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    TRIVANDRUM: T.C.No: 5/1703/30, Golf Links Road, Kowdiar Gardens, H.B. Colony, TVM, 0471-2438271

    KOCHI: Bldg.No.41/352, Mulloth Ambady Lane, Chittoor Road, Kochi - 11, Ph: 0484-2370094

    Todays Mathiitians..... Tomorrows IITiians.....

    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

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    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

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

    S O L U T I O N S

    LEV EL - 1 (Fundamentals of Definite Integration)

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    Trivandrum: 0471-2438271 Kochi: 0484-2370094

    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

    = =

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    Trivandrum: 0471-2438271 Kochi: 0484-2370094

    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

    + + = =

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    Trivandrum: 0471-2438271 Kochi: 0484-2370094

    [ ]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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    Trivandrum: 0471-2438271 Kochi: 0484-2370094

    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