (), we get ( ) tn + +... () + log + 65 ++ tn ( ++ ) + 6 + 5 (ii) where, + This integrl is of the form given in 7. (). Let us epress + A (5 ) + B A ( ) + B Equting the coefficients of n the constnt terms from oth sies, we get A n A + B, i.e., A n B
INTEGRALS 5 Therefore, + 5 ( ) + 5 5 In I, put 5 t, so tht ( ) t. Therefore, I Now consier I Put + t, so tht t. Therefore, I Sustituting () n () in (), we otin I + I... () ( ) t 5 t t + 5 +... () 5 9 ( + ) t sin + [y 7. (5)] t t + + 5 sin + +... () 5 + sin +, where. EXERISE 7. Integrte the functions in Eercises to. 6 +. +. ( ) +. 9 5 5. + 6. 6 7. 8. 66 + 9. sec tn +
6 MATHEMATIS. + +. 9 + 6 + 5. 7 6. ( )( ). 8+ 5. ( )( ) + 6. + + 7. 5 8. + + 9. 67 + ( 5)( ). +. + + + + 5+... 5 + + hoose the correct nswer in Eercises n 5.. equls + + (A) tn ( + ) + (B) tn ( + ) + () ( + ) tn + (D) tn + 5. equls 9 (A) 9 8 sin + 9 8 (B) 8 9 sin + 9 () 9 8 sin + 8 7.5 Integrtion y Prtil Frctions (D) 9 8 sin + 9 Recll tht rtionl function is efine s the rtio of two polynomils in the form P( ), where P () n Q() re polynomils in n Q(). If the egree of P() Q( ) is less thn the egree of Q(), then the rtionl function is clle proper, otherwise, it is clle improper. The improper rtionl functions cn e reuce to the proper rtionl
INTEGRALS 7 functions y long ivision process. Thus, if P( ) Q( ) is improper, then P( ) P( ) T( ) +, Q( ) Q( ) where T() is polynomil in n P( ) is proper rtionl function. As we know Q( ) how to integrte polynomils, the integrtion of ny rtionl function is reuce to the integrtion of proper rtionl function. The rtionl functions which we shll consier here for integrtion purposes will e those whose enomintors cn e fctorise into P( ) liner n qurtic fctors. Assume tht we wnt to evlute, where P( ) Q( ) Q( ) is proper rtionl function. It is lwys possile to write the integrn s sum of simpler rtionl functions y metho clle prtil frction ecomposition. After this, the integrtion cn e crrie out esily using the lrey known methos. The following Tle 7. inictes the types of simpler prtil frctions tht re to e ssocite with vrious kin of rtionl functions. Tle 7. S.No. Form of the rtionl function Form of the prtil frction. p+ q, ( )( ) A B + p+ q. ( ) A + B ( ). p + q + r A + B + ( )( )( c) c. p + q+ r ( ) ( ) A + B + ( ) 5. p + q+ r A B + + ( )( + + c) + + c, where + + c cnnot e fctorise further In the ove tle, A, B n re rel numers to e etermine suitly.
8 MATHEMATIS Emple Fin ( + ) ( + ) Solution The integrn is proper rtionl function. Therefore, y using the form of prtil frction [Tle 7. (i)], we write A + B ( + ) ( + ) + + where, rel numers A n B re to e etermine suitly. This gives A ( + ) + B ( + ). Equting the coefficients of n the constnt term, we get A + B n A + B Solving these equtions, we get A n B. Thus, the integrn is given y ( + ) ( + ) + + +... () Therefore, ( + ) ( + ) + + log + log ++ log + + + Remrk The eqution () ove is n ientity, i.e. sttement true for ll (permissile) vlues of. Some uthors use the symol to inicte tht the sttement is n ientity n use the symol to inicte tht the sttement is n eqution, i.e., to inicte tht the sttement is true only for certin vlues of. Emple Fin + 5+ 6 + Solution Here the integrn 5+ 6 + y 5 + 6 n fin tht is not proper rtionl function, so we ivie
INTEGRALS 9 + 5+ 6 5 5 5 5 + + 5 + 6 ( ) ( ) 5 5 A B Let + ( ) ( ) So tht 5 5 A ( ) + B ( ) Equting the coefficients of n constnt terms on oth sies, we get A + B 5 n A + B 5. Solving these equtions, we get A 5 n B Thus, Therefore, + 5+ 6 + 5+ 6 Emple Fin ( + ) ( + ) 5 + 5 + 5 log + log +. Solution The integrn is of the type s given in Tle 7. (). We write ( + ) ( + ) A + B + + ( + ) + So tht A ( + ) ( + ) + B ( + ) + ( + ) A ( + + ) + B ( + ) + ( + + ) ompring coefficient of, n constnt term on oth sies, we get A +, A + B + n A + B +. Solving these equtions, we get 5 A, B n. Thus the integrn is given y ( + ) ( + ) 5 ( + ) ( + ) ( + ) 5 Therefore, ( + ) ( + ) + ( + ) + 5 log + + log + + ( +) + 5 log + + + ( + )
MATHEMATIS Emple Fin ( + ) ( + ) Solution onsier Then ( + ) ( + ) ( + ) ( + ) n put y. y ( y+ ) ( y+ ) Write y A + B ( y+ ) ( y+ ) y+ y+ So tht y A (y + ) + B (y + ) ompring coefficients of y n constnt terms on oth sies, we get A + B n A + B, which give A n B Thus, ( + ) ( + ) + ( + ) ( + ) Therefore, + ( + ) ( + ) + + tn + tn + tn + tn + In the ove emple, the sustitution ws me only for the prtil frction prt n not for the integrtion prt. Now, we consier n emple, where the integrtion involves comintion of the sustitution metho n the prtil frction metho. ( sinφ ) cosφ Emple 5 Fin φ 5 cos φ sinφ Solution Let y sinφ Then y cosφ φ
INTEGRALS Therefore, ( φ ) sin cosφ φ 5 cos φ sinφ ( y ) y 5 ( y ) y y y y y+ y ( ) y I (sy) Now, we write Therefore, y ( y ) A B + y ( y ) y A (y ) + B [y Tle 7. ()] ompring the coefficients of y n constnt term, we get A n B A, which gives A n B. Therefore, the require integrl is given y y y I [ + ] y + y ( y ) y ( y ) log y + + y log sinφ + + sinφ log ( sin φ+ ) + sin φ (since, sinφ is lwys positive) Emple 6 Fin ++ ( + ) ( + ) Solution The integrn is proper rtionl function. Decompose the rtionl function into prtil frction [Tle.(5)]. Write ++ ( + ) ( + ) A B + + + ( + ) Therefore, + + A ( + ) + (B + ) ( + )
MATHEMATIS Equting the coefficients of, n of constnt term of oth sies, we get A + B, B + n A +. Solving these equtions, we get A, B n 5 5 5 Thus, the integrn is given y ++ ( + ) ( + ) + + 5 5 5 ( + ) + + + 5 ( + ) 5 + Therefore, ++ + + ( +) ( + ) 5 + 5 + 5 + log ++ log ++ tn + 5 5 5 EXERISE 7.5 Integrte the rtionl functions in Eercises to.. ( + ) ( + ). 9. ( )( )( ). ( )( )( ) 5. + + 6. ( ) 7. ( + ) ( ) 8. ( ) ( + ) 9. + 5 +. ( ) (+ ). ( ) ( + ) 6. 5. ( + ) ( ). ( + ). ++ 5. n [Hint: multiply numertor n enomintor y ( + ) n n put n t ] 7. cos ( sin ) ( sin ) [Hint : Put sin t]
INTEGRALS 8. ( + ) ( + ) ( + ) ( + ) 9. ( + ) ( + ). ( ). [Hint : Put e ( e ) t] hoose the correct nswer in ech of the Eercises n... equls ( ) ( ) (A) () ( ) log + log equls ( + ) (A) + log log ( +) + (B) ( ) log + (D) log ( ) ( ) + (B) log + log ( +) + log log ( + +) + () log + log ( +) + (D) 7.6 Integrtion y Prts In this section, we escrie one more metho of integrtion, tht is foun quite useful in integrting proucts of functions. If u n v re ny two ifferentile functions of single vrile (sy). Then, y the prouct rule of ifferentition, we hve v u ( uv ) u + v Integrting oth sies, we get v u uv u + v or v u u uv v... () Let u f () n v g(). Then u f () n v () g
MATHEMATIS Therefore, epression () cn e rewritten s f () g() f () g () [ g () ] f () i.e., f() g () f () g () [ f () g () ] If we tke f s the first function n g s the secon function, then this formul my e stte s follows: The integrl of the prouct of two functions (first function) (integrl of the secon function) Integrl of [(ifferentil coefficient of the first function) (integrl of the secon function)] Emple 7 Fin cos Solution Put f () (first function) n g () cos (secon function). Then, integrtion y prts gives Suppose, we tke cos cos [ ( ) cos ] sin sin sin + cos + f () cos n g(). Then cos cos [ (cos ) ] ( cos ) sin + Thus, it shows tht the integrl cos is reuce to the comprtively more complicte integrl hving more power of. Therefore, the proper choice of the first function n the secon function is significnt. Remrks (i) It is worth mentioning tht integrtion y prts is not pplicle to prouct of functions in ll cses. For instnce, the metho oes not work for sin. The reson is tht there oes not eist ny function whose erivtive is sin. (ii) Oserve tht while fining the integrl of the secon function, we i not ny constnt of integrtion. If we write the integrl of the secon function cos
INTEGRALS 5 s sin + k, where k is ny constnt, then cos (sin + k ) (sin + k ) (sin + k) (sin k (sin + k) cos k+ sin + cos + This shows tht ing constnt to the integrl of the secon function is superfluous so fr s the finl result is concerne while pplying the metho of integrtion y prts. (iii) Usully, if ny function is power of or polynomil in, then we tke it s the first function. However, in cses where other function is inverse trigonometric function or logrithmic function, then we tke them s first function. Emple 8 Fin log Solution To strt with, we re unle to guess function whose erivtive is log. We tke log s the first function n the constnt function s the secon function. Then, the integrl of the secon function is. Hence, (log.) log [ (log ) ] (log ) log +. Emple 9 Fin e Solution Tke first function s n secon function s e. The integrl of the secon function is e. Therefore, Emple Fin sin e e e e e +. Solution Let first function e sin n secon function e First we fin the integrl of the secon function, i.e., Put t. Then t..
6 MATHEMATIS Therefore, Hence, t t t sin ( ) (sin ) ( ) sin ++ sin + Alterntively, this integrl cn lso e worke out y mking sustitution sin θ n then integrting y prts. Emple Fin e sin Solution Tke e s the first function n sin s secon function. Then, integrting y prts, we hve I e sin e ( cos ) + e cos e cos + I (sy)... () Tking e n cos s the first n secon functions, respectively, in I, we get I e sin e sin Sustituting the vlue of I in (), we get I e cos + e sin I or I e (sin cos ) e Hence, I e sin (sin cos ) + Alterntively, ove integrl cn lso e etermine y tking sin s the first function n e the secon function. 7.6. Integrl of the type e [ f ( ) + f ( )] +e f e f... () We hve I e [ f ()+ f ()] e f () + e f () I (),wherei () Tking f () n e s the first function n secon function, respectively, in I n integrting it y prts, we hve I f () e f () e + Sustituting I in (), we get I e f () f () e + e f () + e f () +
INTEGRALS 7 Thus, e [ f ( ) + f ( )] e f( ) + Emple Fin (i) Solution (i) We hve I ( +) e e (tn + ) (ii) + ( +) e (tn + ) + onsier f () tn, then f () + Thus, the given integrn is of the form e [ f () + f ()]. Therefore, I (tn + ) + e e tn + (ii) We hve ( + ) e I ( +) ++) e [ ] ( +) e [ ] e + ( + ) ( +) [ + ] + ( +) onsier f(), then f () + ( + ) Thus, the given integrn is of the form e [f () + f ()]. Therefore, + e e + ( + ) + EXERISE 7.6 Integrte the functions in Eercises to.. sin. sin. e. log 5. log 6. log 7. sin 8. tn cos 9. cos. (sin ).. tn. (log ) 5. ( + ) log. sec
8 MATHEMATIS e 6. e (sin + cos) 7. ( + ) 8. e + sin + cos 9. e ( ) e. ( ). e sin. sin + hoose the correct nswer in Eercises n.. e equls (A) () e + (B) e + (D). e sec (+ tn ) equls (A) e cos + () e sin + 7.6. Integrls of some more types e + e + (B) e sec + (D) e tn + Here, we iscuss some specil types of stnr integrls se on the technique of integrtion y prts : (i) (ii) + (i) Let I (iii) Tking constnt function s the secon function n integrting y prts, we hve I +
INTEGRALS 9 or I I or I log + + Similrly, integrting other two integrls y prts, tking constnt function s the secon function, we get (ii) + + + log + + + (iii) + sin + Alterntively, integrls (i), (ii) n (iii) cn lso e foun y mking trigonometric sustitution secθ in (i), tnθ in (ii) n sinθ in (iii) respectively. Emple Fin Solution Note tht + + 5 + + 5 ( + ) + Put + y, so tht y. Then + + 5 Emple Fin Solution Note tht y + y y y++ log y+ y+ + [using 7.6. (ii)] ( + ) + + 5 + log + + + + 5 + + ( )
MATHEMATIS Put + y so tht y. Thus y y Integrte the functions in Eercises to 9. y y y + sin + [using 7.6. (iii)] + ( + ) + sin + EXERISE 7.7... + + 6. + + 5. 6. + 5 7. + 8. + 9. hoose the correct nswer in Eercises to... + is equl to (A) ( ) (B) (D) + + log + + + ( + ) + () + + log + + + 8 + 7 is equl to + 9 ( ) + + (A) (B) () (D) ( ) 8 7 9log ++ + 8 + 7 + ( ) 8 7 9log + ++ ++ 8 + 7 + ( ) 8 7 log + + 8 + 7 + 9 ( ) 8+ 7 log + 8+ 7 +
7.7 Definite Integrl INTEGRALS In the previous sections, we hve stuie out the inefinite integrls n iscusse few methos of fining them incluing integrls of some specil functions. In this section, we shll stuy wht is clle efinite integrl of function. The efinite integrl hs unique vlue. A efinite integrl is enote y f (), where is clle the lower limit of the integrl n is clle the upper limit of the integrl. The efinite integrl is introuce either s the limit of sum or if it hs n nti erivtive F in the intervl [, ], then its vlue is the ifference etween the vlues of F t the en points, i.e., F() F(). Here, we shll consier these two cses seprtely s iscusse elow: 7.7. Definite integrl s the limit of sum Let f e continuous function efine on close intervl [, ]. Assume tht ll the vlues tken y the function re non negtive, so the grph of the function is curve ove the -is. The efinite integrl f () is the re oune y the curve y f (), the orintes, n the -is. To evlute this re, consier the region PRSQP etween this curve, -is n the orintes n (Fig 7.). Y S Q ( ) y f M DL X' P A B R O r- r n Y' Fig 7. X Divie the intervl [, ] into n equl suintervls enote y [, ], [, ],..., [ r, r ],..., [ n, n ], where, + h, + h,..., r + rh n n + nh or n. We note tht s n, h. h
MATHEMATIS The region PRSQP uner consiertion is the sum of n suregions, where ech suregion is efine on suintervls [ r, r ], r,,,, n. From Fig 7., we hve re of the rectngle (ABL) < re of the region (ABDA) < re of the rectngle (ABDM)... () Eviently s r r, i.e., h ll the three res shown in () ecome nerly equl to ech other. Now we form the following sums. s n h [f( ) + + f ( n - )] n h f( r )... () n n S n h[ f( ) + f( ) + + f( n)] h f( r)... () Here, s n n S n enote the sum of res of ll lower rectngles n upper rectngles rise over suintervls [ r, r ] for r,,,, n, respectively. In view of the inequlity () for n ritrry suintervl [ r, r ], we hve s n < re of the region PRSQP < S n... () As n strips ecome nrrower n nrrower, it is ssume tht the limiting vlues of () n () re the sme in oth cses n the common limiting vlue is the require re uner the curve. Symoliclly, we write lim S n lim sn re of the region PRSQP () n n f... (5) It follows tht this re is lso the limiting vlue of ny re which is etween tht of the rectngles elow the curve n tht of the rectngles ove the curve. For the ske of convenience, we shll tke rectngles with height equl to tht of the curve t the left hn ege of ech suintervl. Thus, we rewrite (5) s h r r f ( ) lim h[ f( ) + f( + h) +... + f ( + ( n ) h] or f ( ) ( ) lim [ f( ) + f( + h) +... + f( + ( n ) h]... (6) n n where h s n n The ove epression (6) is known s the efinition of efinite integrl s the limit of sum. Remrk The vlue of the efinite integrl of function over ny prticulr intervl epens on the function n the intervl, ut not on the vrile of integrtion tht we
INTEGRALS choose to represent the inepenent vrile. If the inepenent vrile is enote y t or u inste of, we simply write the integrl s f () t t or ( ) f u u inste of f ( ). Hence, the vrile of integrtion is clle ummy vrile. + s the limit of sum. Emple 5 Fin ( ) Solution By efinition f ( ) ( ) lim [ f( ) + f( + h) +... + f( + ( n ) h], n n where, h n In this emple,,, f () +, Therefore, + ( ) h n n ( n ) lim [ f() + f( ) + f( ) +... + f( )] n n n n n ( n ) lim [ + ( + ) + ( + ) +... + ] n + n n n n lim [(++... + ) + ( + +... + ( n ) ] n n n n-terms lim [ n+ ( + ++... ( n ) ] n n n ( n ) n( n ) lim [ n+ ] n n n 6 ( n ) ( n ) lim [ n + ] n n n lim [ + ( ) ( )] n n n + [ ]
MATHEMATIS Emple 6 Evlute Solution By efinition e s the limit of sum. e ( ) lim e + e + e +... + e n n n n n n Using the sum to n terms of G.P., where, r e n, we hve e n n e lim [ ] n n e n lim e n n e n ( e ) e n lim n n ( e ) e [using lim ] h h h EXERISE 7.8 Evlute the following efinite integrls s limit of sums.. 5. ( + ).. 5. ( ) e 6. ( + e ) 7.8 Funmentl Theorem of lculus 7.8. Are function We hve efine f ( ) s the re of the region oune y the curve y f (), the orintes n n -is. Let e given point in [, ]. Then f ( ) represents the re of the light she region Fig 7.
INTEGRALS 5 in Fig 7. [Here it is ssume tht f() > for [, ], the ssertion me elow is eqully true for other functions s well]. The re of this she region epens upon the vlue of. In other wors, the re of this she region is function of. We enote this function of y A(). We cll the function A() s Are function n is given y A () f ( )... () Bse on this efinition, the two sic funmentl theorems hve een given. However, we only stte them s their proofs re eyon the scope of this tet ook. 7.8. First funmentl theorem of integrl clculus Theorem Let f e continuous function on the close intervl [, ] n let A () e the re function. Then A () f (), for ll [, ]. 7.8. Secon funmentl theorem of integrl clculus We stte elow n importnt theorem which enles us to evlute efinite integrls y mking use of nti erivtive. Theorem Let f e continuous function efine on the close intervl [, ] n F e n nti erivtive of f. Then f ( ) [F( )] F () F(). Remrks (i) In wors, the Theorem tells us tht f ( ) (vlue of the nti erivtive F of f t the upper limit vlue of the sme nti erivtive t the lower limit ). (ii) This theorem is very useful, ecuse it gives us metho of clculting the efinite integrl more esily, without clculting the limit of sum. (iii) The crucil opertion in evluting efinite integrl is tht of fining function whose erivtive is equl to the integrn. This strengthens the reltionship etween ifferentition n integrtion. (iv) In f ( ), the function f nees to e well efine n continuous in [, ]. For instnce, the consiertion of efinite integrl ( ) is erroneous since the function f epresse y f() ( ) is not efine in portion < < of the close intervl [, ].
6 MATHEMATIS Steps for clculting f ( ). (i) Fin the inefinite integrl f ( ). Let this e F(). There is no nee to keep (ii) integrtion constnt ecuse if we consier F() + inste of F(), we get f ( ) [F ( ) + ] [F( ) + ] [F( ) + ] F( ) F( ). Thus, the ritrry constnt isppers in evluting the vlue of the efinite integrl. Evlute F() F() [F ( )], which is the vlue of f ( ). We now consier some emples Emple 7 Evlute the following integrls: (i) (ii) 9 ( ) (iii) ( + ) ( + ) (iv) sin t cos t t Solution (i) (ii) F( ), Let I. Since Therefore, y the secon funmentl theorem, we get 7 8 9 I F() F() Let 9 I. We first fin the nti erivtive of the integrn. ( ) Put t. Then t or t t Thus, t ( ) t F( ) ( )
INTEGRALS 7 Therefore, y the secon funmentl theorem of clculus, we hve I F(9) F() ( ) 9 ( 7) 8 9 99 (iii) Let I ( + ) ( + ) Using prtil frction, we get + ( + ) ( + ) + + So log + + log + F( ) ( + ) ( + ) Therefore, y the secon funmentl theorem of clculus, we hve I F() F() [ log + log ] [ log + log ] log + log + log log 7 (iv) Let I sin t cos t t. onsier sin t cos t t Put sin t u so tht cos t t u or cos t t u So sin t cos t t u u [ u ] sin t F ( t) sy 8 8 Therefore, y the secon funmentl theorem of integrl clculus I F ( ) F () [sin sin ] 8 8
8 MATHEMATIS EXERISE 7.9 Evlute the efinite integrls in Eercises to.. ( + ).. ( 5 + 69) +. sin 5. cos 6. 5 e 7. tn 8. 6 cosec 9... +. cos.. + + 5. 5 + e 6. 5 7. + + (sec + + ) 8. (sin cos ) 6 + 9.. + ( sin e + ) hoose the correct nswer in Eercises n.. equls + (A) (B) () 6 (D). equls + 9 (A) 6 (B) 7.9 Evlution of Definite Integrls y Sustitution In the previous sections, we hve iscusse severl methos for fining the inefinite integrl. One of the importnt methos for fining the inefinite integrl is the metho of sustitution. () (D)
INTEGRALS 9 To evlute f ( ), y sustitution, the steps coul e s follows:. onsier the integrl without limits n sustitute, y f () or g(y) to reuce the given integrl to known form.. Integrte the new integrn with respect to the new vrile without mentioning the constnt of integrtion.. Resustitute for the new vrile n write the nswer in terms of the originl vrile.. Fin the vlues of nswers otine in () t the given limits of integrl n fin the ifference of the vlues t the upper n lower limits. Note In orer to quicken this metho, we cn procee s follows: After performing steps, n, there is no nee of step. Here, the integrl will e kept in the new vrile itself, n the limits of the integrl will ccoringly e chnge, so tht we cn perform the lst step. Let us illustrte this y emples. Emple 8 Evlute 5 5 +. Solution Put t 5 +, then t 5. Therefore, 5 5 + t t t ( 5 ) + Hence, 5 + 5 ( + ) 5 5 5 ( + ) ( ( ) + ) ( ) Alterntively, first we trnsform the integrl n then evlute the trnsforme integrl with new limits.
MATHEMATIS Let t 5 +. Then t 5. Note tht, when, t n when, t Thus, s vries from to, t vries from to Therefore 5 + 5 t t t ( ) tn Emple 9 Evlute + Solution Let t tn, then t. The new limits re, when, t n + when, t. Thus, s vries from to, t vries from to. Therefore tn + t t t 6 EXERISE 7. Evlute the integrls in Eercises to 8 using sustitution.... + 5 sin φ cos φφ. + (Put + t ) 5. 6. + 7. + + 5 8. hoose the correct nswer in Eercises 9 n. sin + sin + cos e ( ) 9. The vlue of the integrl is (A) 6 (B) () (D). If f () t (A) cos + sin () cos sin t t, then f () is (B) sin (D) sin + cos
INTEGRALS 7. Some Properties of Definite Integrls We list elow some importnt properties of efinite integrls. These will e useful in evluting the efinite integrls more esily. P : f ( ) f () t t P : f ( ) f ( ). In prticulr, ( ) c P : f ( ) f ( ) + f ( ) c P : f ( ) f ( + ) P : P 5 : P 6 : P 7 : f ( ) f ( ) (Note tht P is prticulr cse of P ) f ( ) f ( ) + f ( ) f f ( ) f ( ),if f ( ) f ( ) n (i) if f ( ) f () f( ) f( ), if f is n even function, i.e., if f ( ) f (). (ii) f ( ), if f is n o function, i.e., if f ( ) f (). We give the proofs of these properties one y one. Proof of P It follows irectly y mking the sustitution t. Proof of P Let F e nti erivtive of f. Then, y the secon funmentl theorem of clculus, we hve f ( ) F( ) F( ) [F( ) F( )] f ( ) Here, we oserve tht, if, then f ( ). Proof of P Let F e nti erivtive of f. Then f ( ) F() F()... () c f ( ) F(c) F()... () n f ( ) F() F(c) c... ()
MATHEMATIS c Aing () n (), we get f ( ) + f ( ) F( ) F( ) f ( ) c This proves the property P. Proof of P Let t +. Then t. When, t n when, t. Therefore f ( ) f ( t ) t + f ( + t ) t (y P ) f ( + ) y P Proof of P Put t. Then t. When, t n when, t. Now procee s in P. Proof of P 5 Using P, we hve Let f ( ) f ( ) f ( ) +. t in the secon integrl on the right hn sie. Then t. When, t n when, t. Also t. Therefore, the secon integrl ecomes Hence f ( ) ( ) f t t f ( t ) t f ( ) f ( ) + f ( ) f ( ) Proof of P 6 Using P 5, we hve f( ) f( ) + f( )... () Now, if n if f ( ) Proof of P 7 Using P, we hve f ( ) f (), then () ecomes f ( ) f( ) + f( ) f( ), f( ) f (), then () ecomes f ( ) f ( ) Let f ( ) f ( ) + f ( ). Then t in the first integrl on the right hn sie. t. When, t n when, t. Also t.
INTEGRALS Therefore f ( ) f ( t ) t + f ( ) + f( ) f( ) (y P )... () (i) Now, if f is n even function, then f ( ) f () n so () ecomes f ( ) f ( ) + f ( ) f ( ) (ii) If f is n o function, then f ( ) f () n so () ecomes f ( ) f ( ) + f ( ) Emple Evlute Solution We note tht on [, ] n on [, ] n tht on [, ]. So y P we write + + ( ) ( ) ( ) + + ( ) ( ) ( ) + + + + ( ) Emple Evlute + + + + + sin Solution We oserve tht sin is n even function. Therefore, y P 7 (i), we get sin sin
MATHEMATIS ( cos ) ( cos ) sin sin sin Emple Evlute + cos sin Solution Let I. Then, y P + cos, we hve ( )sin( ) I + cos ( ) ( ) sin sin I + cos + cos sin or I + cos sin or I + cos Put cos t so tht sin t. When, t n when, t. Therefore, (y P ) we get I t + t t t + t (y P + t 7, since t is even function) + Emple Evlute tn t tn tn 5 sin cos Solution Let I 5 sin cos. Let f() sin 5 cos. Then f ( ) sin 5 ( ) cos ( ) sin 5 cos f (), i.e., f is n o function. Therefore, y P 7 (ii), I
INTEGRALS 5 Emple Evlute sin sin + cos Solution Let I sin sin + cos... () Then, y P sin ( ) I sin ( ) + cos ( ) Aing () n (), we get I Hence I cos... () cos sin cos [ ] sin + + cos + sin Emple 5 Evlute + tn 6 Solution Let I cos + tn cos + sin 66... () Then, y P I 6 cos + 6 cos + + sin + 6 6 6 Aing () n (), we get sin... () sin + cos 6 6 I [ ]. Hence I 6 6
6 MATHEMATIS Emple 6 Evlute log sin Solution Let I log sin Then, y P Aing the two vlues of I, we get I log sin log cos I ( log sin + log cos ) ( log sin cos + log log ) (y ing n sutrcting log ) log sin log (Why?) Put t in the first integrl. Then t, when, t n when t. Therefore I log sin t t log, log sin t t log [y P 6 s sin ( t) sin t) log sin log (y chnging vrile t to ) I log Hence log sin log.
INTEGRALS 7 EXERISE 7. By using the properties of efinite integrls, evlute the integrls in Eercises to 9.. cos. sin. sin + cos sin sin + cos. 5 cos 5. 5 5 sin + cos 5 + 6. 5 8 5 7. ( ) n 8. log ( + tn ) 9... (log sin log sin ).. + sin sin cos 5. 6. + sin cos 8. 7 sin. log ( + cos ) 7. sin 5 cos + 9. Show tht f( ) g( ) f( ), if f n g re efine s f() f( ) n g() + g( ) hoose the correct nswer in Eercises n. 5. The vlue of ( + cos + tn + ) is (A) (B) () (D). The vlue of (A) + sin log + cos is (B) () (D)
8 MATHEMATIS Emple 7 Fin cos 6 + sin 6 Miscellneous Emples Solution Put t + sin 6, so tht t 6 cos 6 Therefore Emple 8 Fin cos 6 + sin 6 t t 6 ( ) 5 ( t) + ( + sin 6) + 6 9 Solution We hve ( ) ( ) 5 Put t, so tht t Therefore ( ) t t 5 5 5 t + + 5 5 Emple 9 Fin ( ) ( + ) Solution We hve ( ) ( + ) ( + ) + + Now epress ( ) ( + ) ( + ) + ( ) ( + ) A B + + ( ) ( + )... ()... ()
INTEGRALS 9 So A ( + ) + (B + ) ( ) (A + B) + ( B) + A Equting coefficients on oth sies, we get A + B, B n A, which give A,B. Sustituting vlues of A, B n in (), we get ( ) ( + ) Agin, sustituting () in (), we hve Therefore ( ) ( + + ) ( ) ( + ) ( + ) ( + ) + ( ) ( + ) ( + ) + + log log ( + ) tn + ( ) ( + + ) Emple Fin log (log ) + (log ) Solution Let I log (log ) + (log ) log (log ) + (log )... () In the first integrl, let us tke s the secon function. Then integrting it y prts, we get Agin, consier I log (log ) + log (log ) log (log ) + log (log )... (), tke s the secon function n integrte it y prts, log we hve log log (log )... ()
5 MATHEMATIS Putting () in (), we get I log (log ) + log (log ) (log ) Emple Fin cot tn + log log (log ) + Solution We hve I cot + tn tn ( + cot ) Put tn t, so tht sec t t or t t + t Then t I t + t t ( + t ) + t + t ( t + ) t t t t + t + t + t t Put t y, so tht + t t t y. Then I t y y t tn tn + + y + ( ) Emple Fin t tn tn + tn + t tn sin cos 9 cos ( ) Solution Let sin cos I 9 cos
INTEGRALS 5 Put cos () t so tht sin cos t Therefore t t I sin sin cos + + 9 t Emple Evlute sin ( ) sin for Solution Here f () sin sin for Therefore sin sin + sin Integrting oth integrls on righthn sie, we get sin sin sin cos sin cos sin + + + Emple Evlute cos + sin ( ) Solution Let I cos + sin cos ( ) + sin ( ) (using P ) cos + sin cos + sin I cos + sin Thus I cos + sin
5 MATHEMATIS or I cos + sin cos + sin (using P 6 ) cos + sin cos + sin + sec cosec + tn cot + + t ( put tn t n cot u) + t u + u t u tn tn tn + tn Miscellneous Eercise on hpter 7 Integrte the functions in Eercises to... + + +. [Hint: Put t ]. ( + ) 5. + [Hint: + + 6, put t 6 ] 5 6. ( + ) ( + 9) 7. sin sin ( ) 8. e e 5 log log e e log log 9. cos sin. sin 8 8 cos sin cos. cos ( + ) cos ( + ) e.. 8. ( + e ) ( + e ) ( + ) ( + ) 5. cos e log sin 6. e log ( + ) 7. f ( + ) [f ( + )] n sin cos 8. 9., [, ] sin sin ( +α) sin + cos
INTEGRALS 5. +. + sin e + cos. + + ( + ) ( + ) + log ( ) log. tn. + + Evlute the efinite integrls in Eercises 5 to. 5. 8.. 6 sin e cos sin + cos 9. sin sin cos 6. 7. cos + sin +. sin tn (sin ).. [ + + ] Prove the following (Eercises to 9). + log 5. ( + ) 6. 8. 7 cos 7. tn log 9. e cos cos + sin sin + cos 9 + 6sin tn sec + tn e sin sin. Evlute s limit of sum. hoose the correct nswers in Eercises to.. e + e is equl to (A) tn (e ) + (B) tn (e ) + () log (e e ) + (D) log (e + e ) +. cos (sin + cos ) (A) sin + cos + (B) log sin + cos + () log sin cos + (D) (sin + cos )
5 MATHEMATIS. If f ( + ) f (), then f ( ) is equl to + + (A) f ( ) (B) f ( ) + + () f ( ) (D) f ( ). The vlue of + tn is (A) (B) () (D) Summry Integrtion is the inverse process of ifferentition. In the ifferentil clculus, we re given function n we hve to fin the erivtive or ifferentil of this function, ut in the integrl clculus, we re to fin function whose ifferentil is given. Thus, integrtion is process which is the inverse of ifferentition. Let F( ) f( ). Then we write f ( ) F( ) +. These integrls re clle inefinite integrls or generl integrls, is clle constnt of integrtion. All these integrls iffer y constnt. From the geometric point of view, n inefinite integrl is collection of fmily of curves, ech of which is otine y trnslting one of the curves prllel to itself upwrs or ownwrs long the y-is. Some properties of inefinite integrls re s follows:. [ f ( ) + g ( )] f ( ) + g ( ). For ny rel numer k, k f( ) k f( ) More generlly, if f, f, f,..., f n re functions n k, k,...,k n re rel numers. Then [ k f ( ) + k f ( ) +... + k f ( )] n n n n k f ( ) + k f ( ) +... + k f ( )