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Exercise 3.6 (Solutions) Calculus and Analytic Geometry, MATHEMATICS 12

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Definite Integral Let ( ) ( )f x dx x cϕ= +∫

Then [ ]( ) ( ) ( )b

a

bba af x dx x or xϕ ϕ=∫

( ) ( )b aϕ ϕ= − Also

( ) ( )b a

a b

f x dx f x dx• = −∫ ∫

( ) ( ) ( )b c b

a a c

f x dx f x dx f x dx• = +∫ ∫ ∫

where a c b< <

Question # 1

( )2 2 2

2 2

1 1 1

1x dx x dx dx+ = +∫ ∫ ∫

23 21

13x x= + ( )

3 32 1 2 13 3

= − + −

8 1 13 3

= − + 103

=

Question # 2

1 1

3

1

1x dx−

+

∫

1 113

1 1

x dx dx− −

= +∫ ∫

11 13 1

1

1

1 13

x x+

−

−

= ++

( )

143

1

1 ( 1)43

x

−

= + − −

( )4 43 33 (1) ( 1) 1 14

= − − + +

( )3 1 1 24

= − + 2=

Question # 3

( )

0

22

12 1

dxx− −∫ ( )

02

2

2 1x dx−

−

= −∫

( )( )

02 1

2

2 12 1 2x − +

−

−=

− + ⋅ ( )

01

2

2 1( 1) 2x −

−

−=

−

( ) ( )1 12(0) 1 2( 2) 12 2

− −− − −= −

− −

( ) ( )1 10 1 4 12 2

− −− − −= −

− −

( ) ( )1 11 52 2

− −− −= −

− −

1 1( 2) ( 1) ( 2) ( 5)

= −− − − −

1 12 10

= − 25

=

Question # 4

2

6

3 x dx−

−∫ ( )2 1

2

6

3 x dx−

= −∫

( )2

1 12

6

31 1 ( 1)2

x +

−

−=

+ −

( )2

32

6

33 ( 1)2

x

−

−=

−

( )23

2

6

2 33

x−

= − −

( ) ( )3 32 2

2 3 2 3 63

= − − − +

( ) ( )3 32 2

2 1 93

= − −

( )2 1 27

3= − − 54

3=

Question # 5

( )5

3

1

2 1t dt−∫ ( )5 3

2

1

2 1t dt= −∫

( )5

3 12

1

2 13 1 22

t +−=

+ ⋅

( )5

52

1

2 15 22

t −=

⋅

( )55

2

1

2 15

t −=

( ) ( )5

5222 5 1 2(1) 1

5 5− −

= −

( )522 5 1 1

5 5−

= −

( )52 5 1 1

5 5

−= − Ans.

Question # 6

5

2

2

1x x dx−∫ ( )5 1

2 2

2

1x x dx= − ⋅∫

( )5 1

2 2

2

1 1 22

x x dx= − ⋅∫

( ) ( )5 1

2 22

2

1 1 12

dx x dxdx= − ⋅ −∫

( )51 12 2

2

1112 12

x+

−=

+

( )53

2 2

2

11322

x −=

( ) ( )( )3 3

2 2 2 21 2 5 1 2 12 3

= ⋅ − − −

( ) ( )3 32 21 5 1 4 1

3

= − − −

( ) ( )3 32 21 4 3

3

= −

FSc-II / Ex- 3.6 - 2

( ) ( )3 112 2 21 2 33

+ = −

( ) ( )

13 21 2 3 33

= −

1 8 3 33

= −

Question # 7

2

21 2

x dxx +∫

2

21

1 22 2

x dxx

=+∫

( )22

21

212 2

d xdx dxx

+=

+∫ 22

1

1 ln 22

x= +

( )2 21 ln 2 2 ln 1 22

= + − +

( )1 ln 6 ln 32

= −

1 6ln2 3 =

1 ln 22

=

Question # 8

23

2

1x dxx − ∫

32

22

1 2x dxx

= + − ∫

3 3 32 2

2 2 2

2x dx x dx dx−= + −∫ ∫ ∫

Now do yourself

Question # 9 1

2

1

1 12x x x dx−

+ + + ∫

( )1 1

2 2

1

2 1 12x x x dx

−

+ = + + ∫

( ) ( )1 1

2 2

1

1 1 2 12 x x x dx−

= + + +∫

( ) ( )1 1

2 2

1

1 1 2 12dx x x dxdx

−

= + + +∫

( )11 12 2

1

1112 12

x x+

−

+ +=

+

( )13

2 2

1

11322

x x

−

+ += ( )

132 2

1

1 13

x x−

= + +

( ) ( )3 3

2 22 21 (1) (1) 1 ( 1) ( 1) 13

= + + − − + − +

( ) ( )3 32 21 1 1 1 1 1 1

3

= + + − − +

( ) ( )3 32 21 3 1

3

= −

1 3 3 13

= −

133

= −

Question # 10 3

20 9

dxx +∫

3

2 20 3

dxx

=+∫

31

0

13 3

xTan−=

1 11 3 1 03 3 3 3Tan Tan− − = −

( ) ( )1 11 11 03 3

Tan Tan− −= −

( )1 1 03 4 3π = −

12π=

Question # 11

Let 2

1

lnI x dx= ∫ 2

1

ln 1x dx= ⋅∫

Integrating by parts

2

21

1

1lnI x x x dxx= ⋅ − ⋅∫

221

1

lnx x dx= − ∫

( ) 212 ln 2 1 ln1 x= ⋅ − ⋅ −

( ) ( )2 ln 2 1 (0) 2 1= ⋅ − ⋅ − −

( )2 ln 2 0 1= ⋅ − − 2ln 2 1= −

Question # 12 2

2 2

0

x xe e dx

− −

∫

2 22 2

0 0

x xe dx e dx

−= −∫ ∫

2 2

2 2

0 0

1 12 2

x xe e

−

= −−

2 2

2 2

0 0

2 2x x

e e−

= +

2 0 2 02 2 2 22 2e e e e

− − = − + −

( ) ( )1 0 1 02 2e e e e−= − + −

12 1 1e e = − + −

12 2e e = + −

2 1 22 e ee

+ −=

( )212

ee−

=

Question # 13

Let 4

0

cos sincos 2 1

I dπ

θ θ θθ+=

+∫

4

20

cos sin2cos

dπ

θ θ θθ

+= ∫ 2 1 cos 2cos

2θ

θ+

=∵

4

2 20

cos sin2cos 2cos

dπ

θ θ θθ θ

= + + ∫

NOTE

( ) ( )3 112 2

11 2

3 3

3 3 3 3

+=

= ⋅ =

FSc-II / Ex- 3.6 - 3

4 4

0 0

1 sin2cos 2cos cos

d dπ π

θθ θθ θ θ

= +⋅∫ ∫

4 4

0 0

1 1sec sec tan2 2

d dπ π

θ θ θ θ θ= +∫ ∫

4 400

1 1ln sec tan sec2 2

π πθ θ θ= + +

1 ln sec tan ln sec(0) tan(0)2 4 4

1 sec sec(0)2 4

π π

π

= + − +

+ −

( ) ( )1 1ln 2 1 ln 1 0 2 12 2

= + − + + −

( ) ( )1 1ln 2 1 0 2 12 2

= + − + −

( )1 ln 2 1 2 12

= + + − Ans.

Question # 14

63

0

cos dπ

θ θ∫ 6

2

0

cos cos dπ

θ θ θ= ⋅∫

( )6

2

0

1 sin cos dπ

θ θ θ= −∫

6 62

0 0

cos sin cosd dπ π

θ θ θ θ θ= −∫ ∫

626

00

sin sin sind dd

ππ

θ θ θ θθ

= − ∫

3 6

0

sinsin sin(0)6 3

ππ θ = − −

3 31 10 sin sin (0)2 3 6π = − − −

231 1 1 (0)2 3 2

= − −

1 1 12 3 8

= −

1 12 24

= − 1124

=

Question # 15

4

2 2

6

cos cot dπ

π

θ θ θ⋅∫

( )4

2 2

6

cos cosec 1 dπ

π

θ θ θ= −∫

( )4

2 2 2

6

cos cosec cos dπ

π

θ θ θ θ= −∫

4

2 22

6

1cos cossin

dπ

π

θ θ θθ

= − ∫

4 4

2 2

6 6

cot cosd dπ π

π π

θ θ θ θ= −∫ ∫

( )4 4

2

6 6

1 coscosec 12

d dπ π

π π

θθ θ θ+ = − − ∫ ∫

4 4 4 4

2

6 6 6 6

1 1csc cos 22 2

d d d dπ π π π

π π π π

θ θ θ θ θ θ= − − −∫ ∫ ∫ ∫

4 4

4

666

3 1 sin 2cot2 2 2

dπ π

π

πππ

θθ θ= − − −∫

( ) ( )4

6

3cot cot4 6 2

sin 2 sin 21 4 62 2 2

π

ππ π θ

π π

= − + −

− −

( )33 1 1 21 3

2 4 6 2 2 2π π

= − + − − − −

( ) 3 1 1 31 32 12 2 2 4

π = − + − − −

1 31 3 8 4 8π= − + − − + 5 9 3

4 8 8π= − + −

9 3 108

π− −=

Question # 16

4

4

0

cos t dtπ

∫ ( )4 22

0

cos t dtπ

= ∫

24

0

1 cos 22

t dtπ

+ = ∫

4 2

0

1 2cos 2 cos 24t t dt

π + +=

∫

( )4

2

0

1 1 2cos 2 cos 24

t t dtπ

= + +∫

4

0

1 1 cos 41 2cos 24 2

tt dtπ

+ = + + ∫

4

0

1 2 4cos 2 1 cos 44 2

t t dtπ

+ + + = ∫

( )4

0

1 3 4cos 2 cos 48

t t dtπ

= + +∫

4

0

1 sin 2 sin 43 48 2 4

t ttπ

= + +

FSc-II / Ex- 3.6 - 4

( ) ( )

( ) ( ) ( )

sin 41 43 2sin 2 48 4 4

sin 4 03 0 2sin 2 0

4

ππ π

= + +

− − −

1 3 0 02 0 08 4 4 4π = + + − − −

1 3 28 4

π = +

1 3 88 4

π + =

3 832

π +=

Question # 17

Let 3

2

0

cos sinI dπ

θ θ θ= ∫

Put cost θ= sindt dθ θ⇒ = − sindt dθ θ⇒ − =

When 0θ = then 1t = And when 3

πθ = then 12t =

So 1

22

1

( )I t dt= −∫

1

22

1

t dt= − ∫ 1

3 2

13t= −

( ) ( )3

31 123 3

= − −

1 183 3

= − −

1 124 3

= − −

724

= − −

724

=

Question # 18

( )4

2 2

0

1 cos tan dπ

θ θ θ+∫

( )4 2

22

0

sin1 coscos

dπ

θθ θθ

= +∫

4 22

20

sin sincos

dπ

θ θ θθ

= +

∫

( )4

2 2

0

tan sin dπ

θ θ θ= +∫

42

0

1 cos 2sec 12

dπ

θθ θ− = − + ∫

4 2

0

2sec 2 1 cos 22

dπ

θ θ θ − + −=

∫

( )4

2

0

1 2sec 1 cos 22

dπ

θ θ θ= − −∫

4

0

1 sin 22 tan2 2

πθθ θ= − −

( )

( )

sin 21 42 tan2 4 4 2

sin 2 02 tan(0) 0

2

ππ π

= − −

− + +

1 12(1) 2(0) 0 02 4 2π = − − − + +

1 32 2 4

π = −

1 62 4

π− =

68

π−=

Question # 19

Let 4

0

secsin cos

I dπ

θ θθ θ

=+∫

4

0

secsincos 1cos

dπ

θ θθθθ

= +

∫

( )4 2

0

sectan 1

dπ

θ θθ

=+∫

Put tan 1t θ= + 2secdt dθ θ⇒ = When 0x = then 1t =

Also when 4

x π= then 2t =

So 2

1

dtI t= ∫

21ln t=

ln 2 ln1= − ln 2 0= − ln 2=

Review

If ( ) :

( )( ) :

g x a x bf x

h x b x c≤ ≤

= ≤ ≤

Then

( ) ( ) ( )c b c

a a b

f x dx g x h x= +∫ ∫ ∫

Question # 20

Let 5

1

3I x dx−

= −∫

Since

3x − ( )3 3 0 3

3 3 0 3x if x x

x if x x− − ≥ ⇒ ≥

= − − − < ⇒ <

So [ ] ( )5 3 5

1 1 3

3 ( 3) 3x dx x dx x dx− −

− = − − + −∫ ∫ ∫

( )3 5

1 3

( 3) 3x dx x dx−

= − − + −∫ ∫

3 52 2

1 3

( 3) ( 3)2 2

x x

−

− −= − +

2 2 2 2(3 3) ( 1 3) (5 3) (3 3)

2 2 2 2 − − − − −= − − + −

FSc-II / Ex- 3.6 - 5

0 16 4 02 2 2 2

= − − + −

8 2= + 10=

Question # 21

Let

2131

231

8

2xI dx

x

+ = ∫

1 21 2

3 3

18

2x x dx− = + ∫

Put 13 2t x= +

231

3dt x dx

−⇒ =

233dt x dx

−⇒ =

When 18

x = then 52

t =

And when 1x = then 3t =

So ( )3

2

52

3I t dt= ∫ 33

52

33t=

( )3

3 53 23 3 3

= −

12527 83

3 3

= −

27 1253 3 24 = −

913 24 =

918

=

Question # 22 3 2

1

21

x dxx−+∫

3

1

11 1x dxx = − − + ∫

3 3 3

1 1 11

dxx dx dx x= − −+∫ ∫ ∫

32 331 1

1

ln 12x x x= − − +

( ) ( )2 23 1 3 1 ln 3 1 ln 1 1

2 2

= − − − − + − +

( ) ( )9 1 2 ln 4 ln 22 2 = − − − −

44 2 ln2

= − − 2 ln 2= −

Question # 23

( )( )3 2

22

3 2 11 1

x x dxx x

− +− +∫

3 2

3 22

3 2 11

x x dxx x x

− +=− + −∫

( )3 23

3 22

1

1

d x x xdx dxx x x

− + −=

− + −∫

33 2

2ln 1x x x= − + −

3 2 3 2ln 3 3 3 1 ln 2 2 2 1= − + − − − + −

ln 27 9 3 1 ln 8 4 2 1= − + − − − + −

ln 20 ln 5= − 20ln5

= ln 4=

Question # 24

4

20

sin 1cos

x dxx

π−

∫ 4

2 20

sin 1cos cos

x dxx x

π

= − ∫

4

20

sin 1cos cos cos

x dxx x x

π

= − ⋅ ∫

( )4

2

0

sec tan secx x x dxπ

= −∫

40sec tanx xπ

= −

( )sec tan sec(0) tan(0)4 4π π = − − −

2 1 1 0= − − + 2 2= −

Question # 25

Let 4

0

11 sin

I dxx

π

=+∫

4

0

1 1 sin1 sin 1 sin

x dxx x

π−= ⋅

+ −∫

4

20

1 sin1 sin

x dxx

π−=−∫

4

20

1 sincos

x dxx

π−= ∫

Now same as Question # 24

Question # 26

Let 1

0

34 3

xI dxx

=−∫

Put 4 3t x= − 3 4x t⇒ = −

Also 3dt dx= − 13

dt dx⇒ − =

When 0x = then 4t = And when 1x = then 1t =

So 1

4

4 13

tI dtt

− = − ∫

1

1 12 24

1 43

t dtt t

= − −

∫

4 1 1

2 2

1

1 43 t t dt−

= + −

∫

Now do yourself

Question # 27

Let ( )2

6

cossin 2 sin

xI dxx x

π

π

=+∫

Put sint x= cosdt x dx⇒ =

2

2

11 2

211

xx x

x xxx

− −

−+ −

+

− −− −

−+ +

FSc-II / Ex- 3.6 - 6

When 6

x π= then 12

t =

When 2

x π= then 1t =

So ( )1

12

2dtI

t t=

+∫

Now consider

( )1

22A Bt tt t

= +++

( )1 2A t Bt⇒ = + + …… (i) Put 0t = in (i)

1 (2 0) (0)A B= + + 1 2A⇒ = 12

A⇒ =

Put 2 0t+ = 2t⇒ = − in eq. (i)

1 0 ( 2)B= + − 1 2B⇒ = − 12

B⇒ = −

So ( )1 11 2 2

22 t tt t−

= +++

1 1 1

1 1 12 2 2

1 11 2 2(2 ) 2

dt dt dtt t t t−

⇒ = ++ +∫ ∫ ∫

1 1

1 12 2

1 1 1 12 2 2

dt dtt t= −+∫ ∫

1 11 1

2 2

1 1ln ln 22 2

t t= − +

1 1ln 1 ln2 2

1 1ln 2 1 ln 22 2

= − − + − +

1 1 1 50 ln ln 3 ln2 2 2 2 = − − −

1 1 5ln ln 3 ln2 2 2 = − − +

51 2ln 12 32

= ×

1 5ln2 3 =

Question # 28

Let ( ) ( )2

0

sin1 cos 2 cos

x dxIx x

π

=+ +∫

Put cost x= sindt x dx⇒ = − sindt x dx⇒ − =

When 0x = then 1t =

And when 2

x π= then 0t =

So ( ) ( )0

11 2

dtIt t−=

+ +∫

( )( )0

11 2

dtt t

= −+ +∫ ( )( )

1

01 2

dtt t

=+ +∫

Now consider

( )( )1

1 21 2A B

t tt t= +

+ ++ +

1 (2 ) (1 )A t B t⇒ = + + + …. (i) Put 1 0t+ = 1t⇒ = − in (i)

( )1 2 1 0A= − + 1A⇒ = Put 2 0t+ = 2t⇒ = − in (i)

( )1 0 1 2B= + − 1 B⇒ = − i.e. 1B = − So

( )( )1 1 1

1 21 2 t tt t−= +

+ ++ +

( )( )1 1 1

0 0 0

1 1 11 21 2

dt dt dtt tt t= −

+ ++ +∫ ∫ ∫

1 1

0 0ln 1 ln 2t t= + − +

( )( )

ln 1 1 ln 1 0ln 2 1 ln 2 0

= + − +− + − +

ln 2 0 ln 3 ln 2= − − +

2 2ln 3× =

4ln 3

=

Error Analyst

Usman Khilji – FAZMIC Sargodha 2004-06 Muzammil Ahsan - 2011

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