Get Discount oupons for your oching institute nd FREE Study Mteril t lculus derive nd use the result f ' f ln f stte nd use the me

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1 Get Discount oupons for your oching institute nd FREE Study Mteril t Integrtion lculus 6 INTEGRATION In the previous lesson, you hve lernt the concept of derivtive of function. You hve lso lernt the ppliction of derivtive in vrious situtions. onsider the reverse problem of finding the originl function, when its derivtive (in the form of function) is given. This reverse process is given the nme of integrtion. In this lesson, we shll study this concept nd vrious methods nd techniques of integrtion. OBJETIVES After studying this lesson, you will be ble to : eplin integrtion s inverse process s (nti-derivtive) of differentition; find the integrl of simple functions ns like n,sin, cos, sec,cosec,sectn,coseccot,,cosec,e stte the following results : MATHEMATIS 9 Get Discount oupons for your oching institute nd FREE Study Mteril t etc.; (i) [ ± g ( )] ± f g d f d g d [ ± kf ( )] d ± pickm,s kf d k f d ww. w.pic g.com find the integrls of lgebric, trigonometric, inverse trigonometric nd eponentil functions; find the integrls of functions by substitution method. evlute integrls of the type d d d,,, ± ± d ( p q ) d, b c b c, d d,, b c ( p q ) d b c

2 Get Discount oupons for your oching institute nd FREE Study Mteril t lculus derive nd use the result f ' f ln f stte nd use the method of integrtion by prts; evlute integrls of the type : ± d, d, ( p q ) b c d, sinn cosm d d,, bsin derive nd use the result e sinbd, sin d, d bcos e [ f ( ) f '( )] d ef c; c ; nd integrte rtionl epressions using prtil frctions. e cosb d, cos d, EXPETED BAKGROUND KNOWLEDGE Differentition of vrious functions Bsic knowledge of plne geometry Fctoriztion of lgebric epression Knowledge of inverse trigonometric functions 6. INTEGRATION N Integrtion Integrtion literlly mens summtion. onsider, the problem of finding re of region ALMB s shown in Fig Fig kmy Myo oching.com com We will try to find this re by some prcticl method. But tht my not help every time. To solve such problem, we tke the help of integrtion (summtion) of re. For tht, we divide the figure into smll rectngles (See Fig.6.). Fig MATHEMATIS Get Discount oupons for your oching institute nd FREE Study Mteril t

3 Get Discount oupons for your oching institute nd FREE Study Mteril t Integrtion Unless these rectngles re hving their wih smller thn the smllest possible, we cnnot find the re. This is the technique which Archimedes used two thousnd yers go for finding res, volumes, etc. The nmes of Newton (6-77) nd Leibnitz (66-76) re often mentioned s the cretors of present dy of lculus. The integrl clculus is the study of integrtion of functions. This finds etensive pplictions in Geometry, Mechnics, Nturl sciences nd other disciplines. In this lesson, we shll lern bout methods of integrting polynomil, trigonometric, eponentil nd logrithmic nd rtionl functions using different techniques of integrtion. lculus 6.. INTEGRATION AS INVERSE OF DIFFERENTIATION onsider the following emples : d (i) ( ) d ( sin ) cos d d Let us consider the bove emples in different perspective (i) is function obtined by differentition of d (iii) ( e ) MATHEMATIS 5. is clled the ntiderivtive of cos is function obtined by differentition of sin sin is clled the ntiderivtive of cos (iii) Similrly, e is clled the ntiderivtive e of e Generlly we epress the notion of ntiderivtive in terms of n opertion. This opertion is clled the opertion of integrtion. We write. Integrtion of is. Integrtion of cos is sin. Integrtion of e is e The opertion of integrtion n is denoted by the symbol. Thus. d. cosd sin. d e d Remember tht d is symbol which together with symbol denotes the opertion of integrtion. The function to be integrted is enclosed between nd d. Definition : If [ ] e e d f f ', then f () is sid to be n integrl of f '() nd is written d s f '()d f() The function f '() which is integrted is clled the integrnd. om Get Discount oupons for your oching institute nd FREE Study Mteril t

4 Get Discount oupons for your oching institute nd FREE Study Mteril t lculus onstnt of integrtion If dy y,then d d d d Now consider ( ) or ( c) d Integrtion where c is ny rel constnt. Thus, we see tht d integrl of is not unique. The different vlues of d, where c is clled the constnt of integrtion. Thus e d e, cosd sin c In generl ddiffer by some constnt. Therefore, f '()d f. The constnt c cn tke ny vlue. We observe tht the derivtive of n integrl is equl to the integrnd. Note : f()d, f(y)dy, f(z)dz but not like f(z)d Emple 6. Find the integrl of the following : Solution : (i) (iii) (i) d, since 0 d, since ww n d n n n 0 (iii) n d d 0 d d w.pic ching. g.com wsi d d n d n n d n since 0 n n n Emple 6. (i) If dy cos d, find y. If dy sin d, find y. Solution : (i) dy d cos d d y sin 5 dy d sin d d y cos MATHEMATIS Get Discount oupons for your oching institute nd FREE Study Mteril t

5 Get Discount oupons for your oching institute nd FREE Study Mteril t Integrtion 6. INTEGRATION OF SIMPLE FUNTIONS We hve lredy seen tht if f () is ny integrl of f '(), then functions of the form f () provide integrl of f '(). We repet tht cn tke ny vlue including 0 nd thus f '()d f() which is n indefinite integrl nd it becomes definite integrl with defined vlue of. lculus Emple 6. Write ny different vlues of Solution : d. d, where is constnt. The four different vlues of d my be,, nd etc. Integrls of some simple functions given below. The vlidity of the integrls is checked by showing tht the derivtive of the integrl is equl to the integrnd. Integrl Verifiction. n n d n where n is constnt nd n.. sin d cos MATHEMATIS 5 n d d n d cos sin d y( d sin cos d d tn sec d ( cos ). cos d sin. 5. sec d tn cosec d cot ww ww ot 6. sec tn d sec w 7. cosec cot d w cosec tdww d cot cosec d d sec sec tn d d cosec cosec cot d 8. d sin 9. d tn d d ( sin ) d d ( tn ) 0. d sec My( w.pic ching.com d ( sec ) d Get Discount oupons for your oching institute nd FREE Study Mteril t hin n om

6 Get Discount oupons for your oching institute nd FREE Study Mteril t Integrtion lculus. e d e d d e e. d log d d log if > 0. d log ( log ) d d WORKING RULE n. To find the integrl of, increse the inde of by, divide the result by new inde nd dd constnt to it.. d will be very often written s f d. f ( ) HEK YOUR PROGRESS RESS 6.. Write ny five different vlues of 5 d. Write indefinite integrl of the following : () 5 (b) cos (c) 0. Evlute : () (e) 6 d d (b) (f) 7 9 d d (c) (g) d d (d) (h) 5 d 9 8. Evlute : cos θ d θ sin θ (b) sin θ () d θ cos θ ww.pic w.pic yoching.com d (c) cos θ sin θ d θ (d) cos θ sin dθ θ 6. PROPERTIES OF INTEGRALS If function cn be epressed s sum of two or more functions then we cn write the integrl 7 of such function s the sum of the integrl of the component functions, e.g. if f (), then 5 MATHEMATIS Get Discount oupons for your oching institute nd FREE Study Mteril t

7 Get Discount oupons for your oching institute nd FREE Study Mteril t Integrtion [ 7 ] f d d 7 d d lculus 8 8 So, in generl the integrl of the sum of two functions is equl to the sum of their integrls. Similrly, if the given function [ ] f g d f d g d f 7 we cn write it s ( 7 ) f d d 7 d 8 8 d The integrl of the difference of two functions is equl to the difference of their integrls. i.e. [ ] d g( ( ) f g d f d g d If we hve function f() s product of constnt (k) nd nother function [ g( )] i.e. f() kg(), then we cn integrte f() s ) f d kg( d k g d Integrl of product of constnt nd function is product of tht constnt nd integrl of the function. i.e. f ( ) d kf d k f d Emple 6. Evlute : d ( ) ( ) d ww.p w.pic ww Solution :(i) d log d d ( ) ( ) MATHEMATIS 55 Get Discount oupons for your oching institute nd FREE Study Mteril t

8 Get Discount oupons for your oching institute nd FREE Study Mteril t lculus Remember in it would not be correct to sy tht d d d d log Integrtion Becuse d d log log log Therefore, integrl of product of two functions is not lwys equl to the product of the integrls. We shll del with the integrl of product in subsequent lesson. Emple 6.5 Evlute : (i) d n cos, when n 0 nd n Solution : d d (i) When n 0, n o cos cos d d Now d o cn be written s d. o d d o 0 When n, d d n cos cos sec d tn sin θ cos θ dθ d θ sin θ sin θ cosec θdθ cot θ om sin θ cos sin ching.com θ θ dθ 56 MATHEMATIS Get Discount oupons for your oching institute nd FREE Study Mteril t

9 Get Discount oupons for your oching institute nd FREE Study Mteril t Integrtion Emple 6.6 Evlute : (i) ( ) sin cos d d (iii) d (iv) d Solution : (i) (iii) sin cos d sind cos d cos sin d d d d log log d d ( ) d / d d (iv) d ww Emple 6.7 Evlute : (i) sin θ dθ θ pic ckd tn sin e d w.pickmyoching.com lculus (iii) ( ) tn cot d (iv) 6 d Solution : (i) sin θ cos θ sin θ sin θcos θ sin θ cos θ MATHEMATIS 57 Get Discount oupons for your oching institute nd FREE Study Mteril t

10 Get Discount oupons for your oching institute nd FREE Study Mteril t lculus ( cos sin ) θ θ ( cos sin ) ± θ θ (sign is selected depending upon the vlue of θ ) () If sinθ cos θ sin θ sinθ dθ cos θ sin θ dθ then cos θd θ sin θdθ sin θ cos θ sinθ dθ cos θ sin θ d θ (b) If cos θd θ sin θd θ sin θ cosθ e d e d d d ochi e d e s d ( t e sec (iii) ( ) d ( ) tn cot d tn cot tn cot d ( tn cot ) d ( sec cosec) d tn cot d sec d ww tn cot (iv) 6 d d cosec d (dividing 6 yoch ching ching.com ww ww Emple 6.8 Evlute : y Integrtion by ) d d d d 5 tn 5 58 (i) d 5 e 9e d (iii) e d b MATHEMATIS Get Discount oupons for your oching institute nd FREE Study Mteril t

11 ww Get Discount oupons for your oching institute nd FREE Study Mteril t Integrtion Solution : (i) / d d / / d d d d 5/ / / 5 / lculus (iii) 5 e 9e d e d b MATHEMATIS 59 e d 9e d d e e e e d 9 e d e d e 9e e Let us first rtionlise the denomintor of the integrnd. b b b b b M ( ) ( b) km ck.pi b b d b d b b w.pi ching ing.c.com ww b d d b b ( ) ( b) b b ( ) ( b) ( b ) Get Discount oupons for your oching institute nd FREE Study Mteril t

12 Get Discount oupons for your oching institute nd FREE Study Mteril t lculus HEK YOUR PROGRESS 6. Integrtion. Evlute : () d (b) d (c) 0 d 9 (d) d 6 (e). Evlute : () d d (b) cos d sin (d) (e) d cos cos. Evlute :. Evlute : We know tht (f) m d tn d (c) cos d sin (f) ( ) () cos d (b) cos d (c) () d (b) d d ) n icn 7 ( ) 7 ( b ) ( n ) ( b) ( n ) d n d n b ( b ), n d ( ww.pi w.pic ickmyoching.com ching om ( ) ( n ) n n b b d ) n n cosec cot cosec d cos To find the integrl of ( b, increse the inde by one nd divide the result by the incresed inde nd the coefficient of nd dd constnt of integrtion. Emple 6.9 Integrte the following : (i) ( 9) e 7 (iii) cos ( 5π) (iv) cosec ( ) om d 60 MATHEMATIS Get Discount oupons for your oching institute nd FREE Study Mteril t

13 Get Discount oupons for your oching institute nd FREE Study Mteril t Integrtion Solution : (i) 9 9 d ( 9) e 7d e e7 d e7 e d ( e 7 is constnt quntity) lculus 7 7 e e e cos 5π d sin 5 π (iii) d d (iv) d π π π π d ( sin ( 5 ) ) cos ( 5 ) ( 5 π)π)π)π)π) c cos ( 5 ) cot cosec d c ( ) d cot cosec cosec d Emple 6.0 Evlute : ( 5)( ) 5 d Solution : ( )( ) d ( ) 5 d 5 d 5 d / / / d d 5 d ching.co.com coco cos Emple 6. Evlute : e d MATHEMATIS 6 Get Discount oupons for your oching institute nd FREE Study Mteril t

14 Get Discount oupons for your oching institute nd FREE Study Mteril t Integrtion lculus Solution : e e d d e d d d e d d Emple 6. Integrte sec cosec w.r.t. e log Solution : ( ) ( c co ) Emple 6. Evlute : sec cosec d tn cot d omco sec tn cosec cot nd ( t tn ) ng ( tn ) ( cot ) d ( sec cosec ) d tn cot tn cot d ( tn cot 7 6 My d d d Solution : d 7 6 Emple 6. Evlute : ( ) Solution : cos d cos d d cosd 7tn 6sin w.pick ching.com (com sin 6.5 TEHNIQUES OF INTEGRATION 6.5. Integrtion By Substitution o o ng.coc comco.cc This method consists of epressing f d in terms of nother vrible so tht the resultnt 6 MATHEMATIS Get Discount oupons for your oching institute nd FREE Study Mteril t

15 Get Discount oupons for your oching institute nd FREE Study Mteril t Integrtion function cn be integrted using one of the stndrd results discussed in the previous lesson. First, we will consider the functions of the type f ( b), 0 where f () is stndrd function. Emple 6.5 Evlute : sin b d π cos 7 d (iii) (i) ( ) Solution : (i) ( ) sin b d Put b t. Then or d sin b d sint d π sin d (Here the integrtion fctor will be replced by.) π cos 7 d sint ( cost) ( b) cos π Put 7 t 7d M π cos 7 d cost 7 cost 7 sin t 7 w.pickmyoching.com.com π sin 7 7 lculus (iii) Put sin π d π t MATHEMATIS 6 Get Discount oupons for your oching institute nd FREE Study Mteril t

16 Get Discount oupons for your oching institute nd FREE Study Mteril t lculus Then or d d π sin d sint ( cost) cos t π cos Similrly, the integrls of the following functions will be sin d π sin d π sin d cos π cos cos π sin b cos ( b ) d sin ( ) cos d Emple 6.6 Evlute : n b d, where n (i) ( ) n Solution : (i) ( b d, where n Put b t ) n b d t ) n n ( sin t n n ( ) g.co hing ing ( b) or d d w.pic ching hing.com ww( d Integrtion n b n where n 6 MATHEMATIS Get Discount oupons for your oching institute nd FREE Study Mteril t

17 Get Discount oupons for your oching institute nd FREE Study Mteril t Integrtion ( b) d lculus Put b t d ( ) d b t Emple 6.7 Evlute : (i) e Solution : (i) 5 7 d e 5 7 d log t log b e Put 5 7 t 5 7 e d et 5 e e t 5 e d Put t ww d t e d e ww ww ww e t e d d d 5 ww.p MATHEMATIS 65 Get Discount oupons for your oching institute nd FREE Study Mteril t

18 Get Discount oupons for your oching institute nd FREE Study Mteril t lculus Likewise b e d e b Integrtion Similrly, using the substitution b t, the integrls of the following functions will be : ( b) n d ( b) d n b, n n log b cos b sin ( b ) d sin b ) cos ( b ) d tn b b sec ( b ) d ( b ) cot b sec ( b ) tn ( b dch ) sec b cosec ( b ) d cot ( ) cosec b cosec ( b ) cot ( b ) d Emple 6.8 Evlute te : chc (i) sin d d Mn sin (iii) cos d (iv) sin sin d Solution : We use trigonometricl identities nd epress the functions in terms of sines nd cosines of multiples of (i) cos cos sin d d sin ( cos ) d d cos d sin sin sin sin d d sin sin sin w.pic ching com 66 ( sin sin) d MATHEMATIS Get Discount oupons for your oching institute nd FREE Study Mteril t

19 Get Discount oupons for your oching institute nd FREE Study Mteril t Integrtion (iii) (iv). Evlute : cos cos cos cos cos d d sin sin d ( cos cos ) d sin sin sin sin d [ sinasinb cos ( A B ) cos ( A B) ] ( cos cos5 ) d sin5 sin 5 HEK YOUR PROGRESS 6. cos cos cos () sin ( 5 ) d (b) ( ) π (c) sec d sec d (d) cos ( ) (e) ( ) tn ( 5 5 ) sec 5 tn 5 d (f) ( 5 ) cot ( ) cosec 5 cot 5 d. Evlute : ww d () 5 d ( ) (b) ( ) d (c) ( ) 0 (d) ( ) 5 d (g) ( ). Evlute : () e d d w.pt.p( (e) (h) (b) M(d (b) o d 5 d (f) 7 d ww.pi w.pic d 5 9 e 8 d (c) ( 7 e ) d lculus MATHEMATIS 67 Get Discount oupons for your oching institute nd FREE Study Mteril t

20 Get Discount oupons for your oching institute nd FREE Study Mteril t lculus. Evlute : () cos d (c) (b) sin cos d sin cos d (d) cos cos d 6.5. Integrtion of Function of The Type To evlute f ' f f '( ) f ( ) d, we put f () t. Then f ' () d. f ' d f t log t log f Integrtion Integrl of function, whose numertor is derivtive ive of the denomintor, is equl to the logrithm of the denomintor. Emple 6.9 Evlute : (i) d Solution : d (iii) ( ) 5 (i) Now is the derivtive of. By pplying the bove result, we hve d log derivtive of 5 is the. (iii) 5 is the derivtive of d ( ) Emple 6.0 Evlute : (i) e e e e d d log 5 ww.pickmyoching.com ing.c log d e d e 68 MATHEMATIS Get Discount oupons for your oching institute nd FREE Study Mteril t

21 Get Discount oupons for your oching institute nd FREE Study Mteril t Integrtion Solution : (i) e e is the derivtive of e e Alterntively, e e e e d log e e e e d, For e e Put e e t. Then e e d e e d e e t log t log e e e d e Here e is not the derivtive of e. But if we multiply the numertor nd denomintor by e, the given function will reduce to e e e e d log e e e e e e d l log e e e e e HEK K YOUR PROGRESS 6. ( e e ) is the derivtive of ( e e ). Evlute : () d (b) 9 d (c) d 9 0 d (d) d (e) d (f) 5 ( 5 ) d (g) 8 log w.pic lculus MATHEMATIS 69 Get Discount oupons for your oching institute nd FREE Study Mteril t

22 Get Discount oupons for your oching institute nd FREE Study Mteril t lculus. Evlute : () e d d (b) e e be 6.5. Integrtion by Substitution Emple 6. Integrtion (i) (iv) tnd secd Solution : (i) ( ) ( cos) sin d (v) cosec5 cotd (iii) tn d tn sin (vi) sin sin tnd d cos sin d cos log cos ( sin is derivtive of cos ) tnd log sec Alterntively, sin d tnd cos sin d cos Put cos t. Then sin d log or log sec cos y chi ww tnd t log t log cos log cos log sec och( d 70 MATHEMATIS Get Discount oupons for your oching institute nd FREE Study Mteril t

23 Get Discount oupons for your oching institute nd FREE Study Mteril t Integrtion sec d sec cn not be integrted s such becuse sec by itself is not derivtive of ny function. But this is not the cse with sec nd sec tn. Now sec ( sec tn) ( sec tn) d Put sec tn t. Then ( ) sectn sec d sec d sec sectn d sec tn t log t log sec tn tn cos sin (iii) d d tn cos sin Put cos sin t So tht ( sin cos ) d (iv) Put tn d tn t log t log cos s sin sin sin d d cos cos cos cos sin cos d cos cos ww ww sec d tn d t d sin d sec t tnt cos tnt log cost sec d cn be written s ww.p w.pic lculus MATHEMATIS 7 Get Discount oupons for your oching institute nd FREE Study Mteril t

24 Get Discount oupons for your oching institute nd FREE Study Mteril t lculus (v) tn log cos 5 cosec cot d cosec coseccot d Put cosec t. Then cosec cot d (vi) sin 5 cosec cot d t ( ) sin d t 5 5 ( cosec ) 5 5 Put t Then d nd t ( ) ( t ) sin sin t d sin sintchd Emple 6. Evlute d sintcos costsin sint yosintco ( s ( A B ) sinacosb cosasinb cos sincott (cos nd sin re constnts.) cos t sinlog sint ( ) cos sin log sin ( ) ching com ww Solution : Put sin θ d cos θdθ cos θ d d θ sin θ cos θ d θ sin θ d θ cos θ Integrtion 7 MATHEMATIS Get Discount oupons for your oching institute nd FREE Study Mteril t

25 Get Discount oupons for your oching institute nd FREE Study Mteril t Integrtion sec θ d θ log sec θ tn θ sin θ log cos θ lculus log log log log log Emple 6. Evlute : d Solution : Put sec θ d sec θtn θdθ sec θ tnθ dθ d sec θ ww www ww.pick y secθ tn θ θ tn θ d ( tn θ sec θ ) sec θ d θ tn θ d cosec d θ θ θ sin θ log cosec θ cot θ MATHEMATIS 7 Get Discount oupons for your oching institute nd FREE Study Mteril t

26 Get Discount oupons for your oching institute nd FREE Study Mteril t lculus cos θ log sin θ Integrtion Emple 6. d Solution : Put log log log log tn θ d sec θdθ sec θ d θ d ( tn θ) Emple 6.5 d d θ θ tn Put sin θ d cos θdθ cos θ d d θ sin θ cos θ d θ cos θ d θ Myo yoching.com tn θ tn θ 7 MATHEMATIS Get Discount oupons for your oching institute nd FREE Study Mteril t

27 Get Discount oupons for your oching institute nd FREE Study Mteril t Integrtion θ sin lculus Emple 6.6 d Solution : Let sec θ d sec θtn θdθ sec θ tn θ d θ sec θ sec θθ d Emple 6.7 d log sec θ tn θ log log Solution : Put tn θ d sec θ dθ Emple 6.8 sec θdθ [As emple 6.5] log sec θ tn θ log log now d ckm ww ww.p w.pickmyoching.com Solution : Since is not the derivtive of, therefore, we write the given integrl s d Let t. d MATHEMATIS 75 Get Discount oupons for your oching institute nd FREE Study Mteril t

28 Get Discount oupons for your oching institute nd FREE Study Mteril t lculus Also t t d t t Emple 6.9 Solution : Put d d t. Then d Also t t My. ckmd km km d t tn tn t d ( t) t log t Integrtion 76 MATHEMATIS Get Discount oupons for your oching institute nd FREE Study Mteril t

29 Get Discount oupons for your oching institute nd FREE Study Mteril t Integrtion log lculus Emple 6.0 d Solution : In order to solve it, we will reduce the given integrl to the integrls given in Emples 6.8 nd 6.9. i.e., d d d d tn log Emple 6. d Solution : We cn reduce the given integrl to the following form ( ) ( ) d d d log tn Emple 6. () d Solution : () ching ing.c co MATHEMATIS 77 Get Discount oupons for your oching institute nd FREE Study Mteril t (b) d d g.c hing d d

30 Get Discount oupons for your oching institute nd FREE Study Mteril t Integrtion lculus d tn (b) Put Also pw d d t. d t t Emple 6. km tn d d t t log t log kmyoching.com Solution : Let tn t sec d d t sec t t t 78 MATHEMATIS Get Discount oupons for your oching institute nd FREE Study Mteril t

31 Get Discount oupons for your oching institute nd FREE Study Mteril t Integrtion t tnd t t t t t t t t lculus t t t t Proceed ccording to emple 6.8 nd Emple 6.9 solved before. Emple 6. cot d Solution : Let cot t cosec d t d t cosec t t t cot d t t t t t t t t t t Proceed ccording to Emples 6.8 nd 6.9 solved before. Emple 6.5 ( ) Let sin cos t www wwt ww tn cot d cos sin d Also sincos t t sin cos ww.p w.pick t sin cos MATHEMATIS 79 Get Discount oupons for your oching institute nd FREE Study Mteril t

32 Get Discount oupons for your oching institute nd FREE Study Mteril t Integrtion lculus sin cos d cossin t t Emple 6.6 Evlute : d () 8 (b) Solution : d [ ] sin sin cos d ( ) () 8 8 ( ) (b) d d d sin sin ching.com c d d ( ) d d 6 6 (Using the result of Emple 6.5) 80 MATHEMATIS Get Discount oupons for your oching institute nd FREE Study Mteril t

33 Get Discount oupons for your oching institute nd FREE Study Mteril t Integrtion d sin sin lculus. Evlute : () HEK YOUR PROGRESS 6.5 d 9 (b) e d (c) e (d) d 6 9 (e) d d sin (f) (g) d 6 (h) d d 5 (i) dθ (j) sin θ cos θ d (m) (k) (n) e e d e km 9 6 (p) d sinθ 9 (q) dθ cos θ (s) 6 d ( d (t) d )w) ww INTEGRATION BY PARTS In differentition you hve lernt tht d d d ( fg ) f ( g ) g ( f ) d d d km (l) (o) (r) d d ( ) d sec d tn )ww ing.c MATHEMATIS 8 Get Discount oupons for your oching institute nd FREE Study Mteril t