Definite Solut

8/14/2019 Definite Solut

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Definite - Integration

UNIT - 5

S O L U T I O N S

content

* Level - 1

* Level - 2

* Level - 3

e-Learning Resources

w w w . m a t h i i t . i n


2/46

1. ( )2

13

1 1 12 log log 2

0 0 00

1.

3 3

x x xc e dx e dx x dx

= = = =

2. ( ) ( ) / 4 / 4

2 2

0 0tan sec 1a xdx x dx

=

[ ] [ ] /4 /4 /4 /42

0 00 0sec 1 tan 1 .

4 xdx dx x x

= = =

3. ( )c / 2 / 2 / 2 / 2

2

0 0 0 02

sin sin 1sec tan .

1 cos 2 2 22cos

2

x x x x x xdx dx x dx dx

xx

+ += = +

+

/ 2

0

tan tan .2 2 4 2

xx

= = =

4. ( )/ 2

0sinxb Let I e x dx

= / 2/ 2

0 0cos cos

x xe x e x dx

= + / 2 / 2 / 2

0 0 0cos sin sin x x x

e x e x e x dx

= + ( ) ( )

/ 2/ 2

02 sin cos 1

x I e x x e

= = +

Hence ( )/ 2

/ 2

0

1sin 1 .

2

xe x dx e

= +

5. ( )2 2

2

211

1 1 1.

2

x x ec e dx e e x x x

= =

6. ( )a Put sin cos , x t x dx dt = =

so that reduced integral is

( ) ( )1 1

00

1 1log 1 log 2

1 2dt t t

t t

= + + + + 2 1 4

log log log .3 2 3

= =

7. ( )( )

/ 2

5 / 2/ 3

1 cos 1 cos

1 cos1 cos

x xb I dx

xx

+ =

( )/ 2

3/ 3

sin

1 cos

xdx

x

=

Now put 1 cos x t = Also, when1

,3 2

x t

= = and1

, 12

x t= =

Therefore,

121

31/ 21/ 2

3.

2 2

dt tI

t

= = =

Trivandrum: 0471-2438271 Kochi: 0484-2370094


S O L U T I O N S

LEV EL - 1 (Fundamentals of Definite Integration)


3/46


8. ( )d Put2

1 1,t dt dx

x x= = then it reduces to

1/2 1/ 21/ 2 1

11

1.t t

ee dt e e e

e

= = = 9. (a) Put 2tan sec x dx d = =

As 1 0 04

x and x

= = = = , then

( ) / 4 / 4

/42

0

0 0

2 sec 2 tan 2 tan I d d

= =

10. ( ) ( )2

4 22

3

22

4 .4 2

ax bxc ax bx c dx cx c

+ + = + + =

Hence depends on c.

11. ( ) [ ]

/ 4 / 4

/ 6/ 6

1cos 2 tan2d ec x dx log x

=

1 1

tan tan log 3.2 4 6 2

log log = =

12. ( ) ( )1 1

log log log logb bb

aa ac Let I xdx x x xdx

x x= =

( ) ( ) ( )2 2 21

2 log log log2

b

a I x I b a = =

( )( ) ( )1 1log log log log log log2 2

bb a b a aba

= + = .

13. ( )a Put 2tan sec x dx d = = Also as 0, 0x = = and 1,4

x

= =

Therefore1 / 4

1 2

0 0tan sec xdx d

= 1

log 2 log 2.4 4 2

= =

14. ( )( )

1

20

dxd Let I

a b x b

= +

Put ( ) ( )t a b x b dt a b dx= + =

As 1 0 , x t a and x t b= = = = then

( )21 1 1 1 1 1

aa

bb

a b I dt

a b t a b t a b ab ab

= = = =

15. ( )/ 2

2/ 4

1cos

sina Let d

. Put sin cos ,t dt d = = then we have1

1

21/ 21/ 2

1 1

2 1.dtt t

= =


4/46


16. ( )( )

11 2

3/ 20 2

sin

1

xb I dx

x

=

Put 1 21

sin1

x t dx dt x

= =

and sinx t=

Also 04

t to as

= 102

x to= / 4

2

0

1.sec log 2.

4 2 I t t dt

= =

17. ( )/2

0 2 cos

dxc I

x

=+

/2

0 2 2 2 22 sin 2 cos cos sin

2 2 2 2

dx

x x x x

=+ +

2

/ 2 / 2

0 02 2 2

sec2

sin 3cos 3 tan2 2 2

xdx

dx x x x

= =

+ +

Put 21tan sec ,2 2 2

x xt dt dx= = then

11

20

2 12 tan .

3 3 3

dtI

t

= = +

18. ( )d Put 12

1tan ,

1t x dt dx

x

= =+

then

/41 2 2

1 /4

20 0

tan.

1 2 32

x tdx tdt

x

= = = +

19. ( )d Put2

1 1t dt dx

x x= = as

2t

= and

[ ]2/

2 /21/ / 2

1sin

sin cos cos cos 1.2

xdx t dt t

x

= = = =

20. ( )2

/2

0sin

2 4

x xc Let I e dx

= +

1

00

12 sin 2 sin tan

4 4 11 1

tt e

I e t dt t

= + = + +

[ ]0

2 2sin 0 0.

2 2

te t

= = =

21. ( )b Put 1 ,x xe t e dx dt + = = then we have

( )( )1 11 12 2

1 11 1e e

t dtdt

t t

+ + = =


5/46


[ ]1

1

2

1 1log log 1 1 log 2 2ee et t e

e e

+ = = + + +

1 1log 1.

2e

e

e e

+ = +

22. ( )a Let

/4

0

sin cos

9 16 sin 2

x x

I dxx

+

= + Put sin cos , x x t = then ( )sin cos x x dx dt + =

( )

0 0

221 1 25 169 16 1

dt dt I

tt = =

+ =0

1

1 1 1

10 5 4 5 4dt

t t + +

( ) ( )0

1

1 1log 5 4 log 5 4

10 4t t

+ ( )

1 1log 9 log1 log 3.

40 20= =

23. ( )c Let ( )/2

/4logsin cot

x I e x x dx

= + or

/ 2 / 2

/ 4 / 4log sin cotx x I e x dx e xdx

= + /2 /2

/4/4log sin log sinx xe xdx e x

= +

/2

/4logsinxe xdx

/ 2 / 4 / 41log sin log sin log 2.

2 4 2e e e

= =

24. ( )b Put 12

1sin ,

1t x dt dx

x

= =

then

[ ]1

1/ 2 /6 /6

020 0sin sin cos sin1

x x I dx t tdt t t t x

= = = +

3 1 1 3.

6 2 2 2 12

= + =

.

25. ( )a Put 2 cos 2 sin , x dx d = = then2 0

0 /2

2 1 cos2 sin

2 1 cos

xdx d

x

+ +=

( )

( )

/2

0

cos / 2

4 sin cossin / 2 2 2 d

= ( )/2

02 1 cos d

= +[ ]

/ 2

02 sin 2 1 2

2

= + = + = +

26. ( )c ( )220 0 01 sin

sec sec tan1 sin cos

dx xdx x x x dx

x x

= =

+

[ ] [ ] [ ]0

tan sec tan sec 1 0 1 1 2.x x

= = + = + + =

27.

22 2

00 0

( ) 1 sin sin cos 4 sin cos2 4 4 4 4

x x x x xc dx dx

+ = + = = 4(1-0-0+1) = 8


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28. ( )b 11

1 1 2

0 0

cos cos 1 1. xdx x x x = =

29. ( )c /2

0

cos

1 cos sin

xdx

x x

+ +

( ) ( )

( ) ( ) ( )

2 2/2

20

cos / 2 sin / 2

2 cos / 2 2sin / 2 cos / 2

x xdx

x x x

=

+

( )

( )

2 / 2 / 2

0 0

1 tan / 21 11 tan

2 1 tan / 2 2 2

x xdx dx

x

= = +

=1 1

log log 2.4 4 22

+ =

30. ( )d . Let ( )/ 6

2

02 3 cos 3 I x x dx

= +

( ) ( )/ 6

/ 62 2

00

sin 3 sin 3 12 3 6 . 16

3 3 36

x x x x dx

= + = + .

31. ( )d Put 2sin 2sin cos x t dt x xdx= = .

Now / 2 1 1

1

4 2 00 0

sin cos 1 1 1tan .

1 sin 2 1 2 8

x xdx dt t

x t

= = = + + 32. ( )a Put 2tan sect x dt xdx= = .

Now / 4 1 1

6 2 6 7

00 0

1 1tan sec .

7 7 x xdx t dt t

= = =

33. ( )b Let / 6 / 6

2

30 0

sintan sec

cos

x I dx x xdx

x

= = .

Put 2tan sec ,t x dt xdx= = then we have

11 2 33

00

1.

2 6

t I t dt

= = =

34. ( )b Let/ 2

20

sin cos

cos 3cos 2

x xdxI

x x

=+ + . We put cos sin , x t xdx dt = = then

1 1

20 0

. 2 1

3 2 2 1

t dt I dt

t t t t

= = + + + + = [ ]1

02 log( 2) log( 1) [2 log 3 log 2 2 log 2]t t+ + =

[ ] [ ]9

2 log 3 3log 2 log 9 log8 log .8

= = =

35. ( )d( )

1 1

22 20 02 cos 1 cos 1 cos

dx dx

x x x =

+ + + +

( )

1

1 1

2 200

1 costansin sincos sin

dx xx

+ = = + +1 11 1tan cot tan cot . .

sin 2 2 sin

= =


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36. ( )b 2

31 1

21

1tan tan

1

x x I dx

x x

+ = + +

31 1

2 21tan cot

1 1

x xdx

x x

= + + + 3

12 .

2dx

= =

37. ( )a Put logx u= in 1,I so thatu

dx x du e du= = Also as x e= to2

, 1e u = to 2

Thus,2 2

11 1

.u x

e e I du dx

u x= = Hence 1 2.I I=

38 ( )a ( )

/ 2/ 4

/ 4/ 4

sin sin cos2

xx e

e xdx x x

=

( )/ 2

/ 4

1sin cos

2

xe x x

=

( )/ 2 / 2 / 41 1 11 0 .

2 22 2

ee e = =

39. ( )c ( )

( )

( )

( )

/ 2 / 2

2 20 0

1 2 cos 2 cos 2 3

2 cos 2 cos

x xdx dx

x x

+ + =

+ +

( )

/ 2 / 2

20 02 3

2 cos 2 cos

dx dx

x x

= + + ( )

21 1

220 0 2

14 6

3 3

dt tdt

t t

+=

+ +

2

x put tan t

=

( )1 1

220 0 22 12

3 3dt dt dt

t t= ++ +

1

1 1

2 2 20 00

1 12 123 6 3 6 3

dt t dt t t t

= + + + + + 1

2

0

12 .3 2

tt

= = +

40. ( )c ( )

02 2 2 2 21 cos sin 2 cos sin

2 2 2 2

dx

x x x xa a

+ +

( ) ( )0 2 22 2

1 cos 1 sin2 2

dx

x xa a

= + +

( ) ( ) ( )( )2 2 20

2;

1 1 / 1

dt

a a a t

=

+ + + { where tan

2

xt = }

( )

( )

( )1

2

0

12 1tan

1 11

a at

a aa

+ + = + ( )

1 1

2 2

2tan tan 0

11 aa

= = .

41. ( )b Required value ( )1

10

0

1 1.

10 10

x = =

42. ( )b Required value is/ 3

0

sin30.

3

x

=


8/46


43. ( )a / 4 / 4

0 0

1 tantan

1 tan 4

xdx x dx

x

+ = + /4

0

1log sec log 2.

4 2x

= + =

44. ( )b 1 1

20 0 1

x

x x xdx e dx

e e e =+ + Now put

x xe t e dx dt = =

Also as 0x = to 1, t = 1 to e , then reduced form is

1 1

2 11

1tan tan ,

1 1

e edt et

t e

= = + + 1 1 1

tan tan tan1

x yx y

xy

= + Q

45. ( )a 1 1 1

1 log 1 loge e ex x I dx dx dx

x x x

+= = +

[ ] ( )2

1

1

log 3log .2 2

e

e

exx + =

46. ( )c Integrate it by parts taking log 12

x +

as first function

22 2

1

00

1 1log 1

2 2 2 21

2

x x xdx

x

= + +

21

0

1 3 1log

2 2 2 2

xdx

x=

+ 1 3 1 1 3 3 2

log 2 4 log 3 4 log 2 log2 2 2 2 4 2 3

= + = +

On comparing with the given value3 3

,4 2

a b= = .

47. ( )b ( )

( )( )1 1

0 0

1

1 1 1

x x dxdxI

x x x x x x

+ += =

+ + + +

( )1 1 1

0 0 0

1 4 21 .

1 3

x xdx x dx xdx

x x+ += = + + =+

48. ( )c /4

/4

4 4 4 400

sin2 2sin cos4 4

sin cos sin cos

dd

=+ +

2/ 4

40

2tan sec4

tan 1

d

=

+

(Dividing numerator and denominator by 4cos }.

Now put 2 2tan 2 tan sec ,t d dt = = then the

reduced form is1 1

1

2

00

14 4 tan 4 0 .

1 4

dtt

t

= = =

+


9/46


49. ( )b ( )

( )

( )

1 1

3 30 0

1 2( 1)

1 1

xx e xe xdx dx

x x

+ =

+ +

or( ) ( ) ( )

1

1 1

2 3 20 0

0

2 1.41 1 1

x x xe e e edx dx

x x x

= =

+ + +

50. ( )a Here ( )( )

3

441 1

11xx

x xx x = =

++

( )3

2 2

41 1

1

1

x x dx dx

x x

= +

[ ] ( )2

2 4

11

1 1 32log log 1 log .

4 4 17x x

= + =

51. ( )d

1/ 2

1/ 2 1/ 21

2 2 21/ 4 1/ 4

1/ 4

2 12sin

1/ 21 1

2 2

xdx dx

x xx

= =

( )1/ 2

1

1/ 6sin 2 1 / 6.x = =

52. ( )a Put x t= or1

2dx dt x

= Also as 0x = to 2 so, 0t= to 2 . Therefore

22 2

0 00

3 32 3 2

log3

x ttdx dt

x

= =

( )2

23 1

log3=

53. ( )a ( ) [ ]2 2

00sin cos cos sin 0. x x dx x x

+ = + =

54 ( )a Let( )

/ 4

2 20

cos

cos 1 2 sin

x I dx

x x

=+

( )( )

/ 4

2 20

cos

1 sin 1 2sin

xdx

x x

= +

1/ 2

2 20

1 1 2

3 1 1 2dt

t t

= + + . By partial fractions, where sint x= 1/ 2

1

0

1 1 1 2log tan 2

3 2.1 1 2

tt

t

+ = +

( )( )

1

2 11 1log 2 tan 1

3 2 2 1

+ = +

( ) ( )21 1 1log 2 1 2 log 2 1

3 2 4 3 2 2 = + + = + +

55. ( )b / 2

20

sin

1 cos

x I dx

x

=+ Put cos sin x t xdx dt = = Then

0 1 11

2 2 01 0tan .

1 1 4

dt dt I t

t t

= = = = + +

56. ( )c [ ]2 2

11log log 2 log 2 2 1 xdx x x x= = +

4log 4 1 log 4 log log .ee

= = =


10/46


57. ( )b 5

23

41 .

4 I dx

x

= + Solve further

58. ( )c We have2 2

sin cos1 1

0 0sin cos

x x

I tdt tdt = + Putting 2sint u= in the first integral and 2cost v= in the second integral, we have

0 /2sin 2 sin 2x x I u udu v vdv

=

/2

0 / 2 /2sin 2 sin2 sin 2

x x

u udu u udu v vdv

= +

I

/ 2 / 2 / 2

0 00

cos 2 1sin 2 cos 2

2 2

u uu udu udu

= = +

( )/ 2

/ 2

00

cos 2 1sin 2 .

2 4 4

u uu

= + =

59. ( )b ( ) ( ) ( )1 1 5 0 .....af x bf for each x ix x

+ =

. Replacing ( )1 , x by in ix

we get

( )1

5af bf x xx

+ =

..........(ii)

Eliminating1

fx

from ( )i and ( )ii , we get

( ) ( )2 2 5 5a

a b f x bx a bx

= + ( ) ( ) ( )2

22 2 2

11

log 52

ba b f x dx a x x a b x

=

( ) ( )log 2 2 10 log1 52

ba b a b a a b= + +

( )2

2 21

7 1 7log 2 5 log 2 5 .

2 2a a b f x dx a a b

a b

= + = +

60. ( )d ( )/ 4

2 2

0sec 1 tan

n

nI d

= / 4 / 2

2 2 2

0 0sec tan tan

n n

n I d d

= / 4

1

2 2

0

tan 11

1 1

n

n n n n I I I n n

= + =

. Hence 8 61 1

.8 1 7

I I+ = =

61. ( )b ( )

2 / 3 2 / 3

22 20 0

1

4 9 9 2 / 3

dx dx

x x=

+ +

2/ 3

1

0

1 1 1tan .

9 2 / 3 2 / 3 4 6 24

x = = =

62. ( )a 4 4

1 1 1

2 2 20 0 0

1 12

1 1 1

x x dx I dx dx

x x x

+ = = +

+ + +

( )1 1

2

20 01 2

1

dx I x dx

x = +

+

( )1

31

1

00

3 422 tan

3 3 2 6

x I x x

= + = + =


11/46


63. ( )d 2 30

sina

I x x dx= . Put 3 2 3dt

x t x dx= =

[ ]3 3

3

00

1 1 1sin cos cos 1

3 3 3

a a I t dt t a = = = 3

11 cos .

3a =

64. ( )c / 4 / 4

0 0

sin costan cot

sin cos

x x x x dx dx

x x

+ + =

( )

/ 4

20

sin cos2

1 sin cos

x xdx

x x

+=

Put ( )sin cos , cos sin x x t x x dx dt = + = 0

212

1

dtI

t =

( )0

1

12 sin 2 0 / 2 .

2I t

= = =

65. ( )a 1 1

0 0

1 1 1.

1 1 1

x x x I dx dx

x x x

= =

+ + 1 1 1

2 2 20 0 0

1

1 1 1

x dx x

dx dx x x x

= = =

111 2

0 0sin 1 1.2 I x x

= + =

66. ( )c [ ]11

1log log log1 1.

e e

e I dx x ex

= = = =

67. ( )a 1

2logx x I dx

x= . Let log

dx x t dt

x= =

( )

log2

log 2

00

2 2 log .2

xx t

I t dt x

= = =

68.

/2

2 2 2 2

0

( )cos sin

dxd Ia x b x

= +Dividing the numerator and denominator by cos2x, we get

/ 2 / 22 2

2 2 2 2 2 2 2

0 0

(1/ cos ) sec

(sin / cos ) tan

x dx x dxI

a b x x a b x

= =+ +

Substituting btanx = t and b sec2x dx = dt and limit when x = 0, then t = 0 and when / 2 x then t = =

1

2 2

0 0

/ 1 1tan

dt b t I

a t b a a

= = +

1 11 1tan tan 0 02 2ab ab ab

= = =

69. ( )d ( ) ( ) / 4 5 / 4

0 / 4cos sin sin cos I x x dx x x dx

= + ( )

/ 4

2cos sin x x dx

+

[ ] [ ] [ ]5

44 40 2

4

sin cos sin cos sin cos x x x x x x

= + + + +

1 1 1 1 1 1 1 11 1

2 2 2 2 2 2 2 2I

= + + + +

2 1 2 2 2 1I = + or 2 1 2 2 2 1 4 2 2I = + + =


12/46


70. ( )b 2

04 4

2

a a xdx a a + + 2 22 8 2 8 0a a a a +

( )( )4 2 0 2 4.a a a +

71. ( )b ( )

0 0

221 12 2 1 1

dx dxI

x x x = =

+ + + + ( )

01 1 1

1tan 1 tan 1 tan 0 .

4x

= + = =

72. ( )a 3 3

1

2 11

1tan .

1 3 4 12dx x

x

= = = +

73. ( )d ( )( )( )3

11 2 3 . I x x x dx= or

( )3

4 3 23

3 2

11

6 116 11 6 6 0.

4 3 2

x x x I x x x dx x

= + = + =

74. ( )c

( )

3 3 3

22 2 2

1 1

1 1

dx dx I dx

x x x x x x

= = =

( )( ) [ ]

3 3 3 3

222 2

1 1log 1 log

1dx dx x x

x x= =

= [ ] [ ]4

log 2 log1 log 3 log 2 2 log 2 log 3 log .3

= =

75. ( )a ( )

15

8 3 1

dxI

x x=

+Put 2 1tan tanx x = = , 22 tan secdx d =

( )

1

1

2tan 15

2 2tan 8

2tan sec

tan 3 tan 1I d

= + ( )1

1

2tan 15

2tan 8

2tan sec

sec 4 sec d

= ( )1

1

tan 15

2tan 8

2 tan sec

sec 4d

=

( )( )

1

1

tan 15

tan 8

2 tan sec

sec 2 sec 2d

=

+ ( )

( )

1

1

tan 15

tan 8

sec 21log

2 sec 2

= +

1 2 1 1 5

log log log .2 6 5 2 3

= =

76. ( )c 30

sinI d

= . Since sin is positive in integral ( )0,

( )3 20 0

sin sin 1 cos I d d

= =

( ) 20 0

sin sin cosd d

= + [ ]3

0

0

cos 4cos .

3 3

= + =

77. ( )b 1

1

0

1sin 2 tan

1

xdx

x

+ . Put cos ,x = then 1 1

1 cossin 2tan sin 2tan cot

1 cos 2

+ =

1sin 2 tan tan sin 22 2 2 2

= = ( ) 2 2sin sin 1 cos 1 x = = = =

Now, 1 11 20 0

1sin 2 tan 11

x dx x dxx

+ =

11

2 10

0

1 11 sin .2 2 4

x x x = + =


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78. ( )a 3 3 3

2 2 20 0 0

3 1 3 2

9 2 9 9

x x dxdx dx

x x x

+= +

+ + + ( )3

2 1

0

3 1log 9 tan

2 3 3

xx

= + +

( )3 1

log18 log92 3 4

= +

( )3

log 2 log 2 2 .

2 12 12

= + = +

79. ( )d ( )

2 2

41 15

4

111

dx dx

x xx

x

= + +

. Put 4 51 4

1 z dx dzx x

+ = =

17/1617/16

22

1 1 1 1 17log log 2 log

4 4 4 4 16

dzz

z

= = 1 32

log4 17

I =

80. ( )b ( )

3 3

2 22 2

1

1 1

x A B C I dx dx

x x x x x

+ = = + + or ( ) ( )( ) ( )21 1 1A x B x x C x x + + = +

Put 0,1, 1,x = we get 1, 2, 2 A B C = = =

3 3 3

22 2 22 2

1

dx dx dxI

x x x = +

[ ] ( )3

33

2 22

11 2 log 2 log 1x x

x

= +

1 1 3 16 12 log 2 log 2 log .

3 2 2 9 6I I = + =

81. ( )b 1

1log 1.log

e

e I x dx I x dx= =

[ ] ( ) ( )1

log log 0 1 1 1.e

I x x x e e e = = =

82. ( )c ( ) ( )

( )

2 2 / 2 / 2

20 0

sin cos sin cos

1 sin 2 sin cos

x x x x I dx dx

x x x

+ += =

+ +

( ) ( )/ 2 / 2

00sin cos cos sin I x x dx x x

= + = + or ( )1 1 2I = =

83. ( )d Let / 8 / 8

3 2

0 0cos 4 cos 4 .cos 4 I d d

= =

( )/8

2

01 sin 4 cos 4I d

=

Put sin 4 cos 44

dtt d = = when 0 ,

8to

= then t = 0 to 1

( )1

31

2

00

1 1 11 .

4 4 3 6

t I t dt t

= = =

84. ( )a We have8

3

2 31

1

xdx

x x

=

+ Put2

1 2 x t dx t dt + = =

when 3 8, x to= then 2 3t to= 2

3 3

2 2

2 2

5 3 22 2 3

1 1

t I dt or I dt

t t

= =

3

3

2

2 1 32 log 3 2 log .

2.1 1 2

t I t or I

t e

= = +


14/46


85. ( )a We have,1 11

2 2

00 02 x x x I x e dx I x e x e dx = =

( )1

02 2 0 1

x x I e xe e I e e e = = 2.I e =

86. ( )b We have, ( ) ( )2 2 21 , ; 1 , 1, 2 x x x x x x+ > + >

( )2 2

2 21 1

1 1, 1, 2

1 1

dx dxx

x xx x

<

87. ( )b On integrating both functions, we get

( ) ( )cot

tan2 2

//

1 1log 1 log log 1

2 2

xx

I eI e

t t t = + + +

2 2

2 2

1 1 1 1 1log sec log 1 log cot log log(cos ) log 1

2 2 x x ec x

e e e

= + + +

1log log 1.e

e

= = =

88. ( )a 3 / 4

/ 4 1 cos

dx

x

+3 / 4 3 / 4

2 2 / 4 / 4

1 cos 1 cos

1 cos sin

x xdx dx

x x

=

= ( )3 / 4

2

/ 4cos cot cosec x x ec x dx

( )3 / 4/ 4cot cos 2. x ec x

= + =

89. ( )a ( )

2

21 1

e dxI

x ln x=

+Let ( )

11 ln x t dt dx

x+ = =

Now, when2

1 , x to e= then 1 3t to=3

3

211

1 1 21 .

3 3

dtI

t t

= = = =

90. ( )b [ ]/2 /22

/4/4cos cot cot cot 1.

2 4ec xdx x

= = =

91. ( )c ( )

1/ 2log2 61

x

u

du

e

=

12

21

21

1 6

xe

ut dt as e t t

= =

+

( )

11 1

12 tan tan 16 4 12

xe

x

t e

= =

1 tan 1 3 4.3

x x xe e e

= = =

92. ( )c ( )( ) ( ) ( )2 1

1' ' I f g x f g x g x dx

=

Let ( )( ) ( )( ) ( )' ' f g x z f g x g x dx dz= =

When ( )1 , (1) x z f g= = when ( )2 , (2) x z f g= =

( (2)) ( (2))

( (1))( (1))

1

log

f g f g

f gf g I dz zz = =( ) ( )( )log ( (2)) log ( (1)) 0, 2 1 . I f g f g g g = = =Q


15/46



S O L U T I O N S

LEVEL - 2 (Properties of Definite Integration)

1. ( )b ( ) ( )0 0

sin sin2

xf x dx f x dx

=

Since ( ) ( )0 0

1,

2

a a

xf x dx a f x dx= if ( ) ( ). f a x f x =

2. ( )c ( )/2

0

cot.....

cot tan

x I dx i

x x

=+

/2

0

cot2

cot tan2 2

x

dx

x x

= +

( )/2

0

tan.....

tan cot

xdx ii

x x

=+ . Now adding ( )i and ( )ii , we get

[ ]/2 /2

00

cot tan2 .

4tan cot

x x I dx x I

x x

+= = =+

3. ( )d/ 2 / 2 / 2

0 0 01 tan 1 cot1 tan

2

d d dI

= = =+ + +

. On adding,

/ 2

0

1 12

1 tan 1 cotI d

= + + + [ ]/ 2 / 2

00 .2 4d I

= = = =

4. ( )b ( )0

3 3 3

0

x xt t t

a a f x t e dt t e dt t e dt = = +

( ) ( ) ( )

03 3 3

.x

t t x

a a

df x d dt e dt t e dt x e

dx dx dx = + =

5. ( )b Let ( ) .f x x x= Then ( ) ( ) f x x x x x f x = = = .

Therefore1

10,x x dx

= ( By the property of definite integral ).

6. ( )d / 2 / 2

0 0

sinlog tan log

cos

x xdx dx

x

=

/2 / 2

0 0log sin log cos 0 x dx x dx

= = ( ) ( ){ }0 0 .a a

f x dx f a x dx= Q

7. ( )a / 2 / 2

0 0log sin log cos x dx x dx

= / 2 / 2 / 2

0 0 02 log sin cos log sin 2 log 2 I x x dx xdx dx

= =

0

1log sin log 2,

2 2tdt

= ( Putting 2x t= )

/ 2

0

1.2 log sin log 2

2 2

t dt

=

( ) ( ){ }2 log2 log2 .2 2

b b

a a I I I f x dx f t dt

= = = Q


16/46


8. ( )c ( )/ 2

0

cos sin1 .......

1 sin cos

x xdx i

x x

=

+

Now/ 2

0

cos sin2 2

1 sin cos2 2

x x

I dx

x x

=

+

( )/ 2

0sin cos ......

1 sin cosx x dx ii

x x = +

On adding, 2 0 0.I I= =

9. ( )d Let ( )2

log2

xf x

x

= +

( ) ( )1

2 2log log

2 2

x x f x f x

x x

= = = + +

1

1

2log 0.

2

xdx

x

= +

10. ( )c Let ( ) 17 4cos f x x x=

( ) ( ) ( )( ) ( )417

cos f x x x f x = =

Therefore,1

17 4

1cos 0. x xdx

=

11. ( )d Let ( )3/ 2

/ 2

3/ 2 3/ 20

sin.......

cos sin

xdxI i

x x

=+

3/ 2

/2

0

3/ 2 3/ 2

sin2

cos sin2 2

x

dx

x x

=

+

( )3/ 2

/2

3/ 2 3/ 20

cos.........

sin cos

xdxii

x x

=+

Adding ( )i and ( )ii , we get [ ]/2 /2

00

1 11 .

2 2 4 I dx x

= = =

12. ( )c ( ) / 4 / 4

0 0log 1 tan log 1 tan

4 I d I d

= + = +

/4

0

1 tanlog 1

1 tanI d

= + + ( ) / 4 / 4

0 0log 2 log 1 tan I d d

= + /4 /4

00

1 log 2log 2 log 2.

2 2 8I d

= = =

13. ( )d ( )

( )

2 2

0 0

sin 2 2sin 2

cos cos 2 I d d

a b a b

= =

2

0

sin 2

cosI d

a b

=

2

0

sin22 0 0.

cosI d

a b

= =

14. ( )a Put 1 x t dx dt = = . Also as 0x = to 1, 1t = to 0

Therefore, ( ) ( )( ) ( ) ( )1 0 1 1

0 1 0 01 . f x dx f t dt f t dt f x dx = = =


17/46


15. ( )a ( )1/2

1/2

1cos log ..........( )

1

x I x dx i

x

= +

( )1/2

1/2

1cos log

1

x I x dx

x

+ =

1/2

1/2

1cos log .............( )

1

x I x dx ii

x

= + Adding ( )i and ( )ii , we get

1/ 2 1/2

1/2 1/2

1 12 cos log cos log

1 1

x x I x dx x dx

x x = + +

or 2 0I = or 0I = .

16. ( )d1 / 2

20 0

cos,

sin cos 41

dx d

x x

= =

++ ( Put sin , cos x dx d = = ).

17. ( )b

18. ( )c ( )1

21 1

1 11

1 1 2.2

x x dx x dx x

= = =

19. ( )b Let ( )30

sin ...... I x xd x i

= Also ( ) ( )3

0sin ...... I x xdx ii

=

Adding ( )i and ( )ii , we get ( )30 0

2 sin 3sin sin 34

I x dx x x dx

= =

0

cos 3 1 1 43cos 3 3

4 3 4 3 3 3

xx

= + = + =

. Hence,2

.3

I

=

20. ( )b 2 1 1 2

2 2 2 2

2 2 1 11 1 1 1 x dx x dx x dx x dx

= + +

( ) ( ) ( )1 1 2

2 2 2

2 1 11 1 1 x dx x dx x dx

= +

4 4 44.

3 3 3= + + =

21. ( )c Let ( )/ 2

0

cos......

sin cos

x I dx i

x x

=+

and/ 2

0

cos2

sin cos2 2

x

I dx

x x

=

+

( )/2

0

sin........

cos sin

x I dx ii

x x

=+

Adding ( )i and ( )ii , we get/2

02 1 .

2 4 I dx I

= = =

22. ( )d ( )/ 2

4 40

sin cos.......

cos sin

x x x I dx i

x x

=+

( )/ 2

4 40

cos sin2

..........sin cos

x x x

iix x

=

+


18/46


By adding ( )i and ( )ii , we get/ 2

4 40

cos sin2

2 cos sin

x x I dx

x x

=

+ 2

/ 2

40

tan sec

4 1 tan

x x I dx

x

=

+

Now, Put 2tan ,x t= we get2

1

2 00tan .

8 8 16

dtI t

I t

= = = +

23. ( )b

Let

/ 2

0 sin 4 I x dx

=

4 x is ve

when

4x

and ve+ when

4x

>

/ 4 / 2

0 / 4sin sin 2 2.

4 4 x dx x dx

= + =

24. ( )d

25. ( )b / 2

0sin cos x x dx

( ) ( ) ( ) / 4 / 2

0 / 4sin cos sin cos 2 2 1 . x x dx x x dx

= + =

26. ( )c [ ]/ 2 / 2

00 0cos 2 cos 2 sin 2. x dx x dx x

= = =

27. ( )b 4 4

/ 4 / 4 / 44 4

4 4 / 4 0 0

cos secsin 2 sec 2

sin tan

x xdx xdx xdx

x x

= =

Put tan ,x t= we get2

1

40

12

tdt

t

+

1 11 1

4 2

30 00 0

1 1 82 2 .

3 3t dt t dt

t t

= + = + =

28. ( )b 1.5 1 2 1.5

2 2 2 2

0 0 1 2 x dx x dx x dx x dx = + +

2 1.5

1 20 1 2 2 1 3 2 2 2 2.dx dx= + + = + =

29. ( )d ( ) ( )

( ) ( )0 0tantan

sec tan sec tan

x xx x I dx dx

x x x x

= =

+ +

0 0

tan sin2

2 sec tan 2 1 sin

x x I dx dx

x x x

= =

+ + 0 012 1 sindx

dxx

= +

On solving, we get2

1 .2 2

I

= =

30. ( )a Let( ) ( )

( ) ( )0 0tantan

sec cos sec cos

x xx x I dx dx

x x x x

= =

+ +

It gives20

sin

2 1 cos

x I dx

x

=

+

Now put cosx t= and solve, we get2

.2 2 4

I

= =


19/46


31. ( )a 1

3 2

1sin cos 0. x x dx

= Since the function is an odd function.

32. ( )b Let ( ) ( )2sin 3

0cos 2 1x f x e n x dx

= +Since ( )( ) ( ) ( )cos 2 1 cos 2 1 2 1n x n n x + = + +

( ) ( )

2 2cos 2 1 sin sinn x and x x

= + = .Hence by the property of definite integral,

( )2sin 3

0cos 2 1 0xe n xdx

+ = ( ) ( )2 . f a x f x = 33. ( )b

1

1/ 1/ 1log log log

e e

e e x dx xdx xdx= +

[ ] [ ]1

1/ 1log log

e

e x x x x x x= +

( ) ( )1 1 2 1

1 0 1 1 2 2 1 .e ee e e e

= + + = =

34. ( )a [ ]( ) [ ] /2 /2 / 2

0 0 0sin sin x x dx xdx x dx

=

[ ]/ 2

2 2/ 2

00

, sin 0 .2 8

x x dx

= = =

Q

35. ( )c ( )1

01

n I x x dx=

( ) ( )( )1 1

0 01 1 1 1

n n I x x dx x x dx = =

( ) ( )1 11

0 0

1 1n n

x dx x dx+=

( )

( )

( )

( )

1 12 1

0 0

1 1 1 1

2 1 2 1

n nx x

n n n n

+ + = =

+ + + +

1 1.

1 2I

n n =

+ +

36. ( )c ( /6) ( / 2) ( /2) ( /3)2 2

( /6) ( /2) ( /2) ( /3)

[2 sin ] 1 2 2 1 x dx dx dx dx dx

+ + + +

+ + + +

= + + +

2 2 16 2 6 3 2 3

=

2 2 8 10 5

6 3 3 6 6 6 6 6 3

= = = =

37. ( )d ( )3

/2 /2

3 3 30 0

cos..........

1 tan sin cos

dx x I dx i

x x x

= =+ +

( )3

/ 2

3 30

sin.......................

cos sin

xdx ii

x x

=+

Adding ( )i and ( )ii , we get/ 2

02 .

4 I dx I

= =


20/46


38. ( )a ( )

3 / 4 3 / 4

/ 4 / 41 sin 1 sin I d d

= =

+ + 3

4 4

+ = Q

3 / 4

/ 42

1 sinI d

=

+ .

On simplification, we get ( )2 1 tan .8I

= =

39. ( )b Since ( ) ( ) ( )b b

a a I xf x dx a b x f a b x dx= = + +

( ) ( ) ( )b b

a a I a b f x dx xf x dx = +

( ) ( ){ } f a b x f x given+ =Q

( ) ( ) ( ) ( )2 .2

b b b

a a a

a b I a b f x dx I xf x dx f x dx

+ = + = =

40. ( )a ( )0 0

sin sin I x xdx x x dx

= =

[ ]00

2 sin cos . I xdx x I

= = =41. ( )b

42. ( )a Adding/ 4

0.

4 4 I J dx I J

+ = = =

43. ( )d ( )5 5 5

1 1 13 1 3 1 x x dx x dx x dx + = +

( ) ( )3 5

1 33 3 12. x dx x dx= + =

44. ( )d 3

2............( )

5

x I dx i

x x=

+ Using the property

( ) ( )b b

a a I f x dx f a b x dx= = +

. .,i e change in ( )2 3 5 x x x= + = or dx dx=

( ) ( )2 3

3 2

5 5 .......5 5

x x I dx dx ii x x x x

= =+ +

Adding ( )i and ( )ii ,

3 3

2 2

52 1

5

x x I dx dx

x x

+ = =

+ [ ]3

2

13 2 1 .

2x I= = = =

45. ( )c Let ( )2 2cos 5 cos 5

0 0cos 3 cos 3 ,x x I e x dx e x dx

= =

( ) ( )0 0a a

f x dx f a x dx

= Q

2cos 5

0cos 3 2 0 0.x I e xdx I I I

= = = =


21/46


46. ( )b / 2 / 2

0 0

1 cos.

41 tan cos sin

xdx dx

x x x

= =

+ +

47. ( )c 2 2

1 1 1

1 1 1

sin sin

3 3 3

x x x x I dx dx dx

x x x

= =

Here, ( )sin

3

xf x

x

=

is an odd function but

( )2

3

xf x

x=

is an even function

2 2 21 1 1

1 0 02 2 .

3 3 3

x x x I dx dx dx

x x x

= = =

48. ( )d Since ( ) 11sin f x x= is an odd function, therefore1

11

1sin 0.xdx

=

49. ( )d Given integral ( )2 2

2

2 2 I px s dx q xdx = + + or ( ) ( )

22

0

42 0 4 3

3 I px s dx p s= + + = +

Thus, to find the numerical value of I, it is necessary to know the value of p and s.

50. ( )c If ( )1

log ,1

xf x

x

+ = then ( ) ( )

1log

1

x f x f x

x

= = +

Therefore,1

1

1log 0.

1

xdx

x

+ =

51. ( )a ( ) / 2 0 / 2

/ 2 / 2 0

cos cos cos....

1 1 1 x x x x x x

I dx dx dx ie e e

= = +

+ + +

Putting x t= in0

/ 2

cos,

1 xx

dxe +

we get0 / 2

/ 2 0

cos cos

1 1

x

x x

x e x I dx dx

e e

= =

+ + or / 2 / 2

0 0

cos cos

1 1

x

x x

e x x I dx dx

e e

= ++ +

( )

( )

[ ] / 2 / 2 / 2

00 0

1 coscos sin 1.

1

x

x

e xdx xdx x

e

+= = = =+

52. ( )b [ ] ( ) ( )2

32 1 2

2 2 2

0 0 11

70 1 0 .

3 3

x x x dx x dx x dx

= + = + =

53. ( )b 2

3 32

0 0cos cos 0 xdx xdx

= = ( ){ }3 3cos cos .x x = Q

54.

2 / 2 / 2

0 0 0

( ) 2 log sin 2 log sin 4 log sina x dx x dx x dx

= =

14 log 2 2 log 2 2 log2 2

e e = = =


22/46


23/46


Adding equations ( )i and ( )ii , we get

[ ]sin cos

/ 2 / 2 / 2

sin cos 00 0

2 22 1

2 2 2

x x

x x I dx dx x

+= = = = + Therefore, .

4I

=

64. ( )b ( ) ( )1 1/ 3 1

2 2 2

0 0 1/ 33 1 1 3 3 1 x dx x dx x dx = +

1 3 13 3

0 1/ 3 x x x x = +

1 1 1 1 4.

3 3 3 3 3 3 3 3

= + + =

65. ( )c 2/ 2 cos

2/ 2

sin

1 cos

xx I e dx

x

=

+ 2cos

2

sin

1 cos

xxe

x

+Q is an odd function

0.I =

66. ( )b ( ) ( ) ( )1.5 1.5

0.5 0.5

2 2 , I xf x dx x f x dx= =

( ) ( )

b b

a a f x dx f a b x dx = + Q

( ) ( ) ( ) ( )1.5 1.5 1.5

0.5 0.5 0.52 2 . x f x dx f x dx I I f x dx= = =

67. ( )a2

2

2

/ 2

0

2

x

xx

e dxI

e e

=

+ and

2

2

2

2/ 2

0

2

x

xx

e dxI

e e

=

+

0 0

( ) ( )

a a

f x dx f a x dx

=

Q

( )/ 2 / 2

002 1 .

4 I dx x I

= = =

68. ( )b ( ) ( )5 3 5

1 1 33 3 3 I x dx I x dx x dx= = +

( ) ( ) [ ] [ ]2 3 4 5

1 2 3 43 3 3 3 I x dx x dx x dx x dx = + + +

[ ] [ ]2 3 4 5 2 5

1 41 2 3 40 0 I dx dx dx dx x x = + + + = +

( ) ( )2 1 5 4 2.I = + =

69. ( )d 0 2

2 22 0 2

2 2 02 0

2 2

x x I x dx x dx x dx

= = + = +

( ) ( )2 2 4.= + =

70. ( )d Given ( ) ( ) f x f x = . We know that, ( ) ( ) ( )0

00

a a

a a f x dx f x dx f x dx

= = +

( ) ( ) ( )0 1 0

1 0 10 5 f x dx f x dx f x dx

+ = =

( )0

15. f t dt

=


24/46


71. ( )c ( ) ( ) ( )( )1 sin sin ,a a

a a I xf x dx x f x dx

= = ( ) ( )( )b b

a a f x dx f a b x dx= + Q

( ) ( ) ( ) 1sin sina a

a a x f x dx f x dx I

= =

1 2 2 1

22 . I I I I

= = =

72. ( )a 1/2

1/2

1cos1

x I x ln dxx

+ =

Since1

cos ln1

xx

x

+

is an odd function, ( )( )( ) f x f x = Q

0.I =

73. ( )b 2 2

1 1

1

1

log log loge ee e ee e

x x xdx dx dx

x x x

= +

2

1

21 0

1 10

log loge

e

x x

dx dx zdz zdzx x = + = + (Putting ( )log 1/ e x z x dx dz= = )

0 22 2

1 0

1 52 .

2 2 2 2

z z

= + = + =

74. ( )c ( ) ( ) ( )3 2 3

2 2 2 f x dx f x dx f x dx

= + cos sinxe xQ is an odd function

( ) ( )3 2 3

cos

2 2 2sin 2 0 2 3 2 2.

x f x dx e xdx dx

= + = + =

75. ( )a Let ( ) ( ) ( )( ) ( ) ( )( )x f x f x g x g x = + then,

( ) ( ) ( )( ) ( ) ( )( )x f x f x g x g x = +

( ) ( ) ( )( ) ( ) ( )( )0 0 x dx f x f x g x g x dx

= + = .

76. ( )c ( ) /3 /3

/6 /6

sin.......

1 cot sin cos

dx x I dx i

x x x

= =

+ +

( )/ 3

/ 6

cos.......

cos sin

x I dx ii

x x

=

+

Adding ( )i and ( )ii ,/3

/6

12 ; .

2 3 6 12 I dx I

= = =

77.2/ 3

/ 2

2 / 3 2 / 30

sin

sin cos

x I dx

x x

=+ or

2/ 3

/2

02 / 3 2/ 3

sin2

sin cos2 2

x

I dx

x x

=

+

or2/ 3

/ 2

2 /3 2/ 30

cos

cos sin

x I dx

x x

=+

( )

( )

2/3 2/3/2

2/3 2/30

sin cos2

sin cos

x x I dx

x x

+ =

+[ ]

/2 /2

00

12 .

2 4 I dx I x

= = =


25/46


78. ( )a Let ( ) ( )2log 1 f x x x= + +

Now, ( ) ( ) ( )( )( )

2

2 2

2

1log 1 log 1

1

x x f x x x x x

x x

+ + = + = +

+ +

2 22 2

2

((1 ) )log log1 log( 1 ) log( 1 ) ( )

( 1 )

x x x x x x f x

x x

+ = = + + = + + = + +

( ) ( ) ( )0, .a

a f x if f x f x

= = Q

Hence, ( )1

2

1

log 1 0x x

+ + =

79. ( )d ( )2

cos sin I ax bx dx

= or ( )2 2cos sin 2 cos sin I ax bx ax bx dx

= +

( )2 2cos sin 2 cos sin I ax bx dx ax bx dx

= + or ( )2 202 cos sin 0 I ax bx dx

= +

0

1 cos 2 1 cos 22

2 2

ax bx I dx

+ = + or ( )0 2 cos 2 cos 2 2 . I ax bx dx

= + =

80. ( )b 0 0

1 cos 2cos

2

x I dx x dx

+= =

[ ] [ ]/2 /2

0 /20 /2cos cos sin sin I xdx xdx x x

= =

or sin sin 0 sin sin 1 1 2.2 2

I

= = + =

81. ( )c

82. ( )b 2sin 3

0cosx I e xdx

= ( ) ( ) ( )2sin 3

0cos ........

x I e x dx i

=

( )2sin 3

0cos ..........x I e xdx ii

= Adding ( )i and ( )ii , we get 2 0 0.I I= =

83. ( )a

9 1 4 9

0 0 1 42 2 3 4 x dx dx dx dx + = + +

( ) ( )2 12 3 36 16 2 9 20 31.= + + = + + =84. ( )c

2 1 22 2 2

0 0 1 I x dx x dx x dx = = +

[ ]1 2 2

10 10 2 1.dx dx x= + = =

85. ( )c [ ]x xe is a periodic function with period 1.

[ ] [ ]1000 1

0 01000 ,

x x x xe dx e dx

= [ ] ( )1

00, 0 1 1000 1000 1 .

x x if x e e = < < = = Q

86. ( )c ( )1 1

1/ 1/ ......

ana

n n

n n

x x I dx dx i

a x x a x x

= = + +


26/46


1

1

1 1

1 1 1 1

an

n

a x dxn n

a a x a xn n n n

+

= + + +

( ) ( )

b b

a a f x dx f a b x dx = + Q

( )

1

1 ......

an

n

a x

I dx ii x a x

= + Adding ( )i and ( )ii , we get [ ]

( ) 11/

1/1/2 1

a n an

nn I dx x

= =1 1 2 2

2 .2

na na I a I

n n n n

= = =

87.( )c / 2

0sin 2 log tan , I x x dx

= / 2

0 sin 2 log tan ,2 2 I x x dx

= ( ) ( )

0 0

a a

f x dx f a x dx = Q / 2 / 2

0 0sin 2 log cot sin 2 log tan x x dx x x dx

= = 2 0 0. I I I I = = =

88. ( )a [ ]1/ 2 1/ 2

1/ 2 1/ 2

1log

1

x I x dx dx

x

+ = + . If ( )1

log ,1

xf x

x

+ = then

( ) ( )1 1

log log1 1

x x f x f x

x x

+ = = = +

[ ]1/ 2

1/ 20 I x dx

= + (being integral of odd function)

( )0 1/ 2 0

1/ 21/ 2 0

11 0 .

2dx dx x

= + = =

89. ( )b [ ]2

002sin 0. 2 cos 0 xdx dx x

+ = +

( ) ( )2 cos cos 0 2 1 1 4.= = =

90. ( )c ( )/ 2

3

/ 23sin sin 0, I x x dx

= + = ( Q Function ( )33sin sinx x+ is an odd function).

91. ( )c 1 1/ 2 1

0 0 1/ 2

1 1 1

2 2 2 I x x dx x x x x dx

= = +

1/ 2 12 2

0 1/ 2

1 1

2 2 x x dx x x dx

= +

1/ 2 1

2 3 3 2

0 1/ 24 3 3 4

x x x x = +

1 1 1 1 1 1 1 .

16 24 3 4 16 24 8 = + + =


27/46


92 . ( )a ( ) ( )8 5 8

0 0 55 5 5 17 I x dx x dx x dx= = + = .

93. ( )d ( ) ( )2 1 2

0 0 11 1 1 I x dx x dx x dx= = + +

1 22 2

0 1

1.2 2

x xx x

= + + =

94. ( )d [ ] [ ] [ ] [ ] [ ]2 1 0 1 2

2 2 1 0 1 x dx x dx x dx x dx x dx

= + + +

1 0 1 2

2 1 0 12 1 0 1dx dx dx dx

= + + +

[ ] [ ] [ ]1 0 2

2 1 12 0 x x x

= + + +

( ) ( ) ( )2 1 2 0 1 2 1 2 1 1 4.= + + + + = + + =

95. ( )d ( )1 1 1

1 1 1

20 0 0

1tan tan tan 1

1dx x dx x dx

x x

= +

( )1

11 1 2

00

12 tan 2 tan log 1 log 2.

2 2 xdx x x

= = + =

96. ( )b ( ), 0b b

a a

xdx dx a b

x= <


28/46


1 1 22 2 2

2 1 13 3 3

x x x x x x

= + +

( )2 2 2 1 10 28

2 9 3 1 6 .3 3 3 3 3 3

= + + + = + =

100. ( )a Given ( ) 1 f x x=

( ) ( ) ( )2 2 1 2

0 0 0 11 1 1 . f x dx x dx x dx x dx = = +

( )

1 22 2

0 1

1 11 2 2 1 1.

2 2 2 2

x xx x

= + = + =

101 . ( )b Let ( ) ( )/2

0 0sin sin I xf x dx A f x dx

= =

Now, ( ) ( ) ( )( )0 0

2 sin sin I xf x dx x f x dx

= +

( ) ( )0 0

sin sin f x dx f x dx

= =

( )/ 20

2 2 sin I f x dx

= ( ) ( )/ 2

0 0sin sin . I f x dx A f x dx

= = Hence .A =

102. ( )c Put ( )sin cos sin cos x x t x x dx dt + = = Also as 0x = to2

, t = 1 to 1.

Since here limit is 1 to 1, therefore the value of integral will be zero, ( ){ }0a

a f x dx =Q

103. ( )c Given function ( ) [ ]11

1log log log1

x x L x dt t x

t= = =

( ) log , L x x = Hence ( ) ( ) ( ).L xy L x L y= +

104. ( )c For 0 1,x< < we have2

1 ,x

e e< < so that2 21 1 1 1

0 0 0 01 1 .

x xdx e dx e dx e dx e< < <


29/46


108. ( )a ( ) 4 4 40 0

cos cos cosx x

g x t dt t dt t dt

+ ++ = = + ( ) ( )g f x= +

( ) ( ) ( )40

cos ,x

f x u du g x t u= = = + Q

( ) ( ) ( ).g x g x g + = +

109. ( )d 2 21 1 1

0 0 01 ,x xdx e dx e dx + = + but 21

0

xe dx

is not integrable.

110. ( )a ( )

( ) ( )( )

2

0.........

2

a f x I dx i

f x f a x=

+

( )

( ) ( )( )

2

0

2...........

2

a f a x I dx ii

f a x f x

=

+

Adding ( )i and ( )ii , we get2

02 2 .

a

I dx a I a= = =

111. ( )b 0 0

sin sin sinn n n

n

x dx x dx x dx

+ += +

2 1 cos .n = + .

112. ( )a ( ) / 4 / 4

2 2

0 0tan sec 1 tan

n n

nu xdx x x dx

= = /4

1 / 4 4

2 2 2

20 0

0

tansec tan tan

1

nn n

n

x x x dx x dx u

n

= = 2

1

1n n

u un

+ =

113. ( )a Put2

;2

x dx d

= = As 0 1, 02

x to to

= = . Then it reduces to

/ 2

0

2 2log sin log 2 log 2.

2d

= =

114. ( )c Put sin ,x = we get1 / 2

20 0

log log sin .cos

cos1

xdx d

x

=

/ 2

0log sin log 2.

2d

= =

115. ( )b / 2

0cot I x x dx

= Integrating by parts, we get

( )/ 2/ 2

0 0log sin log sin x x x dx

log 2 log 2.2 2I

= =

116. ( )b Let 1x t+ = when 2 0, x to= then 1 1t to=

( )1

3

12 cos I t t t dt

= + +

Since 3t and cost t are odd functions

[ ]1 1

112 2 4. I dt t

= = =

117. ( )a On differentiating both sides, we get

( )

2

sin sin cos cos xf x x x =

( ) ( )2 2

1 1 1sin 3.

sin 3 f x f x f

x x

= = =


30/46


118. ( )b 2 2

0 0

1 1sin sin sin

2 2

n n

x x dx x dx =

[ ]/ 2 / 2

00

22 sin 2 cos 2 .

2

n x dx n x n

= = =

119. ( )a Since ( ) 0,a

a f x dx

= if ( ) ( ) f x f x = Therefore, 3 0.

a

a

dx

x x=

+

120. ( )a ( ) / 3 / 3

/ 6 / 6

cos........

1 tan cos sin

dx x I dx i

x x x

= =

+ +

( )/3

/6

sin.................

cos sin

x I ii

x x

=

+

(Since ( ) ( )b b

a a f x dx f a b x dx= + )

Adding ( )i and ( )ii we get,/3

/6

12 .

2 3 6 12 I dx I

= = =

121.4

4 4

sin

sin cos

x I dx

x x

=

+

( )4

/ 2

4 40

sin2 2 .......

sin cos

x I dx i

x x

= +

4

/ 2

04 4

sin2

4

sin cos2 2

x

I dx

x x

= +

or

( )4

/ 2

4 40

cos4 ............

cos sin

x I dx ii

x x

=+

Adding ( )i and ( )ii we get,/ 2

02 4 4 2

2 I dx

= = =

122. ( )d Since, f is continuos function. Let 1x t= .dx dt =

When 3 5, x to= then 2 6t to=

Therefore, ( ) ( ) ( )

5 6 6

3 2 21 1 . f x dx f t dt f x dx = =

123. ( )b We have, 2 21

lim ......1 4 2n

n n

n n n

+ + + + +

2 2 21 1 2

2

lim lim

1

n n

n nr r

n n

r n rn

n

= =

= =+

+

1

22 01

2

1lim ,

11

n

nr

dx

xrn

n

=

= =+

+

( )1 1

00

1, lim .

n

nr

r Applying formula f f x dx

n n

=

=

=1

1 1 1

0tan tan 1 tan 0 .

4x

= =


31/46


124. ( )b Let3 3 3 3

1 4 1lim ......

1 2 2nS

n n n= + + +

+ +2

3 3 3 3 3 3

1 4lim ......

1 2n

n

n n n n= + + +

+ + +

2 2

3 3 31 1 3

3

lim lim1

n n

n nr r

r rS r n r

nn

= = = =+ +

2

31

1lim .

1

n

nr

r

nn r

n

=

= +

Applying the formula, we get ( )( )2 2

11 13

3 30 0 0

3 1 1 1log 1 log 2.

1 3 1 3 3e e

x x A dx dx x

x x= = = + =

+ +

125. ( )b 99 99 99 99

100 1001

1 2 .......lim lim

n

n nr

n r

n n =

+ + +=

199 1001

99

01 0

1 1

lim .100 100

n

nr

r x x dx

n n =

= = = = 126. ( )b Let

1/ 1/ ! 1 2 3 4

lim lim . . . ....

n n

nn n

n nP

n n n n n n = =

1 1 2log lim log log ... log

n

nP

n n n n = + + +

1

1log lim log

n

nr

rP

n n == or

( ) ( )1 1

00

1log log log 1 .P x dx x x x P

e= = = =

127. ( )b ( )

2 2

2 201

1 /lim 5 1.

11 /

n

nr

r n x L dx

n xr n

=

= = = ++

128. ( )d 1 1 1 1

lim .....1 2 2n n n n n

+ + + + + +

1 1 1 1lim .....

1 2n n n n n n = + + + + + + +

1 1 1 1lim 1 .....

1 21 1 1

n nn

n n n

= + + + + + + +

1

00

1 1 1lim

11

n

nr

dxrn x

n

=

= = + +

( )1

0log 1 log 2 log 1 log 2.e e e ex= + = =

129. ( )a Let22 2

1 1

1lim lim

1

n n

n nk k

k

k nI

n k n k

n

= =

= =

+ +

( ) [ ]1

12

2 00

1 1log 1 log 2 .

1 2 2

x I dx x

x = = + = +


32/46


130. ( )b( )2 2

1 1 1lim .....

1ny

n n n n n n

= + + + + +

( )

1 1 1lim ....

1 11 1

ny

n nn n

n n

= + + + + +

( )

1 1 1lim 1 ....

1 11 1

ny

n n

n n

= + + + + +

or

( )1

1 1lim ,

11

n

nk

yn

kn

=

=

+

Put1k

xn

= and

1dx

n=

1

1

00

lim lim 2 11

n

nn

n

n n

dxy x

x

= = + +

2 1 2 12 lim 1 2 lim 2

n n

n ny

n n

= =

12 lim 2 2 2 2 2.

ny

n = =

131. ( )a 1 11

1 2 3 .....lim lim

p p p p pn

p p

n n r

n r

n n+ +

=

+ + + +=

1

11

01 0

1 1lim

1 1

p pnp

nr

r x x dx

n n p p

+

=

= = = = + +

132. ( )b ( )1

1/

01 0

1lim . 1

nr n x x

nr

e e dx e en =

= = = =

133.( )d / 2

4 4

0 0sin 2 sin x dx xdx

= Applying gamma function,( ) ( )

( )

/ 24

0

5 / 2 . 1/ 2 32 sin 2 .

2 6 / 2 8xdx

= =

134. ( )a Let ( ) ( )1 12

2 5x

F x y t dt = = and ( )2 2 0 2x

F x y t dt = = .Now point of intersection means those point at which 2

1 2 15 6 y y y y x x= = = + and 2

2.y x=

On solving, we get 2 2 65 65

x x x x= + = and2 36 .

25y x= = Thus point of intersection is

6 36, .

5 25

135. ( )b ( ) ( ) ( ) ( )00

' ' .b c b c

f x a dx f x a f b c a f a

+ = + = +

136. ( )c ( ) 1 1' 0 ,2 2F x x x = >

Hence the function is increasing on 1 1,2 2

and therefore ( )F x has

maxima at the right end point of1 1

,2 2

.

( )1 / 2

1

1 3.

2 8 M a x F x F t dt

= = =


33/46


137. ( )b ( )/ 2

2 2

/ 2sin cos sin cos x x x x dx

+

/ 2 / 23 2 2 3

/ 2 / 2sin cos sin cos x xdx x xdx

= +

/ 22 3

0

2 40 2 sin cos 0 2

15 15 x xdx

= + = + =

138.

( )3

20

( ) tan ,

1

dxa Putting x we get

x x

=+ +

( ) ( )

/ 2 / 22

3 3

0 0

sec cos

tan sec 1 sin

d d

=

+ +

/2

2

0

1 1 1 3

2(1 sin ) 8 2 8

= = + = +

139. ( )d ( )( ) ( ) ( ) ( )

2 22 2 4 221 1 2 2 1

' .2 .2 2 1 x x x x x

f x e x e x xe e + + + += =

( ) ( )' 0, , 0 . f x x >

140. ( )c ( ) ( ) ( ) ( )1

logx x f x dx xe f x f x dx f x

x= + = + .

On differentiating both sides, we get

( ) ( )0 ' f x f x= + . We know ( ) ( ), . x x xd

e e f x ce

dx

= =

141.( )c ( )3 / 2 / 2

3 / 2 3

0 0sin cos sin cosd d

= Applying gamma function,

/23 / 2 3

0

31

3 122 2

sin cos3

3 222

2

d

+ +

= + +

( )

( )

5

5 / 4 2 84.

9 5 52 13 / 4 452. .

4 4 4

= = =

142.20

1 1log

1 I x dx

x x

= + + . Put2

tan sec x dx d = =

( )2

/ 2

20

seclog tan cot

secI d

= + ( )

/ 2

0log tan cotI d

= +( )2/ 2

0

1 tanlog

tanI d

+ =

/ 2 / 2

0 02 log sec log tan I d d

=

{ } / 2 / 2

0 02 log sec log tan 0I d

= = Q / 2

02 log cosI d

=

/2

02 log 2 log cos log 2

2 2I

= = Q log 2.I =


34/46


143. ( )a ( )

20 2

log

1

x x I dx

x

=

+ . Put 2tan sec x dx d = =

( )/ 2 240

tan log tansec

secI d

= ( )

/ 2

0sin cos log tan d

=

( )/2

0

1sin 2 log tan 0

2d

= = { }/2

0sin 2 log tan 0 .d

=Q

144. ( )d Given ( ) 1 12 tan 2 tan1

t t

tt

dx f t x t

x

= = = +

Differentiating with respect to t, ( ) ( )2

2 2' ' 1 1.

1 2 f t f

t= = =

+

145. ( )a ( )3

2log

x

xF x t dt = Applying Leibnitzs theorem,

( ) 3 3 2 2' log logd d

F x x x x xdx dx

= ( )2 23log .3 2 log 2 9 4 log . x x x x x x x= =

146. ( )c ( ) ( )1

1 1 1

2

0

2sin sin 1 sin 0 .1 2

xIx

= = = +

147. ( )a ( ) 2 2 22 2 0 f x x x x = = or4 2 2 0x x+ =

or ( )( )2 22 1 0x x+ = 2 1 0, 1.x x = =

148. ( )c ( )( ) ( ) 220 0 0

1 11

2 22

1 11 1

xdxxdx

dxx xx x

+ = +

+ ++ +

( ) ( )2 1 000

1 1 1 1log 1 log 1 tan2 2 2 2

x x x

= + + + +

10 0 0 .

2 2 4

= + + =

149. ( )d ( )3 3

sin 24 4

sin sin

31 1

3 3x x xd e xF x e dx e dxdx x x x

= = Put 3 23 x t x dx dt = =

( ) ( ) ( ) ( )sin

64 64

1 164 1 ,

te

F t dt F t dt F F

t

= = =

On comparing, 64.k =150. ( )b Since , ( )

0sin .

x

f x t tdt = Now, according to Leibnitzs rule, ( ) ( )' sin . 1 0 sin . f x x x x x= =

151. ( )b 2 2 2 22 2 2 2 2 21 1 2 4 3 9 1

lim sec sec sec .... sec 1n n n n n n n n

+ + + +

=2 2

2 2

2 2 21 1

1lim sec lim sec

n n

n nr r

r r r r

n n n n n = == .

Given limit is equal to the value of integral1

2 2

0sec x x dx

1 12 2 2 2

0 0

1 12 sec sec ,

2 2 x x dx t dt Put x t = = = [ ]10

1 1tan tan1.

2 2t= =


35/46


152. ( )c Given curve logy x= and 1x = , 2x = .

Hence required area ( )2 2

11log log xdx x x x= =

( )2 log 2 1 log 4 1 .sq unit= =

153. ( )c

Required area

4 4

1 1 2

y

x dy dy= = 4

3/ 2

1

1 2 7

.2 3 3y= = sq. unit.

154. ( )b Let the ordinate at x a= divide the area into two equal parts

Area of

44

222

8 81 4 AMNB dx x

x x

= + = =

Area of22

81 2

a

ACDM dxx

= + =

On solving , we get 2 2 ,a = since 0 2 2.a a> =

155. ( )d Required area is2

1 20

4 . . A A y dx y dx sq unit

+ = + =

156. Required Area =

/4/4

0 0

cos 2 sin 2(sin 2 cos 2 )

2 2

x x x x dx

+ = +

1 cos sin cos 0 sin 0 1 .2 2 2

sq unit = + + =

157. ( )d Area ( )

( )

43/2

4

0

0

3 4 2 1123 4 56 .

3. 3/ 2 9 9

x x dx sq unit

+= + = = =

158. ( )a 2y x= and 22 2 0, 2 y x y y y= = =

Required area ( )2

32

2 2

00

42

3 3

y y y dy y

= = =

sq. unit.

159. ( )b Obviously, triangle ACB is right angled at C.

Required area 12

AC BC = 1

2 2 2 2 42

= = sq. unit

Y

C

O AX

B3 4y x= +

Y

XO1A

2A

2

34,

2

( )2,3A

C B

Y

O XM D N

( ), 0a

( )2, 0 ( )4 , 0

Y

C

A

B

X

2y x= +

( )2,4

( )2,0

2y x=

( )0,22

x

=


36/46


160.

/3 /3

1 2

0 0

3 3( ) cos , cos 2

2 4d A x dx A x dx

= = = =

1 2: 2 :1A A =

161. ( )b Shaded area

2 2 23 4

11 1

1 1 15

(16 1) .4 4 4 ydx x dx x A sq unit = = = = =

162. ( )b The curve is symmetric about x axis ,

Shaded area [ ]/ 2 / 2

002 sin 2 cos 2. xdx x

= = =

163. ( )a The curves y x= and sin y x x= + intersect at ( )0,0 and ( ), .

Hence area bounded by the two curves ( )0 0 0

sin sin x x dx x dx x dx

= + =

[ ] ( ) ( )0

cos cos cos 0 1 1 2.x

= = + = + =

164. ( )c Required area [ ] ( )/3 /3

002 tan 2 log sec 2 log 2 . x dx x

= = =

165. ( )b 2

2

0

32 2 1 3 .

log2

kx kdx k= =

Now check from options, only ( )b satisfies the above condition.

166. ( )d ( )2 2 2

1 1

1 2 1 1 1 1bb

f x dx b b x = + = + + = +

( ) 22 2

21

2 1 1

d x x f x x

dx x x = + = =

+ +

167. ( )a Area bounded by the curve ( ), y f x x axis= and the ordinates 1x = and x b= is ( )1

b

f x dx

From the question ( ) ( ) ( )

1

1 sin 3 4b

f x dx b b= +

. Differentiate with respect to b , we get

( ) ( ) ( ) ( ).1 3 1 cos 3 4 sin 3 4 f b b b b= + + +

( ) ( ) ( ) ( )3 1 cos 3 4 sin 3 4 . f x x x x= + + +

168. ( )b Given 2 4x y=

Shaded area2

2 23

00

1 8 2.

4 12 12 3

xdx x = = = =

169. ( )a ( )2

3 222

00

4 164 .

3 2 3

x x x x dx sq unit = =

Y

OX

siny x=2

/ 2

2x =Y

OX

1x =

3y x=

X

Y

2x =


37/46


170. ( )c Given curve is ( )3 10

2 3 102

x y x x y

x

+ = + =

Required area is4 4

3 3

3 10

2

x ydx dx

x

+=

( )4

33 16 log 2 3 16 log 2 . x x sq unit = + = +

171. ( )b Required area = area of OABC - area of OBC

163/ 2

16

00

6416 4 64 .

3 / 2 3

xx dx

= = =

172. ( )d Required area ( )2

2

02 2

x x x dx =

23

2

0

2

log 2 3

xx

x

= +

4 8 1 3 44 .

log 2 3 log 2 log 2 3= + =

173. ( )b Required area / 2 / 2

2

0 0

1 cos 2sin .

2

x A x dx dx

= =

[ ] [ ] / 2 / 2

0 0

1 1sin 2 .

2 4 4x x

= =

174. ( )c Required area 3 2 20 3

43

x dx x dx= + 3 22

2 1

30

1 44 sin

2 2 2 23

x x xx

= + +

3 3 2.

2 2 3 3

= + =

175. ( )c We have 24 y x x= and 0, 0, 4y x= =

Required area ( )42 3

42

00

44

2 3

x x x x dx

= =

64 3232 . .

3 3sq unit= =

176. ( )d Shaded area [ ]/ 2 / 4

00tan log sec x dx x

= =( )log sec / 4 log sec 0 log 2 log1 log 2.= = =

A

O

B

C

Y

( )16,4

2y x=

4y =

x

X

Y3x y=

( )2,0

/4x =

( )/4,0

( )/2,0( )/2,0

Y

X


38/46


177. ( )a Solving 0y = and 24 3 , y x x= + we get 1, 4.x =

Curve does not intersect x axis between 1x = and 4x = .

Area ( )4

2

1

1254 3 .

6 x x dx

= + =

178. 2( ) 4 2c We have y ax y ax= =

We know the equations of lines x = a and x = 4aTherefore, the area inside the parabola between the lines

44 4 4 3/21/2

2 2 23 / 2

aa a a

a a a a

x A y dx ax dx a x dx a

= = = =

1/ 2 3/ 2 3/ 2 1/ 2 3/ 2 24 4 28(4 ) (8 1)

3 3 3a a a a a a = = =

179. ( )b 1,y x= if 1x > and ( )1 ,y x= if 1x .Differentiate both sides . . ' 'w r t t , we get ( ) ( )2 2 4 22 2t f t t t f t t = =

Put2

,5

t = we get 4 2 .25 5

f =

15. ( )b The equation of curve is2

2 2 1 1

2 4y x x x x y x y

= = =

This is a parabola whose vertex is1 1

,2 4

. Hence point of intersection of the curve and the line

( )2 1 0 . ., 0 x x mx x x m i e x = = = or 1x m= 11 2 3 2 2 3 3

2

0 0

9 (1 ) (1 ) (1 )( ) (1 )

2 2 3 2 2 3 6

mm x x mx m m m

x x mx dx m

= = = =

( )3 1/36 9

1 27 1 27 3 22

m m m

= = = = =

Also, ( )3 31 3 0m = ( ) ( ) ( )( )21 3 1 9 1 3 0m m m + + =

( ) ( )2

1 3 1 9 0m m + + = 2 2

2 1 3 3 9 0 5 13 0m m m m m + + + = + =

5 25 52. .,

2m i e m

= is imaginary Hence,m = -2.

16. ( )b Curve ( )2 32 y a x x = is symmetrical about x axis and passes through origin. Also3

02

x

a x and 0x < . So curve does not lie in 2x a> and 0x < , curve lies

wholly on 0 2 .x a Area3/ 2

2 / 22 4

0 08 sin ,

2

a xdx a d

a x

= =

( Put 22 sinx a = )2

2 3 1 38 . .4 2 2 2

aa

= = ( Applying Gamma function )

Y

X

O

2x a=


44/46


Y

C

O BAX

17. ( )b Solving the equations,

( )2 2 9................ x y i+ = 2 8 .................( ) y x ii=

Put ( )2 8 y x in i= , 2 8 9 0x x+ = ( )( )9 1 0x x + =

. . 9 1i e x or =

9x = gives imaginary value of y for equation ( )ii hence neglected.

( )1, 0A and ( )3,0B

Required area ( ) ( )1 3

0 12 y dx for ii y dx for i = +

( )1 31/ 2 2 2

0 12 2 2 3 x dx x dx = +

313/ 2 2

1

0 1

9 92 2 2 sin

3 / 2 2 2 3

x x x x = + +

=14 2 9 8 9

2 sin3 2 2 2 2 3

x

+

12 2 9 19 sin .3 2 3

= +

18. ( )b The lines are 1, 0 y x x=

1, 0, 1, 0, 1, 0y x x y x x y x x= < = + = +

Definite Solut

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