INDEFINITE INTEGR ATION

Transcription

1 J-Mthemtics INDFINIT INTGR ATION If f & F re function of such tht F' () f() then the function F is clled PRIMITIV OR ANTIDRIVATIV OR INTGRAL of f() w.r.t. nd is written symboliclly s d f() F() c {F() c} f(), where c is clled the constnt of integrtion.. GOMTRICAL INTRPRTATION OF INDFINIT INTGR AL : f() F() c y(sy), represents fmily of curves. The different vlues of c will correspond to different members of this fmily nd these members cn be obtined by shifting ny one of the curves prllel to itself. This is the geometricl interprettion of indefinite integrl. Let f(). Then f() c. For different vlues of c, we get different integrls. But these integrls re very similr geometriclly. Thus, y + c, where c is rbitrry constnt, represents fmily of integrls. By ssigning different vlues to c, we get different members of the fmily. These together constitute the indefinite integrl. In this cse, ech integrl represents prbol with its is long y-is. If the line intersects the prbols y, y +, y +, y, y t P 0, P, P, P, P etc., respectively, then dy t these points equls. This indictes tht the tngents to the curves t these points X' Y P P P P 0 P P y+ y+ y+ y y y y X re prllel. Thus, d + c f( ) + c (sy), P NOD6\ :\Dt\0\Kot\J-Advnced\SMP\Mths\Unit#06\ng\0-INDFINIT INTGRATION.p6 implies tht the tngents to ll the curves f() + c, c R, t the points of intersection of the curves by the line, ( R), re prllel.. STANDARD RSULTS : (i) (iii) ( v ) (vii) (i) (i) n n ( b) ( b) c; n (n ) b b e e c (ii) (iv) sin( b) cos( b) c (vi ) tn( b) n sec( b) c (viii) sec ( b) tn( b) c () cosec ( b). cot( b) cosec ( b) c (ii) sec ( + b).tn( + b) sec( b) c n b c b pq Y' p q c, ( 0) p n cos( b) sin( b) c cot( b) n sin( b) c cos ec ( b) cot( b) c

2 J-Mthemtics (iii) sec n sec tn c n tn c (iv) cosec n cosec cot c (v) (vii) (i) sin c sec c n tn c n cosec cot + c (vi) (viii) tn c n + c n + c () (i) n c (ii) (iii) (iv) n c n c n c sin c (v) e e.sin b ( sin b b cos b) c b e b b sin b tn c (vi) e e b e.cos b ( cos b b sin b) c cos b tn c b b. TCHNIQUS OF INTGR ATION : ( ) Substitution or chnge of independent vrible : If () is continuous differentible function, then to evlute integrls of the form f( ()) '(), we substitute () t nd '() dt. Hence I f( ()) '() reduces to f(t)dt. ( i ) Fundmentl deductions of method of substitution : Illustrtion : f() n f '() f '() OR n put f() t & proceed. [f()] vlute cos sin sin ( sin ) cos Solution : I sin cos sin ( sin ) sin Illustrtion : Put sin t cos dt I t dt n t t c n sin sin c t vlute tn NOD6\ :\Dt\0\Kot\J-Advnced\SMP\Mths\Unit#06\ng\0-INDFINIT INTGRATION.p6

3 J-Mthemtics Solution : The given integrl cn be written s I Let tn t. Differentiting we get dt Hence I dt t tn t dt Now mke one more substitution tn t u. Then t Returning to t, nd then to, we hve du nd I du n u c u I n tn t c n tn c Do yourself - : (i) vlute : 6 (ii) vlute : 9 6 cos ( i i) Stndrd substitutions : or ; put tn or cot or ; put sin or cos NOD6\ :\Dt\0\Kot\J-Advnced\SMP\Mths\Unit#06\ng\0-INDFINIT INTGRATION.p6 Illustrtion : Solution : or ; put sec or cosec ; put cos or ( )( ) ; put cos + sin or ( )( ) ; put sec tn ( )( ) ; put t or t. vlute ( )(b ) Put cos + bsin, the given integrl becomes

4 J-Mthemtics I (b ) sin cos d ( cos b sin )(b cos b sin (b ) sin cos d b sin cos Illustrtion : vlute. Solution : Put cos sin cos d b b b d c sin c cos I. ( sin cos )d tn tn d cos cos sin ( / ) cos d d n sec tn c cos cos n cos c Do yourself - : (i) vlute : (ii) vlute : du ( b ) Integrtion by prt : u.v u v. v re commonly designted s first & second function respectively. where u & v re differentible functions nd Note : While using integrtion by prts, choose u & v such tht (i) v & (ii) du. v re simple to integrte. This is generlly obtined by choosing first function s the function which comes first in the word ILAT, where; I-Inverse function, L-Logrithmic function, A-Algebric function, T-Trigonometric function & -ponentil function. Illustrtion : vlute : cos Solution : Consider I cos Let t Illustrtion 6 : vlute : then dt i.e. dt or t dt so I cos t.tdt tking t s first function, then integrte it by prt dt I t cos tdt cos tdt dt dt t sin t.sin tdt t sin t cos t c I sin cos c sin NOD6\ :\Dt\0\Kot\J-Advnced\SMP\Mths\Unit#06\ng\0-INDFINIT INTGRATION.p6

5 Solution : Let I sin ( sin ) sin sec ( sin ) ( sin )( sin ) ( sin ) sec sec tn cos sec tn tn sec sec tn n sec sec n sec tn c J-Mthemtics sec tn sec tn (sec tn ) tn sec n c ( sin ) n sin c sec cos Do yourself - : (i) vlute : e (ii) vlute : sin( ) Two clssic integrnds : (i) e [f() f '()] e.f() c Illustrtion 7 : vlute e Solution : e ( ) e e e c ( ) ( ) ( ) NOD6\ :\Dt\0\Kot\J-Advnced\SMP\Mths\Unit#06\ng\0-INDFINIT INTGRATION.p6 Illustrtion 8 : The vlue of Solution : Let I Do yourself - : (i) vlute : e ( ) ( ) / e / is equl to - ( ) e / ( ) e ( ) ( ) / e ( ) ( ) e ( ) ( ) / / e ( ) ( ) ( ) ( ) e e ( ) ( ) / / / / / / (ii) [f() f '()] f() c Illustrtion 9 : vlute c e { } / c ( ) /. e tn (ii) vlute : sin cos e sin cos (D) none of these Ans. (D)

6 J-Mthemtics sin sin Solution : I cos sec tn cos tn + c Do yourself - : e sec (e ) (ii) vlute : ( n ) (i) vlute : tn(e ) ( c ) Integrtion of rtionl function : (i) Rtionl function is defined s the rtio of two polynomils in the form P(), where P() nd Q() Q() re polynomils in nd Q() 0. If the degree of P() is less thn the degree of Q(), then the rtionl function is clled proper, otherwise, it is clled improper. The improper rtionl function cn be reduced to the proper rtionl functions by long division process. Thus, if P() is improper, Q() then P() P () T() + Q() Q(), where T() is polynomil in nd P () is proper rtionl function. Q() It is lwys possible to write the integrnd s sum of simpler rtionl functions by method clled prtil frction decomposition. After this, the integrtion cn be crried out esily using the lredy known methods. S. No. Form of the rtionl function Form of the prtil frction. p q r ( ) ( b)( c) A + B b + C c. p q r ( ) ( b) A B + ( ) + C b. Illustrtion 0 : Solution : p q r ( ) ( b c) where + b + c cnnot be fctorised further f(). ( )( b c) where f() is polynomil of degree less thn. vlute ( )( ) A B ( )( ) or A( + ) + B( ). by compring the coefficients, we get A /7 nd B /7 so tht 6 A + B C b c A + B C b c + D ( b c) n ( ) n c ( )( ) NOD6\ :\Dt\0\Kot\J-Advnced\SMP\Mths\Unit#06\ng\0-INDFINIT INTGRATION.p6

7 J-Mthemtics Illustrtion : Solution : vlute ( )( ) ( ) ( )( ) ( )( ) Now, So, 6 ( )( ) ( ) 6 ( )( ) ( ) 6 Now, ( ) ( ) 6 tn n n( ) c 0 Do yourself - 6 : (i) vlute : (ii) vlute : ( )( ) ( )( ) (ii),, b c b c b c press + b + c in the form of perfect squre & then pply the stndrd results. p q p q (iii), b c b c press p + q (differentil coefficient of denomintor ) + m. NOD6\ :\Dt\0\Kot\J-Advnced\SMP\Mths\Unit#06\ng\0-INDFINIT INTGRATION.p6 Illustrtion : vlute Solution : I 6 6 ( / ) 9 / 6 ( / ) ( / ) /. log / c using, log c ( / ) / / log / Illustrtion : vlute c log ( ) c 7

8 J-Mthemtics Solution : press + (d.c. of + + ) + m or, + (8 + ) + m Compring the coefficients, we get 8 nd + m /8 nd m / I 8 8 log 8 log tn c 8 8 Ans. Do yourself -7 : (i) vlute : (ii) vlute : (iv) Integrls of the form K OR K where K is ny constnt. Divide N r & D r by & proceed. Note : Sometimes it is useful to write the integrl s sum of two relted integrls, which cn be evluted by mking suitble substitutions e.g. * * These integrls cn be clled s Algebric Twins. Illustrtion : vlute : sin cos Solution : I sin cos sin cos sin cos (tn ) cos (tn ) sec (tn ) cos (tn ) Now, put tn t sec dt t / t I dt dt t t / t Now, put t /t z dt dz t dz z t / t I tn tn tn / tn tn c z Illustrtion : vlute : 8 NOD6\ :\Dt\0\Kot\J-Advnced\SMP\Mths\Unit#06\ng\0-INDFINIT INTGRATION.p6

9 J-Mthemtics Solution : I I {dividing N r nd D r by } / / / / dt du ( / ) ( / ) ( / ) 7 ( / ) t 7 u where t nd u I t u 7 7. tn. tn c / / 7 7 tn tn c Do yourself -8 : (i) vlute : (ii) vlute : ( d ) Mnipulting integrnds : NOD6\ :\Dt\0\Kot\J-Advnced\SMP\Mths\Unit#06\ng\0-INDFINIT INTGRATION.p6 (i) (ii) (iii), n N, tke n common & put + n t. n ( ) (n) n n, n N, tke n common & put + n t n, tke n common nd put + n tn. n n / n ( ) Illustrtion 6 : vlute : n n / n ( ) Solution : Let I n n / n / n ( ) n n Put n t n, then n n t dt n n n t dt n n t n t n n I t dt c c Do yourself -9 : (i) vlute : (ii) vlute : / ( ) (iii) vlute : / ( ) ( ) 9

10 J-Mthemtics ( e) Integrtion of trigonometric functions : (i) b sin OR b cos OR sin b sin cos c cos Divide N r & D r by cos & put tn t. Illustrtion 7 : vlute : sin Solution : Divide numertor nd denomintor by cos I sec sec tn sec tn Let tn t sec dt So dt t t I. tn c Illustrtion 8 : vlute : ( sin cos ) Solution : Divide numertor nd denomintor by cos I sec ( tn ) Let tn t, sec dt dt c c t t ( tn ) I tn tn 6 c Do yourself -0 : (i) vlute : (ii) vlute : sin sin sin cos (ii) OR b sin OR b cos b sin c cos Convert sines & cosines into their respective tngents of hlf the ngles & put tn t In this cse sin Illustrtion 9 : vlute : Solution : I t t,cos t t sin cos sin cos 0, tn dt t; t tn tn tn tn let tn t, sec dt dt dt so I 6t t t t dt t 6 sec 6 tn tn NOD6\ :\Dt\0\Kot\J-Advnced\SMP\Mths\Unit#06\ng\0-INDFINIT INTGRATION.p6

11 J-Mthemtics t tn. n c n +c t tn Do yourself - : (i) vlute : (ii) vlute : sin sin cos (iii) cos b sin c p cos q sin r press Numertor (N r ) (D r ) + m d (Dr ) + n & proceed. Illustrtion 0 : vlute : cos d sin cos Solution : Write the Numertor (denomintor) + m(d.c. of denomintor) + n + cos (sin + cos + ) + m(cos sin) + n. Compring the coefficients of sin, cos nd constnt terms, we get + n, + m, m 0 6/, m / nd n 8/ Hence I 6 cos sin 8 d d d sin cos sin cos 6 8 n sin cos I where I In I, put tn t sec d dt I dt dt. t tn t t (t ) d sin cos tn / tn NOD6\ :\Dt\0\Kot\J-Advnced\SMP\Mths\Unit#06\ng\0-INDFINIT INTGRATION.p6 Do yourself - : (i) vlute : (iv) 6 8 tn / Hence I n sin cos tn c sin (ii) vlute sin cos m n sin cos Cse-I : When m & n nturl numbers. sin cos cos sin * If one of them is odd, then substitute for the term of even power. * If both re odd, substitute either of the term. * If both re even, use trigonometric identities to convert integrnd into cosines of multiple ngles. Cse-II : m + n is negtive even integer. * In this cse the best substitution is tn t.

12 J-Mthemtics Illustrtion : vlute Solution : sin cos Put cos t; sin dt. so tht Illustrtion : vlute I ( t ).t dt t t cos cos (t t )dt c Alternte : Put sin t; cos dt so tht 7 I t ( t ) dt t t t dt 6 8 sin sin sin c 6 8 Note : This problem cn lso be hndled by successive reduction or by trignometric identities. sin cos Solution : cos cos sin cos cos cos cos 8 8 cos cos cos cos cos 8 cos cos cos cos 6 cos cos cos 8 sin 6 sin sin sin c sin 6 sin sin c sin Illustrtion : vlute 9 / cos Solution : Let I / sin 9 / / 9 / cos sin cos Here m + n 9 (negtive even integer). Divide Numertor & Denomintor by cos. I Do yourself - : (i) vlute : tn sec tn ( t ( t )dt (using tn t) tn ) sec 7 7 / 7 / / 7 / t t c tn tn c sin sin (ii) vlute : / cos (iii) vlute : cos sin cos NOD6\ :\Dt\0\Kot\J-Advnced\SMP\Mths\Unit#06\ng\0-INDFINIT INTGRATION.p6

13 J-Mthemtics ( f ) Integrtion of Irrtionl functions : (i) ( b) p q & ;put p q t ( b c) p q (ii),put b ; ( b) p q r t ( b c) p q r, put t Illustrtion : vlute. ( ) Solution : Let, I. Put + t tdt ( ) (t ) I.(t)dt {(t ) (t ) } t / t du.dt (t / t) ( ) u ( ) t / t dt dt t t t / t where u t t u t tn c tn c tn c t ( ) Illustrtion : vlute ( ) Solution : Let, I ( ) put t /t dt I / t dt / t t t dt t t NOD6\ :\Dt\0\Kot\J-Advnced\SMP\Mths\Unit#06\ng\0-INDFINIT INTGRATION.p6 dt log (t / ) (t / ) / c t / log c Illustrtion 6 : vlute ( ) Solution : Let, I ( ) Put t, So tht dt t I / t dt tdt ( / t ) / t (t ) t gin let, t u. So tht t dt du.

14 J-Mthemtics du (u ) u which reduces to the form we put u z so tht du z dz zdz dz I (z ) z (z ) z I tn c u I tn c t P Q where both P nd Q re liner so tht tn c tn c Do yourself - : (i) vlute ( ) (ii) vlute : Miscelleneous Illustrtions : Illustrtion 7 : vlute cos sin {sin cos } cos Solution : I sin {sin cos } Put + cot t cot cosec dt cos 6 sin { cot } cot cosec ( + cot ) / dt / / t c t / Illustrtion 8 : 6 6 is equl to - cos sin n tn cot + c tn (tn cot) + c Solution : Let I 6 6 cos sin If tn p, then sec dp I Illustrtion 9 : vlute : cot c 6 sec 6 tn ( p ) dp ( p ) dp 6 p p p dk n cot tn + c (D) tn ( cot) + c ( tn ) sec p 6 tn p p p p dp tn (k) c k where p k, dp dk p p tn p c tn (tn cot ) c p tn ( cot) + c Ans. (C,D) sin cos 6 cos sin Solution : I sin cos 6 cos sin ( sin ) cos ( sin ) cos 6 ( sin ) sin sin sin NOD6\ :\Dt\0\Kot\J-Advnced\SMP\Mths\Unit#06\ng\0-INDFINIT INTGRATION.p6

15 J-Mthemtics Put sin t, so tht cos dt. ( t )dt I (t t )... (i) Now, let (t ) (t ) + µ Compring coefficients of like powers of t, we get, + µ... (ii), µ 7 (t ) 7 I dt t t {using (i) nd (ii)} t dt log dt 7 t t t t dt t t 7 t t dt log t t 7 log t t tn (t ) + c (t ) () log sin sin tn (sin ) + c. Illustrtion 0 : The vlue of 6.sin, is equl to - cos 9. cos c cos 9. sin c sin 9. sin c (D) none of these NOD6\ :\Dt\0\Kot\J-Advnced\SMP\Mths\Unit#06\ng\0-INDFINIT INTGRATION.p6 Solution : Here, I.sin 6 Put cos 6sind cos.sin cos cos ( 6 sin )d 6 sin.sin (sin ).( 6 sin )d 6.( sin )d cos 6 Illustrtion : vlute : 6 ( cos )d 6 cos d sin sin. d sin cos 6 c cos 9. cos c tn cos tn tn tn Ans.

16 J-Mthemtics Solution : I I tn cos tn tn tn sec tn tn tn tn tn ( tn ) ( tn ) cos tn tn tn let, y tn y dy sec. sec tn tn ydy dy I (y ).y y tn y + c tn tn c tn ANSWRS FOR DO YOURSLF : (i) tn 6 c (ii) sin sin c : (i) sin c (ii) n c : (i) e e + c (ii) cos( ) sin( ) c e sin( ) c : (i) e tn + c (ii) : (i) tn(e ) + c (ii) n + c 6 : (i) 7 : (i) 8 : (i) 9 : (i) 7 n n c (ii) tn c tn n c c tn tn c 0 : (i) : (i) : (i) : (i) : (i) tn / tn c n sin cos c tn c n c n c (ii) (ii) (ii) (ii) (ii) (ii) (ii) (ii) 6 n c tn 6 n / c 8 tn tn c 6 tn / n 6 6 tn / (iii) c n cos sin C tn / c c (iii) / c 7 sin sin sin c. 7 NOD6\ :\Dt\0\Kot\J-Advnced\SMP\Mths\Unit#06\ng\0-INDFINIT INTGRATION.p6

17 J-Mthemtics XRCIS - 0 CHCK YOUR GRASP SLCT TH CORRCT ALTRNATIV (ONLY ON CORRCT ANSWR) sin sin. If f(), 0 then lim f'() is equl to (D) / sin cos cos is equl to - cos cos cos c cos cos cos c (D) 8 is equl to - 7 cos cos cos c cos cos cos c 6 (8 + ) 7 + c (8 + ) c 6 (8 + 9) 7 + c (D) (8 + ) c. 8 8 cos sin equls - sin cos sin c sin c cos c (D) cos c. Primitive of w.r.t. is - c c c (D) c NOD6\ :\Dt\0\Kot\J-Advnced\SMP\Mths\Unit#06\ng\0-INDFINIT INTGRATION.p (...) ( < ) - ( + ) + c ( ) + c ( + ) + c (D) none of these is equl to - n c l c 7 c (D) none of these n n equls - n ( l n ) c n ( ln ) c n ( l n ) c (D) n (n ) c

18 J-Mthemtics 9. If B A n + c, where c is the constnt of integrtion then : A ; B A ; B A ; B (D) A ; B 0. / equls - c c c (D) none of these. sin.cos.cos.cos.cos 8.cos6 equls - sin6 c 0 cos c 0. Identify the correct epression. n n c n e c e e c (D) n. equls - cos c 096 (D) cos c 096 tn c n c. n c. n c (D) n c. If ( ) ( ) n( + ) + btn , b equls - c z equls - 6 0, b c 8 n + + C then- 0, b c tn c cot c cot c (D) cot c (D) 0, b (D) None of these NOD6\ :\Dt\0\Kot\J-Advnced\SMP\Mths\Unit#06\ng\0-INDFINIT INTGRATION.p6

19 J-Mthemtics 7. equls- c c c (D) c ( ) is equl to - 9. If + c 0 ( + ) +c tn b tn cos + c, then- (D) + c 0 ( + ) + c, b, b, b (D), b SLCT TH CORRCT ALTRNATIVS (ON OR MOR THAN ON CORRCT ANSWRS) 0. Primitive of tn sec tn w.r.t. is - n sec n sec tn c n sec tn n sec c n sec tn c (D) n tn (sec tn ) c NOD6\ :\Dt\0\Kot\J-Advnced\SMP\Mths\Unit#06\ng\0-INDFINIT INTGRATION.p6. sin equls -. cos c equls sin c n n c cos n n( ) n c (D) e cos e (A sin B cos ) c c (D) cos c n n c n. If, then - A B A B A B (D) A + B ANSWR KY CHCK YOUR GRASP X R CI S - Que B B A B B B B A C A Que B C A C D A A D B A,B,D Que. A,B,C B,C,D C, D c 9

20 J-Mthemtics XRCIS - 0 BRAIN TASRS SLCT TH CORRCT ALTRNATIVS (ON OR MOR THAN ON CORRCT ANSWRS). z b cot tn cos sin g equls - sec sin cos c sec sin cos c n sin cos sin c (D) n sin cos sin c. z sin sin 6 sin sin 7 equls - sin sin sin 6 sin+c sin+c cos+c (D) cos+c.. 7 equls - 7 ( ) 7 7 n n c 7 7 n n c (D) is equl to - 7 n n c 7 n n c ( + ) + c ( ) + c ( + ) / + c (D) ( ) + c sin ( n ) is equl to - 0 ( + sin(n) + cos(n)) + c ( + sin(n) cos(n)) + c 0 0 ( sin(n) cos(n)) + c (D) ( sin(n) + cos(n)) + c 0 cos.cosec is equl to - cot + tn + c cot tn + c cot tn + c (D) tn cot + c e e. c, equls- e c 6 0 e c 6 (D) e. c NOD6\ :\Dt\0\Kot\J-Advnced\SMP\Mths\Unit#06\ng\0-INDFINIT INTGRATION.p6

21 J-Mthemtics 8. tn e ( ).d(cot ) is equl to - e tn + c e tn + c. e tn + c (D). e tn + c 9. n n ( n. ) e is equl to - n n ( ) e n n n + c e n n + c e n n + c (D) e n + c 0. e ( ) e is equl to - e. e c e c e. e c (D) e. e c. Primitive of w.r.t. is - c c c (D) c. / equls - [ ( )] / c 0 / c / c (D) / c. sin is equl to - sin NOD6\ :\Dt\0\Kot\J-Advnced\SMP\Mths\Unit#06\ng\0-INDFINIT INTGRATION.p6 sin n sin sin n sin. The vlue of integrl. If tn tn c + sin n 8 sin + sin n 8 sin + c + c (D) sin n sin sin n sin sin n 8 sin sin n 8 sin d cn be epressed s irrtionl function of tn s - cos sin tn tn c tn tn c (D) sin cos + bn[sin + cos + c, then - cos sin, b 9 7, b 6 tn tn c, b 9 + c + c (D) 7, b 9

22 J-Mthemtics 6. is equl to - n tn + c n tn + c n sec + c (D) n sec + c 7. ( ) is equl to - ( ) + c ( ) + c ( ) + c (D) ( ) + c 8. Let f'().sin cos, 0, f(0) 0, f 0, then which of the following is/re not correct. f() is continuous t 0 f() is non-differentible t 0 f'() is discontinuous t 0 (D) f'() is differentible t 0 9. n equls - n c n c n c (D) n c 0. equls, where, - sin c sin ( ) c c cos ( ) (D) sin. is equl to - sin cos cos c cot cot c cot tn c ANSWR KY tn tn c (D) tn cos c BRAIN TASRS X R CI S - Que A B C D C C C C B C Que B B D C C D A B,C,D B, D A,B,C,D Que. A,B,C,D NOD6\ :\Dt\0\Kot\J-Advnced\SMP\Mths\Unit#06\ng\0-INDFINIT INTGRATION.p6

23 J-Mthemtics XRCIS - 0 MISCLLANOUS TYP QUSTIONS FILL IN TH BLANKS. If e 6e A + B log(9e ) + C, then A..., B... nd C... 9e e. If the grph of the ntiderivtive F() of f() log(log) + (log) psses through (e, 998 e) then the term independent of in F() is.... Let F() be the ntiderivtive of f() cos sin whose grph psses through the point (/, ). Then F(/).... Let f be function stisfying f"() /, f'() nd f(0) 0. Then f(78) is equl to... MATCH TH COLUMN Following question contins sttements given in two columns, which hve to be mtched. The sttements in Column-I re lbelled s A, B, C nd D while the sttements in Column-II re lbelled s p, q, r nd s. Any given sttement in Column-I cn hve correct mtching with ON sttement in Column-II.. The ntiderivtive of Column-I Column-II f() ( b ) ( b ) cos f() sin b cos f() cos b sin is is (p) b tn tn b + c tn is (q) sin tn sin + c, b cos (r) tn b tn b + c (D) f() b cos is ; ( > b ) (s) b log tn tn b + c. f() when Column-I Column-II NOD6\ :\Dt\0\Kot\J-Advnced\SMP\Mths\Unit#06\ng\0-INDFINIT INTGRATION.p6 f() / ( ) f() f() / ( ) (D) f() ASSRTION & RASON (p) c sin (q) (r) c (s) sin + c In ech of the following questions, sttement of Assertion is given followed by corresponding sttement of Reson (R) just below it. Of the sttements mrk the correct nswer s + c Sttement-I is True, Sttement-II is True ; Sttement-II is correct eplntion for Sttement-I Sttement-I is True, Sttement-II is True ; Sttement-II is NOT correct eplntion for Sttement-I Sttement-I is True, Sttement-II is Flse. (D) Sttement-I is Flse, Sttement-II is True.

24 J-Mthemtics. If D() Sttement - I : f () f () f () b c b c D(), where f, f, f re differentible function nd, b, c,, b, c re constnts. f () f () f () b c b c B e c u s e Sttement - II : Integrtion of sum of severl function is equl to sum of integrtion of individul functions. A B C (D) D. Sttement - I : If > 0 nd b c < 0, then the vlues of integrl b c will be of the type A µ tn + c. where A, B, C, µ re constnts. B B e c u s e Sttement - II : If > 0, b c < 0, then + b + c cn be written s sum of two squres. A B C (D) D. If y is function of such tht y( y). Sttement - I : y log[( y) ] B e c u s e Sttement - II : y log( y) + c. A B C (D) D COMPRHNSION BASD QUSTIONS Comprehension # In clculting number of integrls we hd to use the method of integrtion by prts severl times in succession. The result could be obtined more rpidly nd in more concise form by using the so-clled generlized formul for integrtion by prts u() v() u() v () u'()v () + u"() v ()... + ( )n u n () v n () ( ) n un () v n () + c where v () v(), v () v ()..., v () n v () n Of course, we ssume tht ll derivtives nd integrls ppering in this formul eist. The use of the generlized formul for integrtion by prts is especilly useful when clculting P n () Q(), where P n (), is polynomil of degree n nd the fctor Q() is such tht it cn be integrted successively n + times.. If ( + ) cos. If sin u() + cos 8 v() + c, then - u() + u() + v() + (D) v() 6 8 e e. f() C then f() is equl to (D) + + NOD6\ :\Dt\0\Kot\J-Advnced\SMP\Mths\Unit#06\ng\0-INDFINIT INTGRATION.p6

25 J-Mthemtics Comprehension # Integrls of clss of functions following definite pttern cn be found by the method of reduction nd recursion. Reduction formuls mke it possible to reduce n integrl dependent on the inde n > 0, clled the order of the integrl, to n integrl of the sme type with smller inde. Integrtion by prts helps us to derive reduction formuls. (Add constnt in ech question). If I n then I n n+ + n ( ) n ( ) n n. I n is equl to - ( ) n n. n ( ) (D) n. ( ). If I n, m n sin m then I n, m + n cos m I n, m is equl to- sin cos n m (m ) sin cos n m (n ) n sin m (D) n cos m sin cos n m. If u n n, then (n + )u n+ + (n + )bu n + ncu n is equl to - b c n b c n b c n b c (D) n b c NOD6\ :\Dt\0\Kot\J-Advnced\SMP\Mths\Unit#06\ng\0-INDFINIT INTGRATION.p6 MISCLLANOUS TYP QUSTION ANSWR KY X R CI S - Fill in the Blnks.,, ny rel vlue Mtch the Column. p; r; s; (D) q. s; q,; r; (D) p Assertion & Reson... Comprehension Bsed Questions Comprehensi on # :.. Comprehension # :... (D)

26 J-Mthemtics XRCIS - 0 [A] CONCPTUAL SUBJCTIV XRCIS v l u t e t h e f o l l o wing I ndefi ni t e i ntegrls :... sin( ) sin( b). Integrte n( ) n f '() w.r.t. cos ec cot sec. 7. cos ec cot sec cos sin cos. n d 0. cos sin ( ) ( ) [J 99].. sin sin cos tn. tn. tn.tn ( ), wher e f( ) tn n n ( b) 8. c ( b) ( sin cos ) e n e. n ( ) /. / ( ) / / (sin ) (cos ) tn co t [J 89] / (cos ) [J 87]. [J 9] sin [J 8] ( ) cos sin 7 9 sin CONCPTUAL SUBJCTIV XRCIS ANSWR KY X R C IS - ( A ). sin( b) cos ec(b ). n c. c sin ( ).. 9. n(sec ) n(sec ) n(sec ) + c. ( ). n 9 n c 6. sin sec c 7. 6 sin b c k (sin)n cos sin cos sin n(sec) + c 0. e c e. tn n( ) n( ) c. C ( ). tn n c tn 6. ( tn ) c 8 / 7. 8 (tn ) 8. sin cos c sin cos. rc sec 8. n(cos + sin) (sin + cos) 9. tn cot tn 0.. log log(cot + cot ) + c tn tn. n / c ( sin cos ) n c ( sin cos ) / 7 7/ + / / + / + 6 /6 / + log / + / + c + c + c NOD6\ :\Dt\0\Kot\J-Advnced\SMP\Mths\Unit#06\ng\0-INDFINIT INTGRATION.p6

27 J-Mthemtics XRCIS - 0 [B] BRAIN STORMING SUBJCTIV XRCIS. cos 8 cos 7. cos. sin( ). sin( ) cot ( sin )(sec ). 6. n cos 8. cos 9. sin sec cos ec 7. e ( ) sin sin( ) Let 6 0, R nd f() is differentible function stisfying, f(y) f() + (y ) + (y ) ;, y R nd f(). vlute f(). cot tn. sin sin f(). Let f() is qudrtic function such tht f(0) nd ( ) is rtionl function, find the vlue of f'(0). cos e ( sin cos ). / (7 0 ) sin NOD6\ :\Dt\0\Kot\J-Advnced\SMP\Mths\Unit#06\ng\0-INDFINIT INTGRATION.p6 BRAIN STORMING SUBJCTIV XRCIS ANSWR KY X R C I S - ( B ). ( sin sin ) c 6 /. cos.rc cos sin. n sin sin sin c. cos cos.. rc cos c sin n cot cot cot cot cot c cos 8. sin 0. nf H I K F I HG KJ. ( + ) rc tn cot. n e cos cos c 9. tn c. + c.. C e cos ( + cosec). c n tn sec tn c / sin cos n tn c 8 e c sin tn c sin cos (7 0) c 7

28 J-Mthemtics XRCIS - 0 [A] J-[MAIN] : PRVIOUS YAR QUSTIONS. cos [AI-00] cos () tn + C () + tn + C () tn + C () cot + C (log ). [AI - 00 ] () (log + ) + C () (log + ) + C () (log ) + C () log( + ) + C. If sin A + B logsin( ) + C then vlues of (A, B) is - [AI - 00 ] sin( ) () (sin, cos) () (cos, sin) () ( sin, cos) () ( cos, sin). is equl to- [AI - 00 ] cos sin () log tn 8 + C () log cot + C () log tn 8 + C () log tn 8 + C. (log ) is equls to - [AI - 00 ] (log ) log () (log ) + C () + C () e + C () (log ) + C 6. cos sin equls- [AI ] () logtn + C () logtn + C () logtn + C () logtn 7. The vlue of 8 + C sin is - [AI ] sin () + log cos +c () log sin +c () + log sin +c () log cos 8. If the integrl tn + ln sin cos + k then is equl to : [AI-0] tn () () () () + c NOD6\ :\Dt\0\Kot\J-Advnced\SMP\Mths\Unit#06\ng\0-INDFINIT INTGRATION.p6

29 J-Mthemtics 9. If ƒ() (), then ƒ( ) is equl to : [J (Min)-0] () ( ) ( ) C () ( ) ( ) C () ( ) ( ) C () ( ) ( ) C NOD6\ :\Dt\0\Kot\J-Advnced\SMP\Mths\Unit#06\ng\0-INDFINIT INTGRATION.p6 PRVIOUS YARS QUSTIONS ANSWR KY Q u e Ans XRCIS- [A] 9

30 J-Mthemtics XRCIS - 0 [B] J-[ADVANCD] : PRVIOUS YAR QUSTIONS PRVIOUS YARS QUSTIONS. vlute : sin. [J 00 (Mins) M out of 00] 8 m m m m m m. For ny nturl number m, evlute ( + + )( + + 6) where > 0. ANSWR KY m m m m 6 m. ( ). ( ) tn n( 8 ) c 6 (m ). D. A. D 6. C 7. C 0 [J 00 M out of 60] z is equl to - [J 006, (M, M) out of 8]. Let f() c c c (D) c n / n ( ) n ( n ) n K n(n ) for n nd g() (f f... f) f occurs n times ( n n ) n K n (). Then C n g() equls. [J 007, M] n ( n ) n K n(n ) (D) ( n n ) n K n. Let F() be n indefinite integrl of sin. [J 007, M] Sttement- : The function F() stisfies F( + ) F() for ll rel. b e c u s e Sttement- : sin ( + ) sin for ll rel. Sttement- is True, Sttement- is True ; Sttement- is correct eplntion for Sttement-. Sttement- is True, Sttement- is True ; Sttement- is NOT correct eplntion for Sttement-. Sttement- is True, Sttement- is Flse. (D) Sttement- is Flse, Sttement- is True. 6. Let I e, J e e e e e Then, for n rbitrry constnt c, the vlue of J I equls e e log e e e e log e e 7. The integrl c c sec (sec tn ) 9 / / sec tn K sec tn 7 / sec tn K sec tn 7. [J 008, M, M] (D) e e log e e e e log e e equls (for some rbitrry constnt K) c c [J 0, M, M] / sec tn K sec tn 7 (D) / sec tn K sec tn 7 XRCIS- [B] NOD6\ :\Dt\0\Kot\J-Advnced\SMP\Mths\Unit#06\ng\0-INDFINIT INTGRATION.p6