INDEFINITE INTEGRAL5 DR.HEMA REDDY. Mathematics
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1 INDEFINITE INTEGRAL5 DR.HEMA REDDY
2 METHODS OF INTEGRATION 1)Integration by substitution )Integration by partial fractions 3)Integration by parts
3 Integrals of type a a d b p q b d c c d a cos bsin c a sin csin a a bcos d d cos d b c d bsin p q a b a d d c d bcos a sin bcos c
4 dy If = , then y= d c log a c c 4 log a c Option (3) is correct choice
5 d cos d d 1) sin C Option () is the choice
6 log(1 ) C 1. C (1 ) 3 4 d 3. 1 C C 1 Option (1) is correct
7 If f () d logsin log logsin, then f() is equal to 1)sin )cos 3)log sin 4)cot
8 If f () d logsin 1)sin )cos 3)log sin 4)cot Differentiating both sides, we get f()= cot log logsin f () cot logsin logsin then f()= Option (4) is correct
9 If the primitive o f log[f()] 1 f () +c, then f() is is equal to 1) +d )/+d 3) / +d 4) +d Given 1 d log[f ()] c f () Differentiating 1 1 f ()f '() f () [f ()] 1 f '() f () d Choice () is correct
10 If f ()= 3 cos f() and f(0)=1, then f() is equal to 1) e 3sin ) e 3 cos 3) cos e 3 sin 4) e 3tan
11 If f ()= 3 cos f() and f(0)=1, then f() is equal to 1) e 3sin ) e 3 cos 3) cos e 3 sin 4) e 3tan Ans: f ()=3 cos f() f '() 3cos f () On integrating we have log f() = 3 sin +c Since f(0)=1 c=0 Log f() =3sin f()=e 3sin
12 The value of 1) (ae) +c (ae) log(ae) ) +C 3)e + a +c loga e e d is 4) e log(ae)
13 The value of e loga e d (ae) 1) (ae) ) log(ae) +c 3)e + a +c 4) +c is ae log(ae) Ans: G.E= e loga e d = a e d (ae) log(ae) = (ae) d = +c option () is correct
14 If f()= interval 1)(1,) )(-,-1) 3)(-, -) 4)(, ) e ( 1)( )d decreases in the
15 e ( 1)( )d If f()= decreases in the interval a)(1,) b)(-,-1) c)(-, -) d)(, ) A function y= f() decreases if f ()<0 Here If f() decreases then f ()<0 Hence f ()=e (-1)(-)<0 1
16 If 0 f () 0 0 then f()d= 1. ++c. 3 /3 ++c C Determinant of skew symmmetric matri of odd order is zero Option 4 is correct choice
17 If tan 4 d =Ktan 3 + L tan+ f() then 1)K=1/3 L= -1, f()=+c )K=-1, L=1, f()=+c 3)K=1, L=-1, f()= -+c 4)K=1/3, L=1/3, f()=3+c
18 If tan 4 d =Ktan 3 + L tan+ f() then 1)K=1/3 L= -1, f()=+c )K=-1, L=1, f()=+c 3)K=1, L=-1, f()= -+c 4)K=1/3, L=1/3, f()=3+c Sol: tan d 4 (sec 3 tan tan tan d 1) tan d [tan sec tan ]d (tan ) sec d (sec 1)d tan C 3 1 K,L 1,f () C 3 Option (1) is corrrect choice
19 (1 cos cos... )d = 1. tan (/)+C. cot(/)+c 3. cot(/)+c 4. tan(/)+c
20 (1 cos cos... )d = 1.tan (/)+C. cot(/)+c 3. cot(/)+c 4. tan(/)+c Sol: 1 G.I d( it is a G.P) 1 cos cot d 1 cosec sin C Choice (3) is correct
21 d 1 e is 1. log (1+e )+c. log(e )+c 3. log(1+e - )+c 4. -log(1+e - )+C
22 Sol: d 1 e is 1) log (1+e )+c )log(e )+c 3)log(1+e - )+c 4)-log(1+e - )+C d I 1 e Multiplying N and D bye e d I r r e 1 Put e 1 t e d dt dt So I t log t C log(1 e ) C Option (4) is correct
23 011 (1 log )d is
24 011 (1 log )d is 011 Put t 011 log log t 1 1 (011 log 011)d dt t 1 011(1 log ) dt t 1 1 G.I t dt 011 t 1 1 dt t C Option () is correct
25 If cos -1 =I, 1) - ) - 3) 4) I 4 4 I 4 I 1 1 sin cos 1 1 d sin cos Sol : cos cos cos d I 4 I d Option () is correct choice
26 cot d sin cos Let F() = and F() is its antiderivative if f(/4)=6, then F() is equal to 1. tan 3. cot 8 3. cot 8 4. cot 8 G.I Option (3) is correct choice cot cot cosec d cot cot C sin ce F( ) 6 4 cosec d C 6 C 8
27 4 4 d tan 4 1. tan 1 3. tan 1 4. tan 4 C C C C
28 4 d tan C 4. tan tan 1 C C tan C 4 1 G.I d 4 1 I d 4 I dt t t tan C tan 1 1 C Option () is correct
29 d 9 8 =fog() +C 1) ) 3) 4) 1 5 f () tan, g() f () sin g() f (), g() sin 1 4 f () sin, g() 5
30 a b b c a[ a b c a[( ) a a a( 1 coeff of c ] a b ( ) a ) ] cons coeff tant of 1 [ coeffof ] We can use direct formula 9 8 [ 8 9] {[ ( 4)] 9 ( 4) } 5 ( 4)
31 d 9 8 =fog() 1) ) f () sin g() 5 5 3) 1 4 4) f () sin, g() a b c f () tan, g() f (), g() sin ( 1 cons tant 1 a coeff of ) [ coeffof ] coeff of 9 8 [ 8 9] {[ ( 4)] 9 ( 4) } 5 ( 4) d d ( 4) 1 4 sin C 5 Option (4) is correct choice
32 If d is l log(4 +4+5)+m tan -1 (+1/)+C, then l and m are 1)3/, ½ )1/, 3/ 3)3/, 1/8 4)1/, 1/4
33 d This is integral of the form For integral of the forms p q a b p q a b d c d c Here we epress Numerator = A(derivative of a Where A and B are constants. A and B are determined by equating coefficients of, and constant terms a p q b +b+c)+b coeff of p pb A, B q coeff of a a d c
34 If d is l log( +4+5)+m tan -1 (+1/)+C, then l and m are 1)3/8, ½ )1/, 3/ 3)3/, 1/8 4)1/, 1/4 Sol: 3+=l(derivative of )+m 3+=l(8+4)+m 8l=3, 4l+m=l= 3/8, m=1/ coeff of p pb A, B q coeff of a a Option (1) is correct choice
35 If d 5 4cos 1 K tan (M tan ) C then 1. K=1/, M=/3. K=1/3, M=/3 3. K=/3, M=1/3 4. K=1/3, M=3/
36 d 1 K tan (M tan ) C If 5 4cos then 1. K=1/, M=/3. K=1/3, M=/3 3. K=/3, M=1/3 4. K=1/3, M=3/ Sol: Putting tan(/)=td= Cos= 1 t 1 t dt 1 t I 1t t t 1 t tan C K and M 3 3 dt dt 1 t Option (3) is correct choice
37 If d tan a b 1 sin the following is correct 1. a, b. a, b 3. a, b is arbitrary cons tan t 4 4. a, b is arbitrary cons tan t then which one of
38 If d tan a b 1 sin the following is correct then which one of 1. a, b. a, b 3. a, b is arbitrary cons tan t 4 4. a, b is arbitrary cons tan t Option (3) is correct choice
39 d 1 sin cos e e e e (1) log 1 tan c] ().log 1 tan C 1 (3) log 1 tan c] (4) log 1 tan C
40 d 1sin cos (1) loge 1 tan c] ().log e 1 tan C 1 (3) loge 1 tan c] (4) loge 1 tan C dt G.I 4t 1 t 1 t dt 1 log e (t 1) C t 1 1 log e (1 tan ) C Option (3) is correct choice dt tan t; d 1 t t 1 t sin,cos 1t 1t
41 d a sin a b 1 1) tan ( tan ) C 1 ab b cos 1 ) tan (tan ) C 1 a ab b 1 3) tan ( tan ) C 1 b ab a 1 4) tan ( tan ) C is
42 a sin d b cos a 1 b ab Option (c) is correct. is 1 1 a) tan ( tan ) C b) tan (tan ) C 1 a 1 b ab b ab a 1 1 c) tan ( tan ) C d) tan ( tan ) C r r Dividing N and D by cos I 1 sec b a tan 1 b a a b b 1 dt 1 ab t 1 ab 1 a ab b d Put tan t sec d dt 1 I tan t c 1 tan tan c
43 The result can be used as formula d 1 a a sin b cos ab b 1 tan tan C d 1 3 9sin 4cos 6 1 tan tan C
44 The value of d 3 sin cos 3 1)log e tan +c log tan 3 ) +c 3) tan tan 1 +c ) log tan +c
45 We can also evaluate integrals of the form a sin d bcos These types of integrals can also be evaluated by substituting tan / =t and applying proper method. But This can also be evaluated by substituting a=r cos, b=r sin and so r= and =tan -1 b a b a
46 d 1 d a sin bcos r sin( ) 1 1 b cosec( )d r a b tan 1 r a 1 log tan r 1 1 b log tan tan c 1 a b a
47 d 1 1 b a b 1 log tan tan c a sin bcos a d 1 log tan sin cos 8 d log tan tan 3 sin cos 3 1 log tan 1
48 The value of 1)log e tan 3 +c ) +c 1 3) +c 4) 1 +c By using tan tan 1 d 3 sin cos log tan 3 log tan 1 d 1 1 b a b 1 log tan tan c a sin bcos a d log tan tan 3 sin cos 3 1 log tan 1 Option (4) is correct choice
49 If cos 3sin d 4sin 5cos then A and B are. This is integral of the form =A+B log(4sin-5cos)+c, ,., , 4., a cos bsin d ccos dsin a cos bsin ac bd ad bc d log(ccos dsin ) C ccos dsin c d c d a cos bsin d A Blog(ccos dsin ) C ccos dsin cos 3sin cos 3sin d d 4sin 5cos 5cos 4sin A ;B Option (3) is correct choice
50 If cos 3sin d 4sin 5cos =A+B log(4sin-5cos)+c, then A and B are , , , , 41 41
51 sin cos d cos sin 1. log(sin cos ) 1. log(sin cos ) 3. log(sin cos ) 1 4. log sin cos G.I cos sin d sin cos log(sin cos ) 1 log sin cos Option (4) is the choice
52 If ( 1) 1 d tan g() K ( 1) then g() is equal to 1) ) 1 ( 1) 1 1 3) tan -1 (/) 4) 1 1
53 If equal to 1) ) Here ( 1) 1 d tan g() K ( 1) ( 1) 3) tan -1 (/) 4) Option () is correct choice 1 then g() is ( 1) 1 1 ( 1) ( 1) ( 1) ( 1) 1 ( 1) ( 1) ( 1) 1 1 so d tan C ( 1) 1 1 Therefore g() 1
54 ( 1)d ( )( 3) is ( 3) 1). log C ( ) ).log ( 3) ( ) C ( ) 3) log C ( 3) 4) log ( )( 3) C
55 ( 1)d is ( )( 3) ( 3) ( 3) 1). log C ).log C ( ) ( ) ( ) 3) log C 4) log ( )( 3) C ( 3) 1 A B ( )( 3) 3 Putting, we get A 1 Putting 3, we get B 1 1 ( )( 3) 3 log( ) log( 3) ( 3) log C Option (1) is correct
56 If sin d f () C cos (1 cos ) 1 cos 1)log cos cos )log 1 cos sin 3) log 1 sin 1 sin 4) log sin Put cos t sin d dt dt 1 1 t(1 t) t t 1 t1 log C t Option (1) is correct
57 1) d 4 ( 1) 1 log( 4 1) C 4 is equal to ) 4 1 ( 1) log 4 4 C 3) 1 log C 4) 4 1 ( 1) log 4 C
58 d log( 1) C ( 1) Sol: G.E= = 1) 4 ) 4 3) 1 4) d ( 1) log dt 4 t(t 1) dt 4 t t 1 1 [log t log(t 1)] log C C 4 1 ( 1) log ( 1) log 4 In general d 1 log n n C C n n ( 1) 1 c
59 If nn, and I n = (log) n d then I n + n I n = 1. (log) n-1. (log) n 3. (log) n 4. X (log) n-1 In n (log ) d n (log ).1 d In (log ). n(log ).. d n n1 1 n In.(log ) nin 1 n I ni (log ) n n1 Option (3) is correct choice
60 If d [f ()] d = cos+sin and f(0)=, then f() is equal to 1) sin ) cos+sin+ 3) sin + 4) cos +
61 d If [f ()] d equal to = cos+sin and f(0)=, then f() is 1) sin ) cos+sin+ 3) sin + 4) cos + Sol: Integrating on both sides f()= ( cos sin )d = sin +C Now f(0)= gives c= Therefore f()= sin+ Option (c) is correct choice.
62 If an antiderivative of f() is e of g() is cos, then f(). Cos d + g(). e d= 1)f() g() )f()+g() 3)e cos 4)e cos +f(). g() and that
63 If an antiderivative of f() is e that of g() is cos, then f(). Cos d + g(). e d= 1)f() g() )f()+g() 3)e cos 4)e cos +f(). g() and Sol: f(). Cos d + g(). e d = cos f()d- d/d(cos) f()+ g(). e d =cos e - g() e + g(). e d = e cos Option (3) is correct
64 If g()d =g(), then [f()+f ()] g() d is equal to 1)f() g() ) f (). g () 3)[f()-f ()] g() 4)f(). g ()
65 If g()d =g(), then [f()+f ()] g() d is equal to 1)f() g() ) f (). g () 3)[f()-f ()] g() 4)f(). g () Here I= [f()g()+f ()g()] =I 1 + I I 1 = f()g()=f() g()- g() f () d =f()g()-i So I 1 + I = f() g() Option (1) is correct =f() g()- g() f ()d
66 If 1 sin f () log 1 C, 1 then f()= 1. tan tan tan tan sin tan d 1 1 tan. d tan log 1 C f()= tan -1 Option 1 is correct
67 1) sin d cos = sin sin c 4 ) 3) 4) sin sin cos c 4 cos sin cos c 1 cos sin cos c 4
68 Integrating by Parts A Shortcut: Tabular Integration Tabular integration works for integrals of the form: f g d where: Differentiates to zero in several steps. Integrates repeatedly. Also called tic-tac-toe method
69 Integrating by Parts e d f & deriv. g & integrals e d 0 e e e e e e e C Compare this with the same problem done the other way:
70 Integrating by Parts 3 sin d sin cos sin cos sin 3 cos 3 sin 6cos 6sin + C
71 1) sin d cos = sin sin c 4 ) 3) 4) sin sin cos c 4 cos sin cos c 1 cos sin cos c 4
72 Integrating by Parts sin 0 sin cos 1 sin 1 cos 4 1 cos sin cos c 4
73 The integral to 1) e tan 1 1 c e tan d is equal ) e tan 1 1 c 3) e tan 1 c 4) e tan 1 C
74 The integral e tan is equal to 1 tan 1) ) 3) e c 4) Sol: Put tan -1 =, we have G.I= 1 1 c e tan option (3) tan e c 1 1 e (1 tan tan )d 1 d d d 1 e (tan sec )d 1tan sec e tan 1 C Usin g e [f () f '()]d e f () C tan 1 e tan c e c tan tan
75 e 1 tan sec d is equal to 1)-e - tan+c )-e - sec +c )e - tan +c 4)e - sec +c
76 e 1 tan sec d is equal to 1)-e - tan+c )-e - sec +c 3)e - tan +c 4)e - sec +c Sol: put -= t d= -dt G.I= t e (sec t sec t.tan t)dt t d e {sec t (sec t)}dt dt t e sec t c e sec c option () is correct
77 1 1 log (log ) log 1.. log 3. log 4. log d put log t e t d e dt 1 1 t t t G.I e dt t e C c t log Option () is correct t
78 If sin d sin( ) then value of (A, B) is 1. (-sin, cos ). (cos, sin ) 3. (sin, cos ) 4. (-cos, sin) A+B log sin(-)+c Put t t, d dt sin(t ) G.I dt sin t sin t.cos cos t.sin sin t cos 1d sin cot td cos sin log sin t C Option (b) is correct
79 4 3 d is
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