of 46

• date post

30-May-2018
• Category

## Documents

• view

215

0

Embed Size (px)

### Transcript of Definite Solut

• 8/14/2019 Definite Solut

1/46

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

• 8/14/2019 Definite Solut

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

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)

• 8/14/2019 Definite Solut

3/46

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

= =

• 8/14/2019 Definite Solut

4/46

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

+ + = =

• 8/14/2019 Definite Solut

5/46

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

• 8/14/2019 Definite Solut

6/46

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