integral calculus

Calculus

Integration

Prof. Dr. M. Abul-Ez

Mathematics Department

Faculty of Science

Sohag University

Mathematics For Engineering

Chapter 1

Indefinite Integration

If ( )F x is a function such that ( ) ( )F x f x′′′′ ==== on the interval [ , ]a b Then

( )F x is called an anti-derivative or indefinite of ( )f x . The indefinite integral of the given function is not unique for example

2 2 2, 3, 5x x x+ ++ ++ ++ + are indefinite integral of ( ) 2f x x==== since

2 2 2( ) ( 3) ( 5) 2d d d

x x x xdx dx dx

= + = + == + = + == + = + == + = + = All of indefinite integral of

( ) 2f x x==== include in ( ) 2f x x c= += += += + where c called the constant of integration, is an arbitrary constant. 1.1- Fundamental Integration Formula:

(1) ( ) ( )

(2) ( )

(4) 11

(5) ln

df x f x c

dxu v dx udx v dx

udx udx

xx dx c m

aa dx c

e du e c

α αα αα αα α++++

= += += += +

+ = ++ = ++ = ++ = +====

= + ≠ −= + ≠ −= + ≠ −= + ≠ −++++

= += += += +

∫∫∫∫

∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫

∫ ∫∫ ∫∫ ∫∫ ∫

∫∫∫∫

Integration of Trigonometric functions: (8) sin cos

(9) cos sin

(10) tan ln sec

(11) cot lncos

x dx x c

= − += − += − += − += += += += +

= += += += +

∫∫∫∫

(12) sec ln sec tan

(13) cosec ln csc cot

x dx x x c

dx x x c

= + += + += + += + +

= − += − += − += − +∫∫∫∫

∫∫∫∫

(14) sec tan

(15) cosec cot

(16) sec tan sec

(17) cosec cot cosec

x dx x c

dx x c

x x dx x c

= += += += +

= − += − += − += − += += += += +

= − += − += − += − +

∫∫∫∫

Integration tends to inverse of Trigonometric functions: 1

2 2 2 1

2 2 21

2 2 2 1

dx b abxa b x c

dx ab abxa b x c

dx a adxbxx b x a c

ab adxdx

bx ab x a cab bx a

−−−−

++++==== −−−−−−−− ++++

++++==== −−−−++++ ++++

++++==== −−−−−−−− ++++

++++==== −−−−−−−− ++++++++

∫∫∫∫

ln( )(23)

ab adxa bxa b x c

ab a bx

x x a cdxx

−−−−

++++==== ++++−−−− ++++ −−−−

+ + ++ + ++ + ++ + +==== ++++++++

∫∫∫∫

22 2 2 2 1

ln( )(24)

1(25) sin

1(26) sinh

1(27) cosh

x x a cdxx

a xa x x a x c

a xx a x x a c

−−−−

+ − ++ − ++ − ++ − +==== ++++−−−−

− = − + +− = − + +− = − + +− = − + +

+ = + + ++ = + + ++ = + + ++ = + + +

− = − − +− = − − +− = − − +− = − − +

∫∫∫∫

In the following some lows witch we use to integrate the square of trigonometric functions

[[[[ ]]]]

(1)cos sin 1,

(2)1 tan sec ,

(3)cot 1 csc

1(4) sin (1 cos 2 )

(5)cos (1 cos 2 )2

1(6)cos cos cos( ) cos( )

(7)sin sin cos( ) cos( )21

(8)sin cos sin( ) sin( )2

x y x y x y

+ =+ =+ =+ =

= −= −= −= −

= += += += +

= + + −= + + −= + + −= + + −

= − − += − − += − − += − − +

= + + −= + + −= + + −= + + −

Integration of square of trigonometric functions:

1 1 1(1) sin (1 cos 2 ) cos 2

1 1 1(2) cos (1 cos 2 ) cos 2

(3) sec tan ,

x dx x dx x x c

x dx x c

= − = − += − = − += − = − += − = − +

= + = + += + = + += + = + += + = + +

= += += += +

∫ ∫∫ ∫∫ ∫∫ ∫

∫∫∫∫

(4) cosec cot

(5) tan (sec 1) tan

(6) cot (cosec 1) cot

x dx x c

x dx dx x x c

x dx x dx x x c

= − += − += − += − +

= − = − += − = − += − = − += − = − +

= − = − − += − = − − += − = − − += − = − − +

∫∫∫∫

∫ ∫∫ ∫∫ ∫∫ ∫

[[[[ ]]]]

1(7) cos cos cos( ) cos( )

21 sin( ) sin( )2 ( )

1(8) sin sin cos( ) cos( )

1 sin( ) sin( )2 ( )

1(9) sin cos sin( ) sin( )

ax bx dx a b x a b x dx

a b x a b xc

a b a b

a b x a b xc

a b a b

= + + −= + + −= + + −= + + −

+ −+ −+ −+ −= + += + += + += + + + −+ −+ −+ −

= − − += − − += − − += − − +

− +− +− +− += + += + += + += + + − +− +− +− +

= + + −= + + −= + + −= + + −

∫ ∫∫ ∫∫ ∫∫ ∫

1 cos( ) cos( )2 ( ) ( )

a b x a b xc

a b a b

− + −− + −− + −− + −+ ++ ++ ++ + + −+ −+ −+ −

Solved Examples:

1 43 3 3

1 ( )( )

1 1ln ( )

1 1 ( )( ) , 1

( 1)( )

Exampl(1)

Exampl(2) :

Exampl(3) :

Exampl(4) :

Exampl(5) :

x dx x c

x dx x dx x c

ax bax b dx c

dx ax b cax b a

ax bdx ax b dx c n

a nax b

− +− +− +− +−−−−

= += += += +

= = += = += = += = +

+++++ = ++ = ++ = ++ = +++++

= + += + += + += + +++++

++++= + = + ≠= + = + ≠= + = + ≠= + = + ≠− +− +− +− +++++

∫∫∫∫

∫ ∫∫ ∫∫ ∫∫ ∫

∫∫∫∫

∫ ∫∫ ∫∫ ∫∫ ∫

1 3 1 3 3 52 2 2 2 2 2

23 5 2 3 2

2 2(1 ) ( )

Find ( ) ( 2) , ( ) ( ) 22

Exampl(6) :

Exampl(7) :

Exampl(8) :

dx ax b caax b

x x dx x x dx x dx x dx x x c

x dxi x x dx ii dx iii x x dx

= + += + += + += + +++++

− = − = − = − +− = − = − = − +− = − = − = − +− = − = − = − +

+ ++ ++ ++ +++++

∫∫∫∫

∫ ∫ ∫ ∫∫ ∫ ∫ ∫∫ ∫ ∫ ∫∫ ∫ ∫ ∫

∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫

3 5 2 5 6 3 6

1 3 3234 4 4

3 33 2 32 2

substitute in the integral we have

1 1 1( ) ( 2) ( 2)

3 18 18

1 1 4 4( ) ( 2)

3 3 9 92

1 2 2( ) 2 ( ) ( 2)

let u x du x du

i x x dx u du u c x c

x dx duii dx dx u du u x c

iii x x dx u du u c x c

−−−−

= + ∴ == + ∴ == + ∴ == + ∴ =

+ = = + = + ++ = = + = + ++ = = + = + ++ = = + = + +

= = = = + += = = = + += = = = + += = = = + +++++

+ = = + = + ++ = = + = + ++ = = + = + ++ = = + = + +

∫ ∫∫ ∫∫ ∫∫ ∫

∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫

∫ ∫∫ ∫∫ ∫∫ ∫

3 2 3 2

3 5 32 2 2 22 2 2

1 ( ) 1

( ) 1 1

( 1) 1 1

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

Exampl (9) :

x x dx x x x x dx

x x x dx x x dx

x x dx x x dx x x c

+ = + − ++ = + − ++ = + − ++ = + − +

= + + − += + + − += + + − += + + − +

= + − + = + − + += + − + = + − + += + − + = + − + += + − + = + − + +

∫ ∫∫ ∫∫ ∫∫ ∫

1 2 1ln 2 3

2 3 2 2 3 2

1 6 1ln(1 2 )

6 61 2 1 25 ( 1) 4 ( 1) 4

( 1) ( 1) ( 1) ( 1)

41 4ln 1

Exampl (10) :

Exampl (11) :

Exampl (12) :

dx dxx c

x dx x dxx c

x xx x x

dx dx dx dxx x x x

dxx x c

= = − += = − += = − += = − +− −− −− −− −

− −− −− −− −= − = − += − = − += − = − += − = − +

− −− −− −− −+ + + ++ + + ++ + + ++ + + +

= = += = += = += = ++ + + ++ + + ++ + + ++ + + +

= + = + + += + = + + += + = + + += + = + + + ++++

∫ ∫∫ ∫∫ ∫∫ ∫

∫ ∫ ∫ ∫∫ ∫ ∫ ∫∫ ∫ ∫ ∫∫ ∫ ∫ ∫

∫∫∫∫

2 2 2 2 2

1 1tan 2ln sec

1 1(sec ) sec ( ) ln sec tan

sin cos sin (sin ) sin3

Example(13):

Example(14):

Example(15):

x dx x c

x x dx x d x x x c

x x dx x d x x c

= += += += +

= = + += = + += = + += = + +

= = += = += = += = +

∫∫∫∫

∫ ∫∫ ∫∫ ∫∫ ∫

2 3 2 3 2 4 6

3 7 112 2 2

5 9 132 2 2

(1 ) (1 ) (1 3 3 )

1 3 3( )

6 6 22

5 9 131

( 1) ( 1) ( 1) ( 1)6

Example(16)

Example(17): x x x x x

x x x x xdx dx dx

x x xdx

x x x x

dxx dx x dx x dx

x x x x c

e e dx e d e e c

+ + + + ++ + + + ++ + + + ++ + + + += == == == =

= + + += + + += + + += + + +

= + + + += + + + += + + + += + + + +

+ = + + = + ++ = + + = + ++ = + + = + ++ = + + = + +

∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫

∫∫∫∫

∫ ∫ ∫ ∫∫ ∫ ∫ ∫∫ ∫ ∫ ∫∫ ∫ ∫ ∫

∫ ∫∫ ∫∫ ∫∫ ∫

2 22 5 2

2 5 2 5

1 1(3 )

1 ( 1) 1ln( 1)

2 2( 1) ( 1)

1 ( 1) 1( 1) ( 1)

2 2( 1) ( 1)

1 ( 1)2 4

Example(18):

Example(19):

Example(20):

x xx x

a dx a dx a ca

e d edx dx e c

e dx d ee d e

−−−−

= = += = += = += = +

−−−−= = − += = − += = − += = − +− −− −− −− −

−−−−= = − −= = − −= = − −= = − −− −− −− −− −

−−−−= += += += +−−−−

∫ ∫∫ ∫∫ ∫∫ ∫

∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫

1 1sin4 cos4 , (19) cos 3 sin 3

cos sin cos ( sin )

Example(21):

Example(22):

x dx x c x dx x c

x x dx x x dx

−−−−= + = += + = += + = += + = +

= − −= − −= − −= − −

∫ ∫∫ ∫∫ ∫∫ ∫

3 2 3 4

5 2 5 2

5 2 4 2

1tan sec tan (tan ) tan

cot cosec cot ( cosec )

1cot (cot ) cot

cos sin cos sin (cos )

(1 sin ) sin (cos )

(1 2si

Example(23):

Example(24):

Example(25):

x x dx x d x x c

x x dx x x dx

x d x x c

x x dx x x xdx

x x xdx

= = += = += = += = +

= − −= − −= − −= − −

= − = − += − = − += − = − += − = − +

= −= −= −= −

∫ ∫∫ ∫∫ ∫∫ ∫

∫∫∫∫

∫ ∫∫ ∫∫ ∫∫ ∫

∫∫∫∫

∫∫∫∫2 4 2

2 4 6 3 5 7

4 3 4 2

n sin )sin (cos )

(sin 2sin sin )(cos )

sin cos

1 2 1( 2 )

1 2 1sin sin sin

cos sin cos sin (sin )

cos (1 cos )(sin )

Example(26):

x x x xdx

let y x dy xdx

I y y y dy y y y c

x x x c

x x dx x x x dx

x x dx

= − += − += − += − +

==== ⇒⇒⇒⇒ ====

∴ = − + = − + +∴ = − + = − + +∴ = − + = − + +∴ = − + = − + +

= − + += − + += − + += − + +

= −= −= −= −

∫∫∫∫

∫ ∫∫ ∫∫ ∫∫ ∫

∫∫∫∫

4 6 5 7

(cos cos )(sin )

cos sin

1 1( )

1 1cos cos

sin cos

x x xdx

let y x dy xdx

I y y dy y y c

Try to solve x x dx

x x dx

= −= −= −= −

==== ⇒⇒⇒⇒ = −= −= −= −

∴ = − = − +∴ = − = − +∴ = − = − +∴ = − = − +

= − += − += − += − +

∫∫∫∫

sec tan sec (sec tan )

1sec (sec ) sec

4sin sin 1

tan sec seccos coscos

cos cos 1cot cosec cot

sin sinsin

Example(27):

Example(28):

Example(29):

Example(30)

x x dx x x x dx

x d x c

x xdx dx x x dx x c

= = += = += = += = +

= = = += = = += = = += = = +

= = = − += = = − += = = − += = = − +

∫ ∫∫ ∫∫ ∫∫ ∫

∫∫∫∫

∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫

(1 cos ) (1 cos )1 cos 1 cos sin

sin sin

cosec cot cosec

cot cosec

sec tan sec sec tansec sec

sec tan sec tanln sec tan

Example(31):

Exampl

dx x dx x dxx x x

dx x dx

xdx x x dx

x x x x xx dx x dx dx

x x x xx x c

− −− −− −− −= == == == =++++ −−−−

= −= −= −= −

= −= −= −= −= − + += − + += − + += − + +

+ ++ ++ ++ += == == == =+ ++ ++ ++ +

= + += + += + += + +

∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫

∫ ∫∫ ∫∫ ∫∫ ∫

∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫

cosec cotcosec cosec

cosec cot

cosec cosec cotln cosec cot

cosec cot

( ) sin (2 3cos ) ,

sin sin( ) , ( )

(2 3cos ) (2 3cos )

e(32):

Example(33):

x xx dx x dx

x x xdx x x c

FindThefolowing egrals

i I x x dx

x dx x dxii J iii K

−−−−====−−−−

−−−−= = − += = − += = − += = − +−−−−

= += += += +

= == == == =++++ ++++

∫ ∫∫ ∫∫ ∫∫ ∫

∫∫∫∫

2 3cos 3sin

1 1 2( ) sin (2 3cos ) .

2(2 3cos )

let u x du x dx

i I x x dx u du u

= += += += + ⇒⇒⇒⇒ = −= −= −= −

− −− −− −− − ∴ = + = =∴ = + = =∴ = + = =∴ = + = =

−−−− = + += + += + += + +

∫∫∫∫

∫ ∫∫ ∫∫ ∫∫ ∫

(((( ))))

4 34 4

sin 1 1 2( ) .2 (2 3cos )

3 3 3(2 3cos )

sin 1 1 1 1( ) .

3 3 3 3(2 3cos )

1(2 3cos )

(1 tan ) (1 2tan tan )

1 2tan (sec 1) 2tan

Example(34):

xdx duii J u c x c

xdx duiii K u du u

x dx x x dx

x x dx x

− −− −− −− −

−−−−

− − −− − −− − −− − −= = = + = + += = = + = + += = = + = + += = = + = + +++++

− − − −− − − −− − − −− − − − = = = == = = == = = == = = = ++++

= + += + += + += + +

+ = + ++ = + ++ = + ++ = + +

= + + − == + + − == + + − == + + − =

∫ ∫∫ ∫∫ ∫∫ ∫

∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫

∫ ∫∫ ∫∫ ∫∫ ∫

∫∫∫∫ (((( ))))

(((( ))))(((( ))))

2ln sec tan

(1 cot )

(1 cot ) (1 2cot cot )

1 2cot (cosec 1)

2cot cosec 2ln sin cosec

(tan 3 sec3 ) (tan 3 2tan 3 sec3 sec 3 )

Example(35):

Example(36):

Find x dx

x dx x x dx

x x dx

x x dx x x c

x x dx x x x x dx

= + += + += + += + +

+ = + ++ = + ++ = + ++ = + +

= + + −= + + −= + + −= + + −

= + = − += + = − += + = − += + = − +

+ = + ++ = + ++ = + ++ = + +

∫∫∫∫

∫ ∫∫ ∫∫ ∫∫ ∫

∫∫∫∫

∫ ∫∫ ∫∫ ∫∫ ∫

(((( ))))(((( ))))

sin sin sin

tan 2 tan tan

(sec 3 1) 2tan 3 sec3 sec 3

2sec 3 2tan 3 sec3 1

2 2tan 3 sec3

cos (sin )

1sec (tan )

Example(37):

Example(38):

Example(39):

x x x x dx

x x x dx

x x x c

e x dx e d x e c

a x dx a d x a ca

e x dx

= − + += − + += − + += − + +

= + −= + −= + −= + −

= + − += + − += + − += + − +

= = += = += = += = +

∫∫∫∫

∫ ∫∫ ∫∫ ∫∫ ∫

∫∫∫∫

[[[[ ]]]]

cos cos

(cos )

lnsin )cot lnsin cot

1 1lnsin

Example(40): (

x xe d x e c

x x dx let y x dy xdx

I ydy y c x c

− = − +− = − +− = − +− = − +

==== ⇒⇒⇒⇒ ====

∴ = = + = +∴ = = + = +∴ = = + = +∴ = = + = +

∫∫∫∫

(((( ))))3 2 3

(1 sec ) 1 3sec 3sec sec

1 3sec 3sec 1 tan sec

3 sec 3sec 1 tan (tan )

13ln sec tan tan tan 1 tan

ln(tan 1 tan )2

Example(41):

Example(42):

x dx x x x dx

x x x x dx

dx x dx x dx x d x dx

x x x x x x

+ = + + ++ = + + ++ = + + ++ = + + +

= + + + += + + + += + + + += + + + +

= + + + + += + + + + += + + + + += + + + + +

+ + + ++ + + ++ + + ++ + + +

∫ ∫∫ ∫∫ ∫∫ ∫

∫∫∫∫

sin 21 cos 2sec2 cot 2 1 cos 2

sin 2 sin 21 cos 2 2sin 2

sin 2 1 1 1ln ln 1 cos 2

1 cos 2 2 2 2

dx xdxdx

xx x xx x

let u x du xdx

x dx duu x c

= == == == =− −− −− −− −−−−−

= −= −= −= − ⇒⇒⇒⇒ ====

∴ = = = − +∴ = = = − +∴ = = = − +∴ = = = − +−−−−

∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫

∫ ∫∫ ∫∫ ∫∫ ∫

(((( ))))(((( ))))(((( ))))(((( ))))

find (1 tan )

(1 tan ) (1 3tan 3tan tan )

1 3tan 3(sec 1) tan tan

1 3tan 3sec 3 tan (sec 1)

1 3tan 3sec 3 tan sec tan

2 2tan 3sec tan sec

Example(43): x dx

x dx x x x dx

x x x x dx

x x x x x dx

x x x x dx

+ = + + ++ = + + ++ = + + ++ = + + +

= + + − += + + − += + + − += + + − +

= + + − + −= + + − + −= + + − + −= + + − + −

= − + + += − + + += − + + += − + + +

= −= −= −= −

∫∫∫∫

∫ ∫∫ ∫∫ ∫∫ ∫

∫∫∫∫

2tan 3sec tan sec

12 2ln sec 3tan tan

Example(44):

Example(45):

x x dx x dx x x dx

x x x x c

dx xdx c

−−−−

+ + ++ + ++ + ++ + +

= − + + + += − + + + += − + + + += − + + + +

= += += += +−−−−

= += += += +++++

∫ ∫ ∫ ∫∫ ∫ ∫ ∫∫ ∫ ∫ ∫∫ ∫ ∫ ∫

∫∫∫∫

(1 cos ) (1 3cos 3cos cos )

31 3cos (1 cos 2 ) cos cos

3 31 3cos cos 2 cos (1 sin )

5 34cos cos 2 sin cos

5 3 14sin sin 2 sin

Example(46):

x dx x x x dx

x x x x dx

x x x x c

+ = + + ++ = + + ++ = + + ++ = + + +

= + + + += + + + += + + + += + + + +

= + + + + −= + + + + −= + + + + −= + + + + −

= + + −= + + −= + + −= + + −

= + + − += + + − += + + − += + + − +

∫ ∫∫ ∫∫ ∫∫ ∫

∫∫∫∫

(1 sin ) (1 3sin 3sin sin )

31 3sin (1 cos 2 ) sin sin

3 31 3sin cos 2 sin (1 cos )

5 34sin cos 2 sin cos

5 3 14cos sin 2 cos

Example(47):

x dx x x x dx

x x x x dx

x x x x c

+ = + + ++ = + + ++ = + + ++ = + + +

= + + − += + + − += + + − += + + − +

= + + − + −= + + − + −= + + − + −= + + − + −

= + + −= + + −= + + −= + + −

= − + + += − + + += − + + += − + + +

∫ ∫∫ ∫∫ ∫∫ ∫

∫∫∫∫

(((( ))))

1 12 22 3

4 91 1 2 2 1 2

. tan tan4 4 3 3 6 34 9

Example(48): dxfind

xdx dx x x

c cx x

− −− −− −− −

= = + = += = + = += = + = += = + = +++++ ++++

∫∫∫∫

∫ ∫∫ ∫∫ ∫∫ ∫

(((( ))))1 1

2 2 24 29 3

12 2 2

Another solution:

1 1 1 3 2 1 2. tan tan

9 9 9 2 3 6 34 9 1 1

Another solution:

1 2 1 1 2. tan

2 2 3 34 9 (2 ) 3

dx dx dx x xc

dx dx xc

− −− −− −− −

−−−−

= = = = += = = = += = = = += = = = ++ ++ ++ ++ + ++++

= = += = += = += = ++ ++ ++ ++ +

∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫

∫ ∫∫ ∫∫ ∫∫ ∫

(((( )))) (((( ))))

12 2 2

2 2 31 3

6 3 2 3 2

where f(x)is linear

2 1 2sec

3 34 9 2 (2 ) (3)

( ) 1 ( )According to sec

( ) ( )

1 3 1 ( ) 1sin

3 3 31 1 ( ) 1 ( )

Example(49):

Example(50):

Example(51):

dx dx xc

x x x x

f x f xc

a af x f x a

x dx x dx d xx c

−−−−

= = += = += = += = +− −− −− −− −′′′′

= += += += +−−−−

= = = += = = += = = += = = +− − −− − −− − −− − −

∫ ∫∫ ∫∫ ∫∫ ∫

∫∫∫∫

∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫

4 2 2 2 2

(cos )tan (cos )

1 cos 1

cos sin 1 2cos sin 1 (cos )2 2cos 1 (cos ) 1 (cos ) 1

tan (cos )

Another solution: cos 2cos sin

cos sin 1 1ta

2 2cos 1 1

Example(52):

d xx c

x xdx x xdx d x

let u x du x xdx

x xdx du

−−−−

= − = − += − = − += − = − += − = − +++++

−−−−= == == == =+ + ++ + ++ + ++ + +

= += += += +

==== ⇒⇒⇒⇒ ====−−−−∴ = =∴ = =∴ = =∴ = =

+ ++ ++ ++ +

∫ ∫∫ ∫∫ ∫∫ ∫

∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫

∫ ∫∫ ∫∫ ∫∫ ∫1 1 2

12 2 2

1n tan (cos )

1 4 1 4sin

4 4 525 16 5 (4 )

( 2)sin

24 ( 2) 2 ( 2)

3 4 3 4(3 4)

34 4tan

Example(53):

Example(54):

Example(55):

Example

u c x c

dx dx xc

x x xdx x dx

xx x c

− −− −− −− −

−−−−

−−−−+ = ++ = ++ = ++ = +

= = += = += = += = +− −− −− −− −

++++= = += = += = += = +− + − +− + − +− + − +− + − +

− +− +− +− + = − += − += − += − + + ++ ++ ++ +

= − + += − + += − + += − + +

∫ ∫∫ ∫∫ ∫∫ ∫

2 2 2 2

4 13 ( 4 4) 9 ( 2) 3

1 ( 2)tan

(56): dx dx dx

x x x x x

xc−−−−

= == == == =+ + + + + + ++ + + + + + ++ + + + + + ++ + + + + + +

++++= += += += +

∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫

20 8 20 ( 8 ) 36 ( 8 16 16)

( 4)sin

66 ( 4)

( 3) 1 2 6 1 ( 2 4) 22 25 4 5 4 5 4

1 ( 2 2) 1 22 25 4 5 41 ( 2 2

Example(57):

Example(58):

dx dx dx

x x x x x x

x x xdx dx dx

x x x x x xx

dx dxx x x x

−−−−

= == == == =+ − − − − − + −+ − − − − − + −+ − − − − − + −+ − − − − − + −

−−−−= = += = += = += = +− −− −− −− −

+ − − − − − − −+ − − − − − − −+ − − − − − − −+ − − − − − − −= == == == =− − − − − −− − − − − −− − − − − −− − − − − −

− − − − −− − − − −− − − − −− − − − −= += += += +− − − −− − − −− − − −− − − −

− − −− − −− − −− − −====

∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫

∫∫∫∫

∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫

∫ ∫∫ ∫∫ ∫∫ ∫

5 4 9 ( 2)

1 ( 2).2 5 4 sin

Another solution (5 4 ) 2 4

put 3 ( (5 4 )) ( 2 4)

1 1, 1, 3 ( 2 4) 1,

dx dxx x x

xx x c

dx x x

x A x x B A x Bdx

A B x x

−−−−

+ =+ =+ =+ =− − − +− − − +− − − +− − − +

− +− +− +− += − − + += − − + += − − + += − − + +

− − = − −− − = − −− − = − −− − = − −

+ = − − + = − − ++ = − − + = − − ++ = − − + = − − ++ = − − + = − − +

− −− −− −− −∴ = = + = − − +∴ = = + = − − +∴ = = + = − − +∴ = = + = − − +

∫ ∫∫ ∫∫ ∫∫ ∫

( 2 4) 1( 3)

5 4 5 4

( 2 4) 1 ( 2 2)25 4 5 4 5 4

1 ( 2).2 5 4 sin

2 39 ( 2)

xxdx dx

x x x x

x dx dx x dx

x x x x x xdx x

x x cx

−−−−

− − +− − +− − +− − +++++ ====− − − −− − − −− − − −− − − −

− −− −− −− − − − −− − −− − −− − −= + == + == + == + =

− − − − − −− − − − − −− − − − − −− − − − − −− +− +− +− +

+ = − − + ++ = − − + ++ = − − + ++ = − − + +− +− +− +− +

∫ ∫∫ ∫∫ ∫∫ ∫

∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫

∫∫∫∫

9 12 81 39

2 3 (18 12) , 3 129 9

Example(59): xdx

put x A x B A B A

++++− +− +− +− +

+ = − ++ = − ++ = − ++ = − + ⇒⇒⇒⇒ = = + == = + == = + == = + =

∫∫∫∫

2 3 1 (18 12) 3999 12 8 9 12 8

1 (18 12) 1 399 99 12 8 9 12 81 (18 12) 1 399 99 12 8 (9 12 4) 4

1 (18 12) 1 399 99 12 8 (3 2) 4

1 39 1 1 (3ln 9 12 8 . . tan

9 9 3 2

x xdx dx

x x x xx

dx dxx x x x

xdx dx

x x x x

xdx dx

xx x −−−−

+ − ++ − ++ − ++ − +∴ =∴ =∴ =∴ =− + − +− + − +− + − +− + − +

−−−−= += += += +− + − +− + − +− + − +− + − +

−−−−= += += += +− + − + +− + − + +− + − + +− + − + +

−−−−= += += += +− + − +− + − +− + − +− + − +

−−−−= − + += − + += − + += − + +

∫ ∫∫ ∫∫ ∫∫ ∫

[[[[ ]]]][[[[ ]]]]

2 2 2 2

22 (4 2 )

, 4, 2 (4 2 ) 82 2

(4 2 ) 82 1 1 (4 2 ) 1 82 2 24 4 4 4

1 (4 2 ) 1 82 24 4 (4 4 )

1 (4 2 ) 1 84

2 24 4 ( 2)

Example(60): xdx put x A x B

A B x x

xx xdx dx dx dx

x x x x x x x xx dx dx

x x x x

x dx dxx

++++ + = − ++ = − ++ = − ++ = − +−−−−

− −− −− −− −∴ = = + = − +∴ = = + = − +∴ = = + = − +∴ = = + = − +

− +− +− +− ++ − − −+ − − −+ − − −+ − − −= = −= = −= = −= = −− − − −− − − −− − − −− − − −

− −− −− −− −= −= −= −= −− − − +− − − +− − − +− − − +

− −− −− −− −= − = − −= − = − −= − = − −= − = − −− − −− − −− − −− − −

∫∫∫∫

∫ ∫ ∫ ∫∫ ∫ ∫ ∫∫ ∫ ∫ ∫∫ ∫ ∫ ∫

∫ ∫∫ ∫∫ ∫∫ ∫

∫ ∫∫ ∫∫ ∫∫ ∫2 11 ( 2)

sin2 2

xx c−−−− −−−−− +− +− +− +

Exercise(1)

Integrate the following functions with respect to x :

2 2 3 72

1 2(1) (3 2 4 ) (2)( 3)( 4) (3)( )

1(4)( 2) (5)(3 1) (6)

( 1)(7) (2 3) (8) (3 4) (9)

x x x x x xx

xx x x x

− + − + + +− + − + + +− + − + + +− + − + + +

+ −+ −+ −+ −−−−−+++++ −+ −+ −+ −

+ ++ ++ ++ +

3 sin tan 2

(1 ) 1(10) (11) (12)

2 2 2(13) (14) (15) ( 2)

(16)10 (17) cos (18)5 sec

sec 2(19) sin (3 2cos ) (20) (21)

3 5tan 3

ln( 1) (1 ln ) ( )(22) (23) (24)

x x xxx x

x x xx x

x ex x

x x x e exx

−−−−

+ −+ −+ −+ −++++++++

+ + ++ + ++ + ++ + + ++++++++ + ++ ++ ++ +

−−−−++++−−−− ++++

+ + ++ + ++ + ++ + +++++

5 2 4 22

(25)(cos sin ) (26)sin cos (27)cos sin

sin8(28) tan sec (29)cot cosec (30)

9 sin 4sin 1 cos 2 cot ((1 cosec )

(31) (32) (33)2 sin 2 cosec(1 cos )

sec 3 1 1(34) (35) (36)

1 sin 2 1 cos 2(1 tan 3 )

(1(37)

x x x x x x

xx x x

xx x x x

x x xx

−−−−

+++++ ++ ++ ++ +++++++++

− −− −− −− −++++3 5

cot ) (1 tan) 1(38) (39)

1 sin 2 1 cos 2 5

sec sin cos(40) (41) (42)

1 cos 24 tan 1

xx x x

x e x x

+ ++ ++ ++ +− −− −− −− − ++++

++++− −− −− −− −

tan cot 1(43) (44) (45)

2 8cos 4 sin 41 2 3 1

(46) (47) (48)6 13 3 4 32 8

2 1(49) (50)

27 6 12 4

x xx xx x

x x x xx xx x

x x x x

+ −+ −+ −+ −− −− −− −− −− −− −− −− −

+ + − ++ + − ++ + − ++ + − ++ −+ −+ −+ −−−−−

+ − + −+ − + −+ − + −+ − + −

Integration of Hyperbolic Functions

For x any real number we define Hyperbolic functions as follows:

1 1 2(1) sinh ( ) (4)cosech

2 sinh ( )

1 1 2(2) cosh ( ) (5)sech

2 cosh ( )

sinh ( ) 1 ( )(3) tanh , (6)coth

cosh tanh( ) ( )

x xx x

x x x x

x e e xx e e

x e e e ex x

x xe e e e

−−−−−−−−

− −− −− −− −

= − = == − = == − = == − = =−−−−

= + = == + = == + = == + = =++++

− +− +− +− += = = == = = == = = == = = =+ −+ −+ −+ −

and hyperbolic functions satisfy the following lows: 2 2

(1) cosh sinh 1

(2) 1 tanh sech

(3) coth 1 cosech

(4) sinh( ) sinh cosh cosh sinh

(5) sinh( ) sinh cosh cosh sinh

(6) cosh( ) cosh cosh sinh sinh

(7) cosh( ) cosh cosh sinh sinh

(8) sinh 2

x y x y x y

− =+ = +− = −+ = +− = −

2sinh cosh

(9) cosh 2 cosh sinh

1(10) sinh cosh 2 1

(11) cosh cosh 2 12

(12) cosh sinh

(13) cosh sinh

2tanh(14)tanh 2

1 tanh

we can proof this lows by using the definition in the following we stat integration formula for hyperbolic functions

(1) sinh cosh

(2) cosh sinh

(3) tanh lncosh

(4) coth ln sinh

(5) sech tanh

(6) cosech coth

(7) sech tanh sech

(8) cosech coth cosech

1(9) sinh

x dx x c

x x dx x c

bb x a

−−−−

= += += += += += += += += += += += +

= += += += +

= − += − += − += − += − += − += − += − +

= − += − += − += − +

====++++

∫∫∫∫

∫∫∫∫1

12 2 2

1(10) cosh

Solved Examples:

1 cosh cosh(1) sech

cosh cosh 1 si

dx bxc

b ab x a

dx ab aa bxa b x c

ab a bx

dx ab abx ab x a c

ab bx a

x xx dx dx dx

−−−−

= += += += +−−−−

++++==== ++++−−−− ++++ −−−−

++++==== −−−−−−−− ++++ ++++

= = == = == = == = =++++

∫∫∫∫

∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫1

tan (sinh )nh

1 1 1(2) sinh (cosh2 1) sinh2

(3) cosh 2 (1 sinh 2 )cosh2 cosh2 sinh 2 cosh2

1 1 sinh 2sinh2

dx x cx

x dx x dx x x c

x dx x x dx x dx x x dx

−−−−= += += += +

= − = − += − = − += − = − += − = − +

= + = += + = += + = += + = +

= + += + += + += + +

∫∫∫∫

∫ ∫∫ ∫∫ ∫∫ ∫

∫ ∫ ∫ ∫∫ ∫ ∫ ∫∫ ∫ ∫ ∫∫ ∫ ∫ ∫

5 53 3

8 2 8 2

(4) cosh2 .2

1 1 1( ) ( )

(5) sinh52

1 1 1 1( ) ( )

2 2 8 2

x xx x

x x x x

x xx x

x x x x

e ee x dx e dx

e e dx e e c

e ee x dx e dx

e e dx e e c

−−−−

− −− −− −− −

−−−−

− −− −− −− −

++++====

= + = − += + = − += + = − += + = − +

−−−−====

= − = + += − = + += − = + += − = + +

∫ ∫∫ ∫∫ ∫∫ ∫

∫∫∫∫

∫ ∫∫ ∫∫ ∫∫ ∫

∫∫∫∫

Solved Examples

[[[[ ]]]]

1: sinh3 cosh3

31 cosh cosh

: sechcosh cosh 1 sinh

(sinh )tan (sinh )

1 sinh1 1 1

: sinh cosh2 1 sinh22 2 2

: cosh 3

Example(1)

Example(2)

Example(3)

Example(4)

x dx x c

x xx dx dx dx dx

x x xd x

x dx x dx x x c

−−−−

= += += += +

= = == = == = == = =++++

= = += = += = += = +++++

= − = − += − = − += − = − += − = − +

∫∫∫∫

∫ ∫ ∫ ∫∫ ∫ ∫ ∫∫ ∫ ∫ ∫∫ ∫ ∫ ∫

∫∫∫∫

∫ ∫∫ ∫∫ ∫∫ ∫

∫∫∫∫ [[[[ ]]]]

2 2 2 3

1 1 1cosh6 1 sinh6

1: tanh 5 1 sech 5 tanh5

: sech 1 tanh sech

1sech tanh sech tanh tanh

Example(5)

Example(6)

x x dx x x c

x dx x dx x x c

x dx x x dx

x dx x x dx x x c

= + = + += + = + += + = + += + = + +

= − = − += − = − += − = − += − = − +

= −= −= −= −

= − = − += − = − += − = − += − = − +

∫∫∫∫

∫ ∫∫ ∫∫ ∫∫ ∫

: sinh 4 sinh 4 (sinh4 )

(cosh 4 1)(sinh4 )

(cosh 4 (sinh4 ) (sinh4 )

1 1cosh 4 cosh4

Example(7) x dx x x dx

x x dx

x x dx x dx

= −= −= −= −

= − += − += − += − +

∫ ∫∫ ∫∫ ∫∫ ∫

∫∫∫∫

∫ ∫∫ ∫∫ ∫∫ ∫

3 33 3

4 24 2

3 32 2 2 3 3

( ) 1: cosh3 ( )

1 1( )

2 2 4 2

( ) 1: sinh3 ( )

1 1( )

Example(8)

Example(9)

x xx x x x x

x xx x

x xx x x x x

xx x x

e ee x dx e dx e e e dx

e ee e dx c

e ee x dx e dx e e e dx

ee e dx e c

−−−−−−−−

− −− −− −− −

++++= = += = += = += = +

= + = + += + = + += + = + += + = + + −−−−

−−−−= = −= = −= = −= = −

= + = + += + = + += + = + += + = + +

∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫

∫∫∫∫

∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫

∫∫∫∫

(((( ))))

1 1: sinh

2 21 12 21 12 21 12 2cosh sinh

Try to solve sinh by parts

: sinh16

Example(10)

Example(11)

x x x x

x x dx x e e dx xe xe dx

xe dx xe dx

xe e xe e c

x e e e e c

x x x c

x x dx

− −− −− −− −

−−−−

− −− −− −− −

−−−−

= − = −= − = −= − = −= − = −

= −= −= −= −

= − − − − += − − − − += − − − − += − − − − +

= − − − += − − − += − − − += − − − +

= − += − += − += − +

====++++

∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫

∫ ∫∫ ∫∫ ∫∫ ∫

∫∫∫∫

: cosh525

: sinh 4 sinh 4 (sinh4 )

(cosh 4 1)(sinh4 )

(cosh 4 (sinh4 ) (sinh4 )

1 1cosh 4 cosh4

Example(12)

Example(13)

x dx x x dx

x x dx

x x dx x dx

−−−−

= += += += +−−−−

= −= −= −= −

= − += − += − += − +

∫∫∫∫

∫ ∫∫ ∫∫ ∫∫ ∫

∫∫∫∫

∫ ∫∫ ∫∫ ∫∫ ∫

Exercise(2)

Integrate the following functions: 2 2

3 4 5 2

(1)sinh3 (2)cosesh (3) sinh

(4)sinh cosh (5) sinh3 (6) cosh3

sech(7) sinh (3 2cosh ) (8) (9)(cosh sinh )

3 5tanh

(10)sinh cosh (11)cosh sinh (12)tanh sech

sinh8(13)coth cosech (14)

9 sinh

x x x x

x x e x e x

xx x x x

x x x x x x

+ ++ ++ ++ +−−−−

++++ 2

sinh cosh(15)

4 1 cosh 2sinh 1 cosh2 coth ((1 cosech )

(16) (17) (18)2 sinh2 cosech(1 cosh )

sech 3 1 1(19) (20) (21)

1 sinh2 1 cosh2(1 tanh3 )

(1 coth ) (1 tanh) sec(22) (23) (24)

1 sinh2 1 cosh2 4 tanh

x xx x x x

x x xx

x xx x x

+++++ ++ ++ ++ +++++++++

− +− +− +− +++++

+ ++ ++ ++ +− −− −− −− − −−−−

Methods of Integration: (1) Integration by parts

When u and v are differentiable functions then ( )

d uv udv v du

udv d uv v du

= += += += +

= −= −= −= −

and by integrate ( ) (1)udv d uv v du= −= −= −= −∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫ to apply this rule we refer to our problem by the integral

u dv∫∫∫∫ and we must separate it into two parts one part being u and the other being dv and we find du and v by differentiation u and integrate dv . Note that It is very important how to chose the function to be integrated, and the function to be differentiated such that the integration on the right side in (1) is much easier to evaluate than the one on the left.

Solved Examples: Example (1): Find xx e dx∫∫∫∫ Solution:

If we chose xu e==== to be differentiated and dv xdx==== to be integrated

2 21 12 2

x x xxe dx x e x e dx∴ = −∴ = −∴ = −∴ = −∫ ∫∫ ∫∫ ∫∫ ∫

and its clear that the integration in the R.H.S is more difficult than the given integration then

we use the partation as follows

by substituting in the rule then

u x dv e dx

du dx v e

x e dx x e e dx

= == == == =

= −= −= −= −∫ ∫∫ ∫∫ ∫∫ ∫

Note that : The integral in the right side xe dx∫∫∫∫ is simple than the

integral xx e dx∫∫∫∫ Finally x xI x e e c= − += − += − += − + .

Example (2): Find: 2 lnx x dx∫∫∫∫ Solution: Consider the partition 2lnu x dv x dx= == == == =

Then 31

du dx vx

= == == == =

Substitute in the rule we have: 3 3 3 3 3

21 1ln ln ln

3 3 3 3 3 9x x x x x

I x dx x x dx x cx

∴ = − = − = − +∴ = − = − = − +∴ = − = − = − +∴ = − = − = − +∫ ∫∫ ∫∫ ∫∫ ∫

Try to solve ln ,nnI x x dx n= ∈= ∈= ∈= ∈∫∫∫∫ �

Example (3): Find 1x x dx++++∫∫∫∫ Solution: Let 1u x dv xdx= = += = += = += = +

du dx v x∴ = = +∴ = = +∴ = = +∴ = = +

by using partition rule ( )udv d uv v du= −= −= −= −∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫

3 3 3 52 2 2 22 2 2 2 2

(1 ) (1 ) (1 ) ( )( )(1 )3 3 3 3 5x x

I x x dx x x c∴ = + − + = + − + +∴ = + − + = + − + +∴ = + − + = + − + +∴ = + − + = + − + +∫∫∫∫

(try with thenew partation what hapen)

try tosolve

( ) ,n

u x dv xdx

x ax b dx n

= + == + == + == + =

+ ∈+ ∈+ ∈+ ∈∫∫∫∫ �

Example (4): Find sinx x dx∫∫∫∫ Solution: Let sinu x dv xdx= == == == = cosdu dx v x∴ = = −∴ = = −∴ = = −∴ = = −

by using partition rule ( )udv d uv v du= −= −= −= −∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫

sin cos cos cos sinI x x dx x x xdx x x x c∴ = = − + = − + +∴ = = − + = − + +∴ = = − + = − + +∴ = = − + = − + +∫ ∫∫ ∫∫ ∫∫ ∫

(try with thepartation what hapen)

try to solve is positive integer

u x dv xdx

I x x dx n

= == == == =

==== ∫∫∫∫

Example (5): Find cosx x dx∫∫∫∫ Solution: let cos

u x dv x dx

du dx v x

= == == == =∴ = =∴ = =∴ = =∴ = =

sin sin sin cosI x x x dx x x x c∴ = − = − +∴ = − = − +∴ = − = − +∴ = − = − +∫∫∫∫

try tosolve is positive integercosnnI x x dx n==== ∫∫∫∫

Example (6): Find 2sinx x dx∫∫∫∫ Solution:

let u x dv xdx

du xdx v x

= == == == =∴ = = −∴ = = −∴ = = −∴ = = −

by substituting in the rule

2 cos 2 cos (1)

cos (5)

cos cos

I x x x x dx

we can solve x xdx by parts as in example

x x dx let u x dv x dx

du dx v x

∴ = − +∴ = − +∴ = − +∴ = − +

= == == == =∴ = =∴ = =∴ = =∴ = =

∫∫∫∫

from(2) in (1)

sin sin sin cos (2)I x x x dx x x x∴ = − = −∴ = − = −∴ = − = −∴ = − = −∫∫∫∫

cos 2 cos cos 2( sin cos )

cos 2 sin 2cos

I x x x x dx x x x x x c

x x x x x c

∴ = − + = − + − +∴ = − + = − + − +∴ = − + = − + − +∴ = − + = − + − +

= − + − += − + − += − + − += − + − +

∫∫∫∫

Example (7): Find 2cosx x dx∫∫∫∫ Solution:

u x dv x dx

du xdx v x

= == == == =∴ = =∴ = =∴ = =∴ = =

sin 2 sin sin 2( cos sin )

sin 2 cos 2sin

I x x x x dx x x x x x c

x x x x x c

∴ = − = − − + +∴ = − = − − + +∴ = − = − − + +∴ = − = − − + +

= + − += + − += + − += + − +

∫∫∫∫

Example (8): Find 2 xx e dx∫∫∫∫ Solution:

2 2( )

x x x x x

u x dv e dx

du xdx v e

I x e x e dx x e xe e c

= == == == =

∴ = =∴ = =∴ = =∴ = =

∴ = − = + − +∴ = − = + − +∴ = − = + − +∴ = − = + − +∫∫∫∫

where is a positive integerm xmtry to solve I x e dx m==== ∫∫∫∫

Example (9): Find 1sin x dx−−−−

∫∫∫∫ Solution:

u x dv dx

dxdu let v x

−−−−= == == == =

∴ = =∴ = =∴ = =∴ = =−−−−

1 12 2

1 2sin sin

x dx x dxI udv uv v du x x x x

− −− −− −− −= = − − = −= = − − = −= = − − = −= = − − = −− −− −− −− −

∫ ∫ ∫ ∫∫ ∫ ∫ ∫∫ ∫ ∫ ∫∫ ∫ ∫ ∫

1 1 22

1 2 1sin sin .2 1

x dxx x x x x c

− −− −− −− −−−−−= + = + − += + = + − += + = + − += + = + − +−−−−

∫∫∫∫

try tosolve where is a positive integer1sinx x dx m−−−−∫∫∫∫

Example (10): Find 1tan x dx−−−−

∫∫∫∫ Solution

u x dv dx

dxdu let v x

−−−−= == == == =

∴ = =∴ = =∴ = =∴ = =++++

1 1 12 2

1 2tan tan tan

x dx x dxudv x dx x x x x

x x− − −− − −− − −− − −∴ = = − = −∴ = = − = −∴ = = − = −∴ = = − = −

+ ++ ++ ++ +∫ ∫ ∫ ∫∫ ∫ ∫ ∫∫ ∫ ∫ ∫∫ ∫ ∫ ∫

1 21tan ln(1 )

2x x x c−−−−= − + += − + += − + += − + +

Example (11): Find sin , cosax axI e bx dx J e bx dx= == == == =∫ ∫∫ ∫∫ ∫∫ ∫ Solution: These integrals are of importance in the theory of electric currents, if each integral is evaluated by parts, the other one is obtained.

let u e dv bx dx

du a e dx v bxb

= == == == =

−−−−∴ = =∴ = =∴ = =∴ = =

(((( ))))

1 1sin cos cos

cos cos

cos (1)

ax ax ax

I e bx dx e bx bx a e dxb b

e abx e bx dx

e aI bx J

J e bx dx

− −− −− −− − = = −= = −= = −= = −

−−−−= += += += +

−−−−∴ = +∴ = +∴ = +∴ = +

∫ ∫∫ ∫∫ ∫∫ ∫

∫∫∫∫

Similarly taking the second integral J let cos

1 1 1sin sin . sin sin

1sin (2)

ax ax ax ax

u e dv bx dx

du ae dx v bxb

aJ e bx bx a e dx e bx e bx dx

b b b b

aJ e bx I

= == == == =

∴ = =∴ = =∴ = =∴ = =

∴ = − = −∴ = − = −∴ = − = −∴ = − = −

= −= −= −= −

∫ ∫∫ ∫∫ ∫∫ ∫

from (1),(2) we get 2

1cos sin cos sin

cos sin

ax axax ax

e a a e a aI bx e bx I bx e bx I

b b b b b b b

a e aI I bx e bx

− −− −− −− − ∴ = + − = + −∴ = + − = + −∴ = + − = + −∴ = + − = + −

−−−−+ = ++ = ++ = ++ = +

2 2 2 2

(1 ) ( ) cos sin

cos sin

a a b e aI I bx e bx

bb b b

b e aI bx e bx

bb a b

b ae bx e bx c

b a b a

+ −+ −+ −+ −+ = = ++ = = ++ = = ++ = = +

−−−−∴ = +∴ = +∴ = +∴ = + ++++

−−−− = + += + += + += + + + −+ −+ −+ −

[[[[ ]]]]

and from(2)

2 2 2 2

sin cos sin

1 1sin sin cos

1sin cos

11 sin cos

axax ax

eI e bx dx b bx a bx c

a a e aJ e bx I e bx bx J

b b b b b b

ae ae bx bx J

a a a b aeJ J J J e bx bx

bb b b b

= = − + += = − + += = − + += = − + +++++

−−−−= − = − += − = − += − = − += − = − +

= + −= + −= + −= + −

+++++ = + = = ++ = + = = ++ = + = = ++ = + = = +

∴ =∴ =∴ =∴ =

∫∫∫∫

[[[[ ]]]]2 2s sin cosaxe

bx dx b bx a bx cb a

= + += + += + += + +++++

∫∫∫∫

in the integrals sinh , coshax axI e bx dx J e bx dx= == == == =∫ ∫∫ ∫∫ ∫∫ ∫ we use the diffination of the hyperbolic functions sinh , coshbx bx as a functions

of xe then ( ) ( )

( ) ( )

sinh ,2 2

cosh ,2 2

bx bx a b x a b xax ax

e e e eIh e bx dx e dx dx

e e e eJh e bx dx e dx dx

− + −− + −− + −− + −

− −− −− −− −= = == = == = == = =

+ ++ ++ ++ += = == = == = == = =

∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫

Example (12): Find 5 22(1 )

++++∫∫∫∫

Solution:

consider

5 25 2

3 23 2

(1 )(1 )

dxI x dx

dxJ x dx

−−−−

= = += = += = += = +++++

∫ ∫∫ ∫∫ ∫∫ ∫

3 2 5 2

3 2 3 2 5 2

3 2 3 2 5 2 3 2

2 2 2 2

3 (1 )

(1 ) 3 (1 )

(1 ) 3 (1 1)(1 )

(1 ) 3 (1 ) (1 )

(1 ) 3 (1 ) (1 ) (1 ) 3 3

u x dv dx

du x x dx v x

J x x x x dx

x x x x dx

x x x x dx x x J I

−−−−

− −− −− −− −

− − −− − −− − −− − −

− − − −− − − −− − − −− − − −

= + == + == + == + =

∴ = − + =∴ = − + =∴ = − + =∴ = − + =

= + + += + + += + + += + + +

= + + + − += + + + − += + + + − += + + + − +

= + + + − + = + + −= + + + − + = + + −= + + + − + = + + −= + + + − + = + + −

∴ =∴ =∴ =∴ =

∫∫∫∫

3 22) 2x J−−−−+ ++ ++ ++ +

To solve consider with the partation

3 2 1 2

1 2 1 2

2 2 2 3 2 2 2 2 3 2

2 2 1 2 2

(1 ) 2(1)

(1 ) (1 )

(1 ) (1 ) (1 ) (1 1)(1 )

(1 ) (1 ) (1 )

x xI J

dx dxJ K

u x dv dx

du x x dx v x

K x x x x dx x x x x dx

x x x x

−−−−

− −− −− −− −

−−−−

− −− −− −− −

−−−−

++++= += += += +

= == == == =+ ++ ++ ++ +

= + == + == + == + =

= − + == − + == − + == − + =

= + + + = + + + − += + + + = + + + − += + + + = + + + − += + + + = + + + − +

= + + + − += + + + − += + + + − += + + + − +

∫ ∫∫ ∫∫ ∫∫ ∫

∫∫∫∫

Substitute in(1)

1 21 2

(1 )(1 )

(1 ) 23 3(1 )

dx x x K J

xJ x x c

x x xI c

−−−−

−−−− = + + −= + + −= + + −= + + −

= + = += + = += + = += + = +++++

++++∴ = + +∴ = + +∴ = + +∴ = + +++++

Example (13): Find 3sec x dx∫∫∫∫ Solution:

sec sec sec

sec sec

sec tan tan

sec sec tan sec tan sec tan sec (sec 1)

sec tan (sec sec ) sec tan (sec sec

2 sec sec tan sec sec tan

x dx x x dx

u x dv x dx

du x x dx v x

x dx x x x x dx x x x x dx

x x x x dx x x x dx x dx

x dx x x x dx x x

= == == == =

∴ = − = − −∴ = − = − −∴ = − = − −∴ = − = − −

= − − = − += − − = − += − − = − += − − = − +

= + == + == + == + =

∫ ∫∫ ∫∫ ∫∫ ∫

∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫

∫ ∫∫ ∫∫ ∫∫ ∫

ln sec tan

1 1sec sec tan ln sec tan

x dx x x x x c

+ + ++ + ++ + ++ + +

∴ = + + +∴ = + + +∴ = + + +∴ = + + +∫∫∫∫

Example (14): Find 2 2I x a dx= += += += +∫∫∫∫ Solution:

2 2 2 2

2 2 2 22 2 2 2 2 2

2 2 2 2

2 2 22 2

2 2 2 2

2 2 2 2 2 1

2 sinh

I x a dx let u x a dv dx

x dxdu v x

x dx x a a dxI x a dx x x a x x a

x a x a

x a dx a dxx x a

x a x a

xx x a x a dx a

xx a dx x x a a

xx a dx x

−−−−

= + = + == + = + == + = + == + = + =

= == == == =++++

+ −+ −+ −+ −= + = + − = + −= + = + − = + −= + = + − = + −= + = + − = + −+ ++ ++ ++ +

++++= + − += + − += + − += + − ++ ++ ++ ++ +

= + − + += + − + += + − + += + − + +

∴ + = + +∴ + = + +∴ + = + +∴ + = + +

∴ + =∴ + =∴ + =∴ + =

∫∫∫∫

∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫

∫ ∫∫ ∫∫ ∫∫ ∫

∫∫∫∫

∫∫∫∫2

2 1sinh2

a xa c

a−−−−+ + ++ + ++ + ++ + +

Example (15): Find 2 2I a x dx= −= −= −= −∫∫∫∫ Solution:

2 2 2 22 2 2 2 2 2

2 2 2 2

2 2 22 2

2 2 2 2

2 2 2 2 2 1

22 2 2 2 1

sin2 2

let u a x dv dx

x dxdu v x

x dx a x a dxI x a dx x a x x a x

a x a x

a x dx a dxx a x

a x a x

xx a x a x dx a

xa x dx x a x a

x a xa x dx a x

−−−−

= − == − == − == − =

−−−−= == == == =−−−−

− − −− − −− − −− − −= + = − − = − −= + = − − = − −= + = − − = − −= + = − − = − −− −− −− −− −

−−−−= − − += − − += − − += − − +− −− −− −− −

= − − − += − − − += − − − += − − − +

∴ − = − +∴ − = − +∴ − = − +∴ − = − +

∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫

∫ ∫∫ ∫∫ ∫∫ ∫

∫∫∫∫

∫∫∫∫ c++++

Exercise(3)

Integrate the following function with respect to x : 2 3

4 2 3 2

1 1 1 1

(1) (i) sin (ii) sin3 (iii) sin (iv) cos

(2) (i) ln (ii) ln (iii) ln (iv) ln

(3) (i) (ii) (iii) (iv) sin

(4) (i) cos (ii) sin (iii) tan (iv) cot

(5) (i) sin (ii) sin cos (

x x x x

x x x x x x x x

x e x e x e e x

x x x x

x x x x x

−−−−

− − − −− − − −− − − −− − − −

2 1 23

1 1 1 3

3 2 5 5 3

iii) sec (iv) sinh

ln(6) (i) sin (ii) 4 (iii) (iv)sin sin3

(7) (i) cos (ii) sin (iii) tan (iv)sin

(8) (i) sin cos (ii) cos (iii) sec (iv)cosec

x x x x

xx x x x x

x x x x

x x x x x

−−−−

− − −− − −− − −− − −

Reduction Formula A Reduction Formula succeeds if ultimately it produces an integral which can be evaluated. We use the partition of integration to prove the following reduction formulas:

1m m m-2

1 1If I sin I sin cos I(1) then show thatm m m

x dx x xm m

−−−−− −− −− −− −= = += = += = += = +∫∫∫∫

proof: 1 1

1 2 2m

I sin sin (sin ) sin ( cos )

sin ( cos )

( 1)sin (cos ) , cos

I sin cos ( 1) sin (cos )

sin cos ( 1) sin (1 sin )

sin cos ( 1) sin

x dx x xdx x d x

let u x dv d x

du m x x dx v x

x x m x x dx

− −− −− −− −

−−−−

− −− −− −− −

−−−−

= = = −= = = −= = = −= = = −

= = −= = −= = −= = −

∴ = − = −∴ = − = −∴ = − = −∴ = − = −

= − + −= − + −= − + −= − + −

= − + − −= − + − −= − + − −= − + − −

= − + −= − + −= − + −= − + −

∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫

∫∫∫∫

(((( ))))

1m 2 m

( 1) sin

I sin cos ( 1) ( 1)I

1 ( 1) I sin cos ( 1)

1 ( 1)I sin cos

xdx m xdx

x x m I m

m x x m I

mx x I

−−−−

−−−−−−−−

− −− −− −− −

= − + − − −= − + − − −= − + − − −= − + − − −

∴ + − = − + −∴ + − = − + −∴ + − = − + −∴ + − = − + −

− −− −− −− −= += += += +

∫∫∫∫

similarly we can prove that

1m m m-2

1 1If I cos I cos sin I(2) then show thatm m m

x dx x xm m

−−−− −−−−= = += = += = += = +∫∫∫∫

[[[[ ]]]] [[[[ ]]]]

m m m-1

1If I I I

If I log I log I

(3) then show that

(4) then show that

(5) then show that

m ax m ax

m x m x

mx e dx x e

mx a dx x a

x dx x x m

= = −= = −= = −= = −

∫∫∫∫

1m m-22

m m m-2

If I sin

( 1)I cos sin I

tanIf I tan I I

sec tan 2If I sec I I

then show that

(7) then show that

(8) then show that

x ax dx

x m m max x ax

x x mdx

−−−−

− −− −− −− −= + −= + −= + −= + −

= = −= = −= = −= = −−−−−

−−−−= = += = += = += = +− −− −− −− −

∫∫∫∫

If I cosec

cosec cot 2I I

If I cos , sin

I sin . cos

(9) then show that

show that

n nn n n

x x mm m

x bx dx J x bx dx

x bx nJ J x bx n I

−−−−

− −− −− −− −= += += += +− −− −− −− −

= == == == =

= − = −= − = −= − = −= − = −

∫∫∫∫

∫ ∫∫ ∫∫ ∫∫ ∫

Exercise(4)

if and prove that

findFind the reduction formula connecting

given that

Show that

1 1 4 4

, 2, 2

cosh sinh

sinh , cosh ,

sin sin

cos , sin

n nn n

n nn n n n

m n m m

m nm n

n nn n

I x x dx J x x dx

I x x n J J x x n I I J

I and I

I x x dx

I x dx J x dx

− −− −− −− −

− +− +− +− +

= == == == =

= − = −= − = −= − = −= − = −

= == == == =

∫ ∫∫ ∫∫ ∫∫ ∫

∫∫∫∫

∫ ∫∫ ∫∫ ∫∫ ∫

prove that

Find thereducion formula for

are positive integer.

sin cos ( 1) .

cos sin ( 1) .

1-1sin sin( 1) sin sin

I x x n I

nJ x x n J

n nx n x dx x nxn

x x dx

−−−−−−−−

= + −= + −= + −= + −

= − + −= − + −= − + −= − + −

+ =+ =+ =+ =∫∫∫∫

++++ ∫∫∫∫

Find thereducion formula for

4cos cos

3sin sin

3cos cos

sinh cosh tanh

(10) (11) (12)

ax n axxdx xdx

ax n axxdx e xdx

n x dx x x dx

n n nx dx x dx x dx

∫∫∫∫

∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫

∫ ∫∫ ∫∫ ∫∫ ∫

∫∫∫∫

∫ ∫∫ ∫∫ ∫∫ ∫

cosech sech

Show thatIf

2(2 3) ( 2)sin 1 21 2 2 2 2( cos ) ( 1)( ) ( 1)( )

( cos )

) (14) (15)

n n n x dx

a n I n Im x n nIn na b x n a b n a b

x dx x dx

a b xI

∫∫∫∫

− −− −− −− −− −− −− −− −= + −= + −= + −= + −−−−−+ − − − −+ − − − −+ − − − −+ − − − −

∫ ∫∫ ∫∫ ∫∫ ∫

==== ∫∫∫∫++++

Trigonometric Integrals The following identities are employed to find the Trigonometric Integrals

[[[[ ]]]]

(1) sin cos 1

(2) tan 1 sec

(3)1 cot csc

1(4) sin 1 cos 2

1(5) cos 1 cos 2

(6) sin 2 2sin cos

(7) 1 cos 2 2sin

(8)1 cos 2 2cos

1(9) sin cos sin( ) sin( )

1(10) sin sin cos( ) cos( )

x y x y x y

+ =+ =+ =+ =

= −= −= −= −

= += += += +

− =− =− =− =

+ =+ =+ =+ =

= − + += − + += − + += − + +

= − − += − − += − − += − − +

[[[[ ]]]]111) cos cos cos( ) cos( )

2x y x y x y= − + += − + += − + += − + +

(1)Integrals in the form sinm x dx∫∫∫∫

if m is odd positive integer

1sin sin (sin )m mx dx x x dx−−−−====∫ ∫∫ ∫∫ ∫∫ ∫ and use 1

1 2 2sin (1 cos )m

m x x−−−−

−−−− = −= −= −= − and take cos siny x dy xdx==== ⇒⇒⇒⇒ = −= −= −= − if m is even odd positive integer use

[[[[ ]]]] [[[[ ]]]]2 21 1sin 1 cos2 ,cos 1 cos2

2 2x x x x= − = += − = += − = += − = +

Also we can use the following reduction formula

1m m-2

1 1I sin sin cos Im m m

x dx x xm m

−−−−− −− −− −− −= = += = += = += = +∫∫∫∫

Example (1): Find 3sin x dx∫∫∫∫ Solution:

3 2 3 3

sin sin (sin ) (1 cos )(sin )

cos sin

1 1sin (1 ) cos cos

x dx x x dx x x dx

let y x dy x dx

x dx y dy y y x x c

= = −= = −= = −= = −

∴ = − − = − − = − − +∴ = − − = − − = − − +∴ = − − = − − = − − +∴ = − − = − − = − − +

∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫

∫ ∫∫ ∫∫ ∫∫ ∫

5 4 2 2

5 2 2 2 2 3

sin sin (sin ) (1 cos ) (sin )

cos sin

1sin (1 ) (1 2 )

1cos cos cos

x dx x x dx x x dx

let y x dy x dx

x dx y dy y y dy y y y

x x x c

= = −= = −= = −= = −

∴ = − − = − − + = − − +∴ = − − = − − + = − − +∴ = − − = − − + = − − +∴ = − − = − − + = − − +

= − − + += − − + += − − + += − − + +

∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫

4 2 2 2

1(3) sin sin sin (1 cos2 )

1 1 11 2cos2 cos 2 1 2cos2 (1 cos4 )

1 3 1 1 3 12cos2 cos4 sin2 sin4

4 2 2 4 2 8

x dx x x dx x dx

x x dx x x dx

xx x dx x x c

= = −= = −= = −= = −

= − + = − + −= − + = − + −= − + = − + −= − + = − + −

= − − = − − += − − = − − += − − = − − += − − = − − +

∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫

∫ ∫∫ ∫∫ ∫∫ ∫

∫∫∫∫

(2)Integrals in the form cosm x dx∫∫∫∫

(i) if m is odd positive integer 1cos cos (cos )m mx dx x x dx−−−−====∫ ∫∫ ∫∫ ∫∫ ∫ and use

1 2 2cos (1 sin )m

m x x−−−−

−−−− = −= −= −= − and take sin cosy x dy xdx==== ⇒⇒⇒⇒ ==== (ii) if m is even odd positive integer use

[[[[ ]]]] [[[[ ]]]]2 21 1sin 1 cos2 ,cos 1 cos2

2 2x x x x= − = += − = += − = += − = +

(iii) we can use the successive formula.

1m m-2

1 1I cos cos sin Im m m

x dx x xm m

−−−− −−−−= = += = += = += = +∫∫∫∫

Example (4): Find 3cos x dx∫∫∫∫ Solution:

3 2 3 3

cos cos (cos ) (1 sin )(cos )

sin cos

1 1cos (1 ) sin sin

x dx x x dx x x dx

let y x dy x dx

x dx y d y y y c x c

= = −= = −= = −= = −

= == == == =

∴ = − = − + = − +∴ = − = − + = − +∴ = − = − + = − +∴ = − = − + = − +

∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫

∫ ∫∫ ∫∫ ∫∫ ∫

5 4 2 2

5 2 2 2

2 3 2 3

cos cos (cos ) (1 sin ) (cos )

sin cos

cos (1 ) (1 2 )

1 1sin sin sin

x dx x x dx x x dx

let y x dy x dx

x dx y dy y y dy

y y y x x x c

= = −= = −= = −= = −

= == == == =

∴ = − = − +∴ = − = − +∴ = − = − +∴ = − = − +

= − + = − + += − + = − + += − + = − + += − + = − + +

∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫

4 2 2 21cos cos cos (1 cos 2 )

4x dx x x dx x dx= = += = += = += = +∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫

21 1 1

1 2cos 2 cos 2 1 2cos 2 (1 cos4 )4 4 2

1 3 1 1 3 12cos 2 cos4 sin2 sin4

4 2 2 4 2 8

x x dx x x dx

xx x dx x x c

= + + = + + += + + = + + += + + = + + += + + = + + +

∫ ∫∫ ∫∫ ∫∫ ∫

∫∫∫∫

(3)Integrals in the form tanm x dx∫∫∫∫

(i) if m is odd positive integer then m-1 is even

1 2 ( 1) 2tan tan (tan ) (sec 1) (tan )

sec sec tan

m m mx dx x x dx x x dx

put y x dy x xdx

− −− −− −− −= = −= = −= = −= = −

= → == → == → == → =∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫

(ii) if m is even odd positive integer use 2 2 2 2 2tan tan (sec 1) tan sec tanm m m mx dx x x dx x x dx x dx− − −− − −− − −− − −= − = −= − = −= − = −= − = −∫ ∫ ∫ ∫∫ ∫ ∫ ∫∫ ∫ ∫ ∫∫ ∫ ∫ ∫ (reduction formula) Example (7): Find 5tan x dx∫∫∫∫ Solution:

5 3 2 3 2

2tan , sec

tan tan (tan ) tan (sec 1)

tan sec tan

tan sec tan (sec 1)

tan sec (tan sec tan )

1 1tan tan ln sec

let y x dy xdx

x dx x x dx x x dx

x x dx x dx

x x dx x x dx

x x dx x x x dx

x x x c

= == == == =

= = −= = −= = −= = −

= −= −= −= −

= − −= − −= − −= − −

= − + += − + += − + += − + +

∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫

∫ ∫∫ ∫∫ ∫∫ ∫

Example (8):Find 6tan x dx∫∫∫∫ Solution:

6 4 2 4 2tan tan (tan ) tan (sec 1)x dx x x dx x x dx= = −= = −= = −= = −∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫

4 2 2 2

4 2 2 2 2

tan sec tan

tan sec tan (sec 1)

tan sec tan sec tan

tan sec tan sec (sec 1)

tan sec tan sec sec

1 1tan tan tan

x x dx x dx

x x dx x x dx x dx

x x dx x x dx x dx dx

x x x x c

= −= −= −= −

= − −= − −= − −= − −

= − += − += − += − +

= − + −= − + −= − + −= − + −

= − + − += − + − += − + − += − + − +

∫ ∫∫ ∫∫ ∫∫ ∫

∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫

∫ ∫ ∫ ∫∫ ∫ ∫ ∫∫ ∫ ∫ ∫∫ ∫ ∫ ∫

In general we use the reduction formula

m m m-2tan

If I tan Then I I( 1)

mm axdx

−−−−= = −= = −= = −= = −

−−−−∫∫∫∫

(4)Integrals in the form: cotm x dx∫∫∫∫

1 2 1 2cot cot (cot ) (cosec 1) (cot )

cosec cosec cot

put y x dy x xdx

− −− −− −− −= = −= = −= = −= = −

= → = −= → = −= → = −= → = −∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫

(ii) if m is even odd positive integer use integration by parts

cot cot (cosec 1)

cot cosec cot

x dx x x dx

x x dx x dx

−−−−

− −− −− −− −

= −= −= −= −

∫ ∫∫ ∫∫ ∫∫ ∫

In general we can use the reduction formula

m m m-2cot

If I cot Then I I( 1)

mm ax dx

−−−−= = − −= = − −= = − −= = − −

−−−−∫∫∫∫

Example (9): Find 4cot 3x dx∫∫∫∫ Solution:

4 2 2 2 2

cot 3 cot 3 (cot 3 ) cot 3 (cosec 3 1)

cot 3 cosec 3 cot 3

x dx x x dx x x dx

x x dx x dx

= = −= = −= = −= = −

= −= −= −= −

∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫

∫ ∫∫ ∫∫ ∫∫ ∫

cot 3 cosec 3 (cosec 3 1)

cot 3 cosec 3 cosec 3

1 1cot 3 cot 3

x x dx x dx

x x dx x dx dx

x x x c

= − −= − −= − −= − −

= − += − += − += − +

= + + += + + += + + += + + +

∫ ∫∫ ∫∫ ∫∫ ∫

∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫

(5)Integrals in the form: secm x dx∫∫∫∫

1 2 ( 1) 2sec sec (sec ) (tan 1) (sec )

sec sec tan

put y x dy x xdx

− −− −− −− −= = −= = −= = −= = −

= → == → == → == → =∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫

(ii) if m is even odd positive integer use

2 ( 2) / 2

sec sec (sec )

tan sec (tan 1)

x dx x x dx

put y x and x x

−−−−

= = += = += = += = +

∫ ∫∫ ∫∫ ∫∫ ∫

to get the reduction formula

m m-2sec tan 2

I sec I1 1

mm x m

−−−− −−−−= = += = += = += = +− −− −− −− −∫∫∫∫

Example (10): Find 4sec 2x dx∫∫∫∫ Solution:

4 2 2 2 2

2 2 2 3

(10) sec 2 sec 2 (sec 2 ) sec 2 (1 tan 2 )

1 1sec 2 sec 2 tan 2 tan2 tan 2

x dx x x dx x x dx

x dx x x dx x x c

= = += = += = += = +

= + = + += + = + += + = + += + = + +

∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫

∫ ∫∫ ∫∫ ∫∫ ∫

(6)Integrals in the form: cosecm x dx∫∫∫∫

1 2 ( 1) 2cosec cosec (cosec ) (cot 1) (cosec )

cosec cosec cot

put y x dy x xdx

− −− −− −− −= = += = += = += = +

= → = −= → = −= → = −= → = −∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫

(ii) if m is even odd positive integer use

2 2/ 2

cosec cosec (cosec )

cot cosec (cot 1)

x dx x x dx

put y x and x x

−−−−

= = += = += = += = +

∫ ∫∫ ∫∫ ∫∫ ∫

to get the reduction formula Example (11): Find 6cosecax dx∫∫∫∫ Solution:

6 4 2 2 2 2

2 2 2 4 2

(11) cosec cosec (cosec ) (1 cot ) (cosec )

(1 2cot cot )(cosec )

(cosec 2cot cosec cot cosec )

1 2 1cot cot cot

ax dx ax ax dx ax ax dx

ax ax ax dx

ax ax ax ax ax dx

ax ax ax ca a a

= = += = += = += = +

= + += + += + += + +

−−−−= − − += − − += − − += − − +

∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫

∫∫∫∫

(7)Integrals in the form cos sinm nx x dx∫∫∫∫ , , are positive integersm n

If m is an odd

1cos sin cos sin (cos ) and put sinm n m nx x dx x x x dx y x−−−−= == == == =∫ ∫∫ ∫∫ ∫∫ ∫ If n is an odd

1cos sin cos sin (sin ) andput cosm n m nx x dx x x x dx y x−−−−= == == == =∫ ∫∫ ∫∫ ∫∫ ∫

if m and n are both an even we use

[[[[ ]]]] [[[[ ]]]]2 21 1sin 1 cos2 ,cos 1 cos2

2 2x x x x= − = += − = += − = += − = +

Example (12): Find 2 2sin cosx x dx∫∫∫∫

Solution:

(((( ))))2

22 2 1(12) sin cos sin cos sin 2

2x x dx x x dx x dx

= == == == =

∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫

21 1 1 1 1sin 2 (1 cos4 ) ( sin4 )

4 4 2 8 4x dx x dx x x c= = − = − += = − = − += = − = − += = − = − +∫ ∫∫ ∫∫ ∫∫ ∫

Example (13): Find 3 4sin cosax ax dx∫∫∫∫

Solution: 3 4 2 4

(13) sin cos sin cos (sin )

(1 cos )cos (sin )

(cos cos )(sin )

(cos sin cos sin )

1 1cos cos

ax ax dx ax ax ax dx

ax ax ax dx

ax ax dx ax ax dx

ax ax ca a

= −= −= −= −

−−−−= + += + += + += + +

∫ ∫∫ ∫∫ ∫∫ ∫

∫∫∫∫

Example (14): Find 4 2sin cosax ax dx∫∫∫∫

Solution:

[[[[ ]]]]

1 1(14) sin cos (1 cos2 ) (1 cos2 )

1(1 cos2 )(1 cos2 )(1 cos2 )

1(1 cos2 )(1 cos 2 )

1(1 cos2 )sin 2

1sin 2 cos2 sin 2

1 1(1 cos4

ax ax dx ax ax dx

ax ax ax dx

ax ax dx

ax dx ax ax dx

= − += − += − += − +

= − − += − − += − − += − − +

= − −= − −= − −= − −

= −= −= −= −

∫ ∫∫ ∫∫ ∫∫ ∫

∫∫∫∫

∫ ∫∫ ∫∫ ∫∫ ∫

) cos2 sin 2

1 1 1 1 1( . cos4 ) . sin 2

8 2 4 3 2

1 1 1( sin4 sin 2 )16 64 48

ax dx ax ax dx

x ax ax ca a

−−−−

= − − += − − += − − += − − +

∫ ∫∫ ∫∫ ∫∫ ∫

The reduction formula for cos sinm nx x dx∫∫∫∫

Case (1): if m and n are positive integer:

sin .cos

sin .( 1).cos .( sin ) cos .( 1)sin (cos )

( 1)sin .cos ( 1)cos sin

( 1)sin .cos sin ( 1)cos sin

( 1)sin .cos (1 cos ) ( 1)cos sin

m n n m

x n x x x m x x

n x x m x x

n x x x m x x

n x x x m x

+ −+ −+ −+ −

+ − −+ − −+ − −+ − −

+ −+ −+ −+ −

−−−−

= − − + += − − + += − − + += − − + +

= − − − + += − − − + += − − − + += − − − + +2( 1)sin .cos ( )cos sinm n n m

n x x m n x x−−−−= − − + += − − + += − − + += − − + +

by integrating both sides w.r.tox we have

are positive integers

,( 2) ,

1 1, ,( 2)

sin .cos

( 1)sin .cos ( )cos sin

( 1) ( )

1 ( 1)cos sin sin .cos

( ) ( )

m n n m

m n m n

m n m nm n m n

n x x dx m n x x dx

n I m n I

nI x x dx x x I

m n m n

+ −+ −+ −+ −

−−−−

+ −+ −+ −+ −−−−−

= − − + += − − + += − − + += − − + += − − + += − − + += − − + += − − + +

−−−−= = −= = −= = −= = −+ ++ ++ ++ +∴∴∴∴

∫ ∫∫ ∫∫ ∫∫ ∫

∫∫∫∫

Example(15):

4 6 5 54,6 4,4

5 5 5 34,2

5 5 5 3 54,0

1 1sin cos sin cos

1 1 1 3sin cos sin cos

10 2 8 8

1 1 1 3 1 1sin cos sin cos sin cos

10 2 8 8 6 6

3 1 1sin sin 2 sin4

8 4 23

I x x dx x x I

x x x x I

x x x x x x I

I x dx x x x

= = −= = −= = −= = −

= − −= − −= − −= − −

= − − −= − − −= − − −= − − −

= = − += = − += = − += = − +

∫∫∫∫

5 54,6

5 5 5 3 5

1sin cos

1 1 3 1 1 3 1 1sin cos sin cos sin 2 sin4

2 8 8 6 6 8 4 23

1 1 3 3sin cos sin cos sin cos

10 16 32 256

1 3sin 2 sin4

128 1204

x x x x x x x

∴ =∴ =∴ =∴ =

− − − − +− − − − +− − − − +− − − − +

= − + −= − + −= − + −= − + −

+ −+ −+ −+ −

Case (2): if m and n are negative integers :

sin .cos

( 1)sin .cos ( 1)sin cos

( 1)sin .cos ( 1)sin cos (1 sin )

( 1)sin .cos ( 1)sin cos ( 1)

m n n m

m n m n

x n x x x m x x

n x x m x x

n x x m x x x

n x x m x x m

+ ++ ++ ++ +

= + − + += + − + += + − + += + − + +

= − − + += − − + += − − + += − − + +

= − − + + −= − − + + −= − − + + −= − − + + −

= − − + + − += − − + + − += − − + + − += − − + + − + 2

sin cos

m n m n

m n x x m x x

++++= − + + + += − + + + += − + + + += − + + + +

by integrating both sides w.r.to x we have

are negative integars

1 1( 2), ,

1 1, ( 2),

sin .cos ( 2)sin .cos ( 1) sin cos

sin .cos ( 2) ( 1)

1 ( 2)sin cos sin .cos

( 1) ( 1)

m n m n m n

m nm n m n

m n m nm n m n

x x m n x x dx m x x dx

x x m n I m I

m nI x x dx x x I

+ − ++ − ++ − ++ − +

+ −+ −+ −+ −++++

+ ++ ++ ++ +++++

= − + + + += − + + + += − + + + += − + + + +

+ ++ ++ ++ += = += = += = += = ++ ++ ++ ++ +∴∴∴∴

∫ ∫∫ ∫∫ ∫∫ ∫

∫∫∫∫

Example(16):

3 24, 3 ( 2, 3)4 3

( 2, 3)3 2

1 5sin cos

3 3sin cos

1 533sin cos

dxI x x I

− −− −− −− −− − − −− − − −− − − −− − − −

− −− −− −− −

−−−−= = += = += = += = +

−−−−= += += += +

∫∫∫∫

1 22, 3 0, 3 0, 32

3 2 20, 3 3

1(sin ) cos 3 3

sin cos

sec sec sec sec tan sec tancos

sec tan sec (1 sec ) sec tan sec sec

1 1sec tan ln sec tan

I x x I Ix x

dxI x dx x x dx x x x x dx

x x x x dx x x xdx x dx

I x x x x

− −− −− −− −− − − −− − − −− − − −− − − −

−−−−

− −− −− −− −

−−−−= − + = += − + = += − + = += − + = +

= = = = −= = = = −= = = = −= = = = −

= − + = − −= − + = − −= − + = − −= − + = − −

= −= −= −= −

−−−−∴ =∴ =∴ =∴ =

∫ ∫ ∫ ∫∫ ∫ ∫ ∫∫ ∫ ∫ ∫∫ ∫ ∫ ∫

∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫

( 4, 3) 3 2 2

3 3sec tan ln sec tan

2 2cos

1 5 5 5sec tan ln sec tan

3 33sin cos 3sin cos

x x x xx x

I x x x x cx x x x

− −− −− −− −

+ −+ −+ −+ −

−−−−∴ = − + − +∴ = − + − +∴ = − + − +∴ = − + − +

Case (3): if m is positive and n is negative integers:

sin .cos

m n n m

m n m n

x n x x x m x x

n x x m x x

− +− +− +− +

− + −− + −− + −− + −

− +− +− +− +

= + − + −= + − + −= + − + −= + − + −

= − + + −= − + + −= − + + −= − + + −

is positive and is negative integars

, ( 2),( 2)

1 1, ( 1),( 2

sin .cos

( 1) sin .cos ( 1) sin cos

( 1) ( 1)

m n m n

m n m nm n m n

n x x dx m x x dx

n I m I

mI x x dx x x I

− +− +− +− +

− +− +− +− +− =− =− =− =

= − + + −= − + + −= − + + −= − + + −= − + + −= − + + −= − + + −= − + + −

− −− −− −− −= = += = += = += = ++ ++ ++ ++ +∴∴∴∴

∫ ∫∫ ∫∫ ∫∫ ∫

∫∫∫∫

Example(17): 6 5

5 7(6, 8) (4, 6) (4, 6)8 7

(4, 6) (2, 4) (2, 4)5

3(2, 4) (0, 2) (0, 2)3

(0, 2) 2

sin 1 5 1 sin 5sin cos

7 7 7 7cos cos

1 3 sin 3sin (cos )

5 5 55cos

1 1 sin 1sin (cos )

3 3 33cos

x dx xI x x I I

xI x x I I

−−−−− − −− − −− − −− − −

−−−−

= = − + = −= = − + = −= = − + = −= = − + = −− −− −− −− −

−−−−= + = −= + = −= + = −= + = −− −− −− −− −

∫∫∫∫

(2, 4) (4, 6)3 5 3

(6, 8)8 7 5 3

sec tan

sin 1 sin sin 1tan , tan

3 53cos 5cos 5cos

sin 1 sin sin sin 1tan

7 7cos cos 7cos 7cos

x dx x

x x xI x I x

x dx x x xI x

x x x x

− −− −− −− −

−−−−

= == == == =

∴ = − = − +∴ = − = − +∴ = − = − +∴ = − = − +

−−−−= = − + −= = − + −= = − + −= = − + −

∫ ∫∫ ∫∫ ∫∫ ∫

∫∫∫∫

we can solve this example without reduction formulas

66 2 7

8sin 1

tan sec tan7cos

x dxx xdx x c

x= = += = += = += = +∫ ∫∫ ∫∫ ∫∫ ∫

Case (4): if m is negative and n is positive integers:

sin .cos

sin .( 1).cos .( sin ) cos .( 1)sin cos

m n n m

m n m n

x n x x x m x x

n x x m x x

+ −+ −+ −+ −

+ − −+ − −+ − −+ − −

+ −+ −+ −+ −

= − − + += − − + += − − + += − − + +

is negative and is positive integars

1 1 2 2

( 2),( 2) ,

1 1, ( 2),( 2)

sin .cos ( 1) sin .cos ( 1) sin cos

( 1) ( 1)

m n m n m n

m n m n

m n m nm n m n

x x n x x dx m x x dx

n I m I

nI x x dx x x I

+ − + −+ − + −+ − + −+ − + −

+ −+ −+ −+ −

+ −+ −+ −+ −+ −+ −+ −+ −

= − − + += − − + += − − + += − − + +

−−−−= = += = += = += = ++ ++ ++ ++ +∴∴∴∴

∫ ∫∫ ∫∫ ∫∫ ∫

∫∫∫∫

Example(18): 4 3

3 5( 6,4) ( 4,2) ( 4,2)6 5

3( 4,2) ( 2,0) ( 2,0)3

2( 2,0) 2

( 4,2) 3

( 6,46

cos 1 3 cos 3cos sin

5 5 5sin 5sin

1 1 cos 1cos sin

3 3 33sin

csec cotsin

cos 1cot

x dx xI x x I I

xI x x I I

dxI x dx x

−−−−− − −− − −− − −− − −

−−−−

−−−−= = − − = −= = − − = −= = − − = −= = − − = −

−−−−= + = −= + = −= + = −= + = −− −− −− −− −

= = = −= = = −= = = −= = = −

−−−−∴ = +∴ = +∴ = +∴ = +

∫∫∫∫

∫ ∫∫ ∫∫ ∫∫ ∫

) 5 3cos cos 1

cot55sin 5sin

−−−−= + − += + − += + − += + − +∫∫∫∫

we can solve this example without reduction formulas

44 2 5

6cos 1

cot cosec cot5sin

x dxx x dx x c

−−−−= = += = += = += = +∫ ∫∫ ∫∫ ∫∫ ∫

(8)Integrals in the form sec tanm nx x dx∫∫∫∫ , , are positive integersm n

If n is an odd and m either an even or odd

1 1sec tan sec tan (sec tan )

andput sec

m n m nx x dx x x x xdx

− −− −− −− −====

====∫ ∫∫ ∫∫ ∫∫ ∫

If m is an even ,n either an even or odd 2 2sec tan sec tan (sec ) andput tanm n m nx x dx x x x dx y x−−−−= == == == =∫ ∫∫ ∫∫ ∫∫ ∫

Example (19): Find 4 3sec 3 tan 3x x dx∫∫∫∫

Solution: 4 3 3 2 2

sec 3 tan 3 tan 3 sec 3 (sec 3 )

tan 3 (tan 3 1)(sec 3 )

(tan 3 tan 3 )(sec 3 )

x x dx x x x dx

x x x dx

= −= −= −= −

∫ ∫∫ ∫∫ ∫∫ ∫

∫∫∫∫

5 3 6 4 6 4

tan 3 3sec3

1 1 1 1 1 1 1( ) tan 3 tan 3

3 3 6 4 3 6 4

put y x dy x dx

I y y dy y y x x c

= ∴ == ∴ == ∴ == ∴ =

∴ = − = − = − +∴ = − = − = − +∴ = − = − = − +∴ = − = − = − +

∫∫∫∫

Example (20): Find 3 2sec tanx x dx∫∫∫∫

Solution: 3 2 3 2 5 3

5 3 2 35

3 3 2 3 3 2

(16) sec tan sec (sec 1) (sec sec )

sec sec sec sec tan

sec tan tan (3sec tan )

sec tan 3 sec tan sec tan 3 sec (sec 1)

sec tan 3 (

x x dx x x dx x x dx

let I x dx x x dx x d x

x x x x x dx

x x x x dx x x x x dx

= − = −= − = −= − = −= − = −

∴ = −∴ = −∴ = −∴ = −

= = == = == = == = =

= −= −= −= −

= − = − −= − = − −= − = − −= − = − −

= −= −= −= −

∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫

∫∫∫∫

∫ ∫∫ ∫∫ ∫∫ ∫

sec sec )

1sec tan (2)

sec sec sec sec tan

sec tan tan (sec tan ) sec tan sec tan

sec tan sec (sec 1) sec tan (sec sec )

1sec tan ln sec tan

x x dx

I x x I

and I x dx x x dx x d x

x x x x x dx x x x x dx

x x x x dx x x x x dx

−−−−

∴ = +∴ = +∴ = +∴ = +

= = == = == = == = =

= − = −= − = −= − = −= − = −

= − − = − −= − − = − −= − − = − −= − − = − −

= + += + += + += + +

∫∫∫∫

∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫

∫ ∫∫ ∫∫ ∫∫ ∫

(((( ))))

3 33 3 3

1 1(1),(2) (3) sec tan sec tan 3

1 3sec tan sec tan ln sec tan

from and I x x I I x x I

x x x x x x

x x x x x x c

= + − = −= + − = −= + − = −= + − = −

= − + += − + += − + += − + +

= − + + += − + + += − + + += − + + +

Example (21): Find 3 3sec tanax ax dx∫∫∫∫

Solution: 3 3 2 2

4 2 5 3 5 3

(17) sec tan sec tan (sec tan )

sec (sec 1)(sec tan )

(sec sec )(sec tan )

sec sec tan

1 1 1 1 1 1 1( ) sec sec

5 3 5 3

ax ax dx ax ax ax ax dx

ax ax ax ax dx

put y ax dy a ax ax

I y y dy y y ax ax ca a a

= −= −= −= −

= ∴ == ∴ == ∴ == ∴ =

∴ = − = − = − +∴ = − = − = − +∴ = − = − = − +∴ = − = − = − +

∫ ∫∫ ∫∫ ∫∫ ∫

∫∫∫∫

(9)Integrals in the form cosec cotm nx x dx∫∫∫∫ , , are positive integersm n

If n is an odd and m either an even or odd

1 1cosec cot cosec cot (cosec cot )

andput cosec

m n m nx x dx x x x xdx

− −− −− −− −====

====∫ ∫∫ ∫∫ ∫∫ ∫

If m is an even ,n either an even or odd

2 2cosec cot cosec cot (cosec )

andput cot

m n m nx x dx x x x dx

−−−−====

====∫ ∫∫ ∫∫ ∫∫ ∫

Example (22): Find 3 3cot cosecax ax dx∫∫∫∫

Solution: 3 3 2 2

3 3 2 4 3 5

cot cosec cot cosec (cosec cot )

(1 cosec ) cosec (cosec cot )

(cosec cosec ) (cosec cot )

cosec cosec cot

1 1cot cosec ( )

ax ax dx ax ax ax ax dx

ax ax ax ax dx

if y ax dy a ax ax dx

ax ax dx y y dy y ya a

= −= −= −= −

= = −= = −= = −= = −

∴ = − − = − +∴ = − − = − +∴ = − − = − +∴ = − − = − +

= −= −= −= −

∫ ∫∫ ∫∫ ∫∫ ∫

∫∫∫∫

∫ ∫∫ ∫∫ ∫∫ ∫

3 51cosec cosec

3 5ax ax c

a a+ ++ ++ ++ +

Example (23): Find 3 4cot cosecax ax dx∫∫∫∫

Solution: 3 4 3 2 2

3 4 3 5 4 6

(19) cot cosec cot cosec (cosec )

cot (1 cot )(cosec )

(cot cot )(cosec )

cot cosec

1 1 1 1cot cosec ( y ) y

1 1cot co

ax ax dx ax ax ax dx

ax ax ax dx

if y ax dy a ax dx

ax ax dx y dy ya a

= += += += +

= = −= = −= = −= = −

− −− −− −− − ∴ = + = +∴ = + = +∴ = + = +∴ = + = +

−−−−= −= −= −= −

∫ ∫∫ ∫∫ ∫∫ ∫

∫∫∫∫

∫ ∫∫ ∫∫ ∫∫ ∫

try to solve cot cosec

ax ax dx

∫∫∫∫

As similar we can integrate the following

sinh , cosh , tanh

cosech , sech , coth

sinh cosh

sech tanh

cosech coth

x dx x dx x dx

x x dx x dx

x x dx

∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫

∫∫∫∫

Exercise(5)

Find 3 2

4 4 2 5

3 2 3 3

(1) sinh (2) sin 4

(3) cot cosec (4) sin cos

(5) sinh cosh (6) sin cos

(7) sin2 cos4 (8) cos

(9) cos (10) sin cos

(11) cosh(2 )cosh(3 ) (12) cos sin

(13) sin3 cos5

x dx x dx

x x dx x x dx

x x dx x dx

x dx x x dx

x x dx x dx

∫ ∫∫ ∫∫ ∫∫ ∫

∫ ∫∫ ∫∫ ∫∫ ∫3 5

3 5 3 4

2 4 3 5

2 4 2 3

(14) sin cos

(15) sin cos (16) sin5 cos

cos(17) (18) tan

(19) tan sec (20) tan sec

(21) tan sec (22) cot cosec

(23) cot cosec (24) cot cosec

(25) cot

x dx x x dx

x x dx x x dx

xdx x dx

x x dx x x dx

∫ ∫∫ ∫∫ ∫∫ ∫

∫ ∫∫ ∫∫ ∫∫ ∫4 3 3

5 3 2 4

cosec (26) cot cosec

(27) sec (28) cosec

(29) sec (30) cot

(31) coth (32) cosech

(33) coth (34) cosech

(35) coth cosech (36) coth cosech

(37) coth cosech (3

x dx x x dx

x dx x dx

x x dx x x dx

x x dx

∫ ∫∫ ∫∫ ∫∫ ∫

∫∫∫∫4 3

8) coth cosech

(39) coth cosech

x x dx

∫∫∫∫

Trigonometric Substitutions: An Integrand which contains one of the form

2 2 2 2 2 2 2 2 2, ,a b x a b x b x a− + −− + −− + −− + − may be transformed into another simple integrals contains trigonometric functions of new variable. The substituting according the following rules:

22 2 2 2 2

22 2 2 2 2 2 2

For sin , sin cos

( sin ) 1 sin cos

(i) a a aa b x put x x dx d

aa b x a b a a

θ θ θ θθ θ θ θθ θ θ θθ θ θ θ

θ θ θθ θ θθ θ θθ θ θ

− = = =− = = =− = = =− = = =

∴ − = − = − =∴ − = − = − =∴ − = − = − =∴ − = − = − =

22 2 2 2 2 2

22 2 2 2 2 2 2 2

For tan , tan sec

( tan ) 1 tan sec

(ii) a a aa b x put x x dx d

aa b x a b a a

+ = = =+ = = =+ = = =+ = = =

∴ + = + = + =∴ + = + = + =∴ + = + = + =∴ + = + = + =

22 2 2 2 2

22 2 2 2 2 2 2

For sec , sec sec tan

( sec ) sec 1 tan

(iii) a a ab x a put x x dx d

ab x a b a a a

θ θ θ θ θθ θ θ θ θθ θ θ θ θθ θ θ θ θ

− = = =− = = =− = = =− = = =

∴ − = − = − =∴ − = − = − =∴ − = − = − =∴ − = − = − =

(iv)For , tansin cos 2

2sin 2sin cos ,

1cos cos sin ,

1 1 1sec (1 tan ) (1 )

2 2 2 2 2

dx dx xput u

a b x a b x

x x ux

x xdu dx dx u dx

====± ±± ±± ±± ±

∴ = =∴ = =∴ = =∴ = =++++

−−−−= − == − == − == − =++++

= = + = += = + = += = + = += = + = +

∫ ∫∫ ∫∫ ∫∫ ∫

2 2 22 2 1

, sin , cos1 1 1

du u udx x x

−−−−∴ = = =∴ = = =∴ = = =∴ = = =+ + ++ + ++ + ++ + +

(v)For , tanhsinh cosh 2

2sinh 2sinh cosh ,

1cosh cosh sinh ,

1 1 1sech (1 tanh ) (1 )

2 2 2 2 2

2 2 1, sinh , cosh

dx dx xput u

a b x a b x

x x ux

x xdu dx dx u dx

du u udx x x

====± ±± ±± ±± ±

∴ = =∴ = =∴ = =∴ = =−−−−

++++= + == + == + == + =−−−−

= = − = −= = − = −= = − = −= = − = −

++++∴ = = =∴ = = =∴ = = =∴ = = =− − −− − −− − −− − −

∫ ∫∫ ∫∫ ∫∫ ∫

Example(1): Find 2

x −−−−∫∫∫∫

Solution:

(((( ))))

2 2 2 2

2sec 2sec tan ,

14 4sec 4 2tan , tan sec 1 4

4sec (2sec tan )4sec

2sec tan 2ln sec tan

1 1 12. . 4 2ln 4 4 2ln 4

2 2 2 4 2 2 4

put x dx d

x dx dd

x x x xx x x x c

θ θ θ θθ θ θ θθ θ θ θθ θ θ θ θ θθ θθ θθ θθθθθ

= ∴ == ∴ == ∴ == ∴ =

− = − = = − = −− = − = = − = −− = − = = − = −− = − = = − = −

∴ = =∴ = =∴ = =∴ = =−−−−

= + += + += + += + +

= − + + − = − + + − += − + + − = − + + − += − + + − = − + + − += − + + − = − + + − +

∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫

Example(2): Find 29

x−−−−∫∫∫∫

Solution:

[[[[ ]]]]

(((( ))))

21 1 2

3sin 3cos , 9 9 (3sin ) 3 1 sin 3cos

9sin 3cos 19 sin 9 1 cos2

3cos 29

9 1 9 1sin2 2sin cos

2 2 2 2

9 9 9sin sin 9

2 3 3 3 2 3 2

let x dx d x

x dx dd d

x x x x xx c− −− −− −− −

==== ⇒⇒⇒⇒ = − = − = − == − = − = − == − = − = − == − = − = − =

∴ = = == −∴ = = == −∴ = = == −∴ = = == −−−−−

= − = −= − = −= − = −= − = −

−−−− = − = − − += − = − − += − = − − += − = − − +

∫ ∫ ∫ ∫∫ ∫ ∫ ∫∫ ∫ ∫ ∫∫ ∫ ∫ ∫

θ θ θ θ θ θθ θ θ θ θ θθ θ θ θ θ θθ θ θ θ θ θ

θ θ θθ θ θθ θ θθ θ θ θ θ θ θθ θ θ θθ θ θ θθ θ θ θθθθθ

θ θ θ θ θθ θ θ θ θθ θ θ θ θθ θ θ θ θ

Example(3): Find 29 4

x x++++∫∫∫∫

Solution: 2

put 2 3tan 2 3sec

9 4 9 9tan 3sec

(3 / 2)sec 1 sec 1 cos(3 / 2)tan 3sec 3 tan 3 cos sin9 4

1 1cosec ln cosec cot (1)

2 3 9 42 3tan tan ,cot , cosec

1ln cosec

x dx d

dx d d d

==== ⇒⇒⇒⇒ ====

+ = + =+ = + =+ = + =+ = + =

∴ = = =∴ = = =∴ = = =∴ = = =++++

= = −= = −= = −= = −

++++==== ⇒⇒⇒⇒ = = == = == = == = =

∴ =∴ =∴ =∴ =

∫ ∫ ∫ ∫∫ ∫ ∫ ∫∫ ∫ ∫ ∫∫ ∫ ∫ ∫

∫∫∫∫

θ θθ θθ θθ θ

θ θ θ θ θ θθ θ θ θ θ θθ θ θ θ θ θθ θ θ θ θ θθ θ θ θ θθ θ θ θ θθ θ θ θ θθ θ θ θ θ

θθθθ21 9 4 3

cot ln3 2 2

x x++++− = − +− = − +− = − +− = − +θθθθ

Example(3): Find 5 4cos

dxx++++∫∫∫∫

Solution:

(((( ))))

2 2 22

1cos cos sin ,

2 25 4cos 1 5(1 ) 4(1 )

(1 ) 5 41

tan2 2 2 2tan tan3 3 3 39

xput u

x x ux

dx du dux u u u

− −− −− −− −

−−−−∴ = − =∴ = − =∴ = − =∴ = − =++++

====++++

= == == == =++++ −−−− + + −+ + −+ + −+ + −+ ++ ++ ++ + ++++

= = = += = = += = = += = = + ++++

∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫

∫∫∫∫

Example(3): Find 12 13sin

dxx++++∫∫∫∫

Solution:

2 2tan sin 2sin cos ,

2 2 2 1 1

2 212 13sin 2 12(1 ) 13(2 )1 12 13

2(2 3)(3 2)12 12 26 6 13 6

2 1 1 1ln 2 3

15 2 3 5 3 2 15 1

x x x u duput u x dx

dx du dux u u uu

du du duu uu u u u

du duu

= ∴ = = == ∴ = = == ∴ = = == ∴ = = =+ ++ ++ ++ +

= == == == =++++ + ++ ++ ++ + + ++ ++ ++ + ++++

= = == = == = == = =+ ++ ++ ++ + + + + ++ + + ++ + + ++ + + +

− −− −− −− −= + = + += + = + += + = + += + = + ++ ++ ++ ++ +

∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫

∫ ∫∫ ∫∫ ∫∫ ∫ ln 3 25

1 1ln 2tan 3 ln 3tan 2

15 2 15 2

−−−−= + + + += + + + += + + + += + + + +

Exercise(6)

Prove that

2 3/ 2 2

22 2 2

2 2 3/ 2 2 2

2 22 2

(4 ) 4 4

25 5 25(2) 5ln 25

(4) 4 4 2ln( 4)2

(5) sin( ) ( )

(6) 4 4 2ln( 4)2

x xdx x c

a xdx c

a xx a xx

x dx x x x c

x x xdx c

aa x a x

xx dx x x x c

x adx x a

−−−−

= += += += +−−−− −−−−

− − −− − −− − −− − −= + − += + − += + − += + − +

−−−−= − += − += − += − +−−−−

+ = + + + + ++ = + + + + ++ = + + + + ++ = + + + + +

= − += − += − += − +− −− −− −− −

− = − + + − +− = − + + − +− = − + + − +− = − + + − +

++++ = += += += +

∫∫∫∫

∫∫∫∫2 2

2 5/ 2 2 3/ 2

2 2 3/ 2 2 2 2

(8)(4 ) 12(4 )

(9)( ) ( )

a x a ac

x dx xc

a x a a x

+ −+ −+ −+ −+ ++ ++ ++ ++ ++ ++ ++ +

= += += += +− −− −− −− −

= += += += +++++ ++++

−−−−= − += − += − += − +−−−−

∫∫∫∫

22 2 2 2 2 5/ 2 2 2 3/ 2

1(11) 16 8ln 16

1(12) ( ) ( )

x dxx x x x c

ax a x dx a x a x c

= − + + − += − + + − += − + + − += − + + − + −−−−

− = − − − +− = − − − +− = − − − +− = − − − +

∫∫∫∫

2 3/ 2 2

(13) ln( 2 4 13)4 13

(4 ) 4 4

dxdx x x x c

x xdx x

cx x x x

= − + − + += − + − + += − + − + += − + − + +− +− +− +− +

−−−−= += += += +−−−− −−−−

∫∫∫∫

12 2 2

1 2 1 2

1(15) tan

54 3(9 ) 18(9 )

1 1(16) sin (2 1)sin 1

4 41 1

(17) cos (2 1)cos 14 4

dx x xx

x x dx x x x x c

−−−−

− −− −− −− −

= + += + += + += + ++ ++ ++ ++ +

= − + − += − + − += − + − += − + − +

= − − − += − − − += − − − += − − − +

∫∫∫∫

sech 12 2 2

12 2 2

1(19) tan

1 3 5 5(20) ln

5 33 5

1 2 3(21) 2 3 ln

2 1(22) 8 2 2 2 sin ( ) ( 2) 8 2

(23)2 sin

dx bxc

a ax a b xdx bx

cab aa b x

xdx x x xx x c

xx x dx x x x c

−−−−

−−−−= += += += +−−−−

= += += += +++++

+ −+ −+ −+ − = += += += + ++++

− + − −− + − −− + − −− + − − = − − + += − − + += − − + += − − + + − −− −− −− −

−−−−− = + − − +− = + − − +− = + − − +− = + − − +

====++++

∫∫∫∫

∫∫∫∫12 1

tan (1 2tan )23 3

x−−−− ++++

2 3/ 2 2

22 3/ 2

(24) (25)(2 4) 4

(26) (27) 4 13( 4 5)

x dx dx

x x xdx

x x dxx x

++++ −−−−

+ ++ ++ ++ +− −− −− −− −

∫ ∫∫ ∫∫ ∫∫ ∫

2 3/ 24 2(28) (29)

(9 4 )1(1 cos )

(30) (31)3 2tan (1 sin )

dx x dx

xx xdx x dx

++++++++−−−−

− +− +− +− +

∫ ∫∫ ∫∫ ∫∫ ∫

sec(32) (33)

cosh 2sinh 3 2tansec (1 sin )

(34) (35)2tan sec 1 (1 cos )

(36) (37)3 2cosh 5 4cosh

(38) (39)3 2sinh 5 4sinh

(40) (41)3sinh 2cosh 5cosh 4cosh

(42) (434 5cos

dx x dxx x x

xdx x dxx x x

dx dxx x

dx dxx x x x

+ −+ −+ −+ −++++

+ − ++ − ++ − ++ − +

+ −+ −+ −+ −

∫ ∫∫ ∫∫ ∫∫ ∫

∫∫∫∫ )5 3cos

(44) (45)4 5sin 5 3sin

(46) (47)4 5cosh 5 3cosh

(48) (49)4 5sinh 5 3sinh

dx dxx x

−−−−

+ −+ −+ −+ −

∫∫∫∫

∫ ∫∫ ∫∫ ∫∫ ∫

Integration by Partial Fraction:

A function ( )( )

( )f x

F xg x

==== where ( )f x and ( )g x are polynomials, is

called a rational fraction . If the degree of ( )f x is less than the degree of ( )g x , ( )F x is called proper ; otherwise ,( )F x is called improper. An improper rational fraction can be expressed as the sum of a polynomial and a proper rational fraction .For example

2 21 1

x x= −= −= −= −

+ ++ ++ ++ +

Every proper rational fraction can be expressed as a sum of simpler fractions (partial fractions)whose denominators are of the form

2( ) ( )n nax b and ax bx c+ + ++ + ++ + ++ + + , n being a positive integer. Four cases ,depending upon the nature of the factors of the denominator, arise: CASE 1.Distinct linear factors To each linear factor ax b++++ occurring once in the denominator of a proper rational fraction, there corresponds a single partial fraction of

the form Aax b++++

, where A is a constant to be determined.

CASE 2.Repeated linear factors To each linear factor ax b++++ occurring n times in the denominator of a proper rational fraction, there corresponds sum of n partial fraction of the form

1 22 ...

( ) ( )n

ax b ax b ax b+ + ++ + ++ + ++ + +

++++ + ++ ++ ++ +

where , 1,2,3,...,kA k n==== are a constants to be determined. CASE 3.Distinct quadratic factors To each quadratic factor 2ax bx c+ ++ ++ ++ + occurring once in the denominator of a proper rational fraction, there corresponds a single

partial fraction of the form 2Ax B

ax bx c

+++++ ++ ++ ++ +

, where ,A B is a constants to

be determined. CASE 4.Repeated linear factors To each linear factor 2ax bx c+ ++ ++ ++ + occurring n times in the denominator of a proper rational fraction, there corresponds sum of n partial fraction of the form

1 1 2 22 2 2 2...

( ) ( )n n

nA b xA b x A b x

ax bx c ax bx c ax bx c

+++++ ++ ++ ++ ++ + ++ + ++ + ++ + ++ + + + + ++ + + + + ++ + + + + ++ + + + + +

where , , 1,2,3,..., .k kA B k n==== are a constants to be determined. Solved Examples

(1) Find 2 4

x −−−−∫∫∫∫

First factor the denominator : 2 4 ( 2)( 2)x x x− = − +− = − +− = − +− = − +

Then the fraction 21

(1)2 24

A Bx xx

= += += += +− +− +− +− +−−−−

and clear the fraction to obtain

2 21 ( 2) ( 2)

2 24 4

A B A x B xx xx x

+ + −+ + −+ + −+ + −= + == + == + == + =− +− +− +− +− −− −− −− −

then we have 1 ( 2) ( 2) (2)A x B x= + + −= + + −= + + −= + + − we determined the constants ,A B by one of two methods method 1:general method : equate the coefficients of like powers of x in (2) and solve simultaneously for the constants then equation (2) becomes 1 ( ) (2 2 ) (3)A B x A B= + + −= + + −= + + −= + + − thus 1 (2 2 )A B= −= −= −= − 0 ( )

= += += += +

= = −= = −= = −= = −

method 2:Short method : Substitute in (2) the value of 2x ==== to find

A ,where 11 4

4A A= → == → == → == → =

Substitute in (2) the value of 2x = −= −= −= − to find B ,where 11 4

−−−−= − → == − → == − → == − → =

substitute in (1) the value of ,A B we have:

1 (1/ 4) ( 1/ 4) 1 1 12 2 4 2 24

1 1 1 1 1and

4 2 2 4 2 4 241 1 2

ln ( 2) ln ( 2) ln4 4 2

x x x xxdx dx dx

x x x xx

xx x c

−−−− = + = −= + = −= + = −= + = − − + − +− + − +− + − +− + − + −−−− = − = − == − = − == − = − == − = − = − + − +− + − +− + − +− + − + −−−−

−−−− = − − + = += − − + = += − − + = += − − + = + ++++

∫ ∫ ∫ ∫∫ ∫ ∫ ∫∫ ∫ ∫ ∫∫ ∫ ∫ ∫

(2) Find 3 2( 1)

+++++ −+ −+ −+ −

∫∫∫∫

Solution: First factor the denominator : 3 2 6 ( 2)( 3)x x x x x x+ − = − ++ − = − ++ − = − ++ − = − + Then the fraction

3 2( 1) ( 1)

(1)( 2)( 3) 2 36

x x dx A B Cx x x x x xx x x

+ ++ ++ ++ += = + += = + += = + += = + +− + − +− + − +− + − +− + − ++ −+ −+ −+ −

and clear the fraction to obtain ( 1)

( 2)( 3) 2 3

( 2)( 3) ( 3) ( 2)( 2)( 3)

x A B Cx x x x x x

A x x Bx x Cx xx x x

++++ = + += + += + += + +− + − +− + − +− + − +− + − +

− + + + + −− + + + + −− + + + + −− + + + + −====− +− +− +− +

then we have 1 ( 2)( 3) ( 3) ( 2) (2)x A x x Bx x Cx x+ = − + + + + −+ = − + + + + −+ = − + + + + −+ = − + + + + −

we determined the constants ,A B by one of two methods method 1:general method : equate the coefficients of like powers of x in (2) and solve simultaneously for the constants then equation (2) becomes

21 ( ) ( 3 2 ) 6

0, 3 2 1

6 1 1/ 6, 3 /10, 2 /15

x A B C x A B C x A

A B C A B C

A A B C

+ = + + + + − −+ = + + + + − −+ = + + + + − −+ = + + + + − −∴ + + = + − =∴ + + = + − =∴ + + = + − =∴ + + = + − =− =− =− =− = ⇒⇒⇒⇒ = − = = −= − = = −= − = = −= − = = −

method 2:Short method : Substitute in (2) the value of 0x ==== to find A ,where 6 1 1/ 6A A− =− =− =− = ⇒⇒⇒⇒ = −= −= −= − Substitute in (2) the value of 2x ==== to find B ,where3 10 3/10B B==== ⇒⇒⇒⇒ ==== Substitute in (2) the value of 3x = −= −= −= − to find C ,where 2 15 2/15C B− =− =− =− = ⇒⇒⇒⇒ = −= −= −= − substitute in (1) the value of ,A B we have:

1/ 6 2/15

( 1) 1 3 26 10 2 15 36

21 3 2ln ln 2 ln 3 ln

6 10 15 3

x dx dx dx dxx x xx x x

xx x x c

+ −+ −+ −+ −= + −= + −= + −= + −− +− +− +− ++ −+ −+ −+ −

−−−−−−−−= + − − + = += + − − + = += + − − + = += + − − + = +++++

∫ ∫ ∫ ∫∫ ∫ ∫ ∫∫ ∫ ∫ ∫∫ ∫ ∫ ∫

(3) Find 3 2(3 5)

++++− − +− − +− − +− − +

∫∫∫∫

Solution: First factor the denominator :

3 2 3 2 2

1 ( ) ( 1) ( 1) ( 1)

( 1)( 1) ( 1)( 1)( 1) ( 1) ( 1)

x x x x x x x x x

x x x x x x x

− − + = − − − = − − −− − + = − − − = − − −− − + = − − − = − − −− − + = − − − = − − −

= − − = − − + = − += − − = − − + = − += − − = − − + = − += − − = − − + = − +

3 2 2(3 5)

(1)1 11 ( 1)

x A B Cx xx x x x

++++ = + += + += + += + ++ −+ −+ −+ −− − + −− − + −− − + −− − + −

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

1 11 ( 1) ( 1)( 1)

x A B C A x B x x C xx xx x x x x x

+ − + − + + ++ − + − + + ++ − + − + + ++ − + − + + += + + == + + == + + == + + =+ −+ −+ −+ −− − + − + −− − + − + −− − + − + −− − + − + −

then we have 2(3 5) ( 1) ( 1)( 1) ( 1) (2)x A x B x x C x+ = − + − + + ++ = − + − + + ++ = − + − + + ++ = − + − + + +

For 1 2 8 4x C C==== ⇒⇒⇒⇒ ==== ⇒⇒⇒⇒ ==== For 1 4 2 1/ 2x A A= −= −= −= − ⇒⇒⇒⇒ ==== ⇒⇒⇒⇒ ====

And for 0x ==== ⇒⇒⇒⇒ 5 1/ 2A B C B− + =− + =− + =− + = ⇒⇒⇒⇒ = −= −= −= −

3 2 2(3 5) 1/ 2 1/ 2 4

1 11 ( 1)

xx xx x x x

+ −+ −+ −+ −∴ = + +∴ = + +∴ = + +∴ = + ++ −+ −+ −+ −− − + −− − + −− − + −− − + −

3 2 2(3 5) 1/ 2 1/ 2 4

1 11 ( 1)

1 1 4 1 1 4ln 1 ln 1 ln

2 2 ( 1) 2 1 1

x dxand dx dx dx

x xx x x x

xx x c

+ −+ −+ −+ −= + += + += + += + ++ −+ −+ −+ −− − + −− − + −− − + −− − + −

++++= + − − − = − += + − − − = − += + − − − = − += + − − − = − +− − −− − −− − −− − −

∫ ∫ ∫ ∫∫ ∫ ∫ ∫∫ ∫ ∫ ∫∫ ∫ ∫ ∫

(4) Find 4 3

3 2( 1)x x x dx

− − −− − −− − −− − −−−−−

∫∫∫∫

Solution: The integrand is an improper fraction. By division

3 2 3 2 2

( 1) 1 1(1)

( 1)( 1)

x x x x xx x

x x x x x x

x A B Cx xx x x

− − − + +− − − + +− − − + +− − − + += − = −= − = −= − = −= − = −− − −− − −− − −− − −

++++ = + += + += + += + +−−−−−−−−

1 ( 1) ( 1)

( 1) ( 1)

1 ( 1) ( 1) (3)

x Ax x B x Cx

x x x x

x Ax x B x Cx

+ − + − ++ − + − ++ − + − ++ − + − +====− −− −− −− −

∴ + = − + − +∴ + = − + − +∴ + = − + − +∴ + = − + − +

For 0 1 1x B B==== ⇒⇒⇒⇒ = −= −= −= − ⇒⇒⇒⇒ = −= −= −= − For 1 2 2x C C==== ⇒⇒⇒⇒ ==== ⇒⇒⇒⇒ ==== For 2x ==== ⇒⇒⇒⇒ 3 2 4 2A B C A= + += + += + += + + ⇒⇒⇒⇒ = −= −= −= −

2 21 2 1 2

( 1)( 1)

xx xx x x

+ − −+ − −+ − −+ − −= + += + += + += + +−−−−−−−−

3 2 2 2

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

2 1 2 1 12ln 2ln 1

( 1) 2

1 12ln

x x x xdx x dx x dx

x xx x x x x

x dx x x xx x xx

− − − + − −− − − + − −− − − + − −− − − + − −= − = − + += − = − + += − = − + += − = − + + −−−−− −− −− −− −

−−−−+ + + = + − − −+ + + = + − − −+ + + = + − − −+ + + = + − − − −−−−

= − + += − + += − + += − + +−−−−

∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫

∫∫∫∫

(5) Find 3 2

4 2( 2)

x x x dx

+ + ++ + ++ + ++ + ++ ++ ++ ++ +

∫∫∫∫

Solution: First factor the denominator : 4 2 2 23 2 ( 1)( 2)x x x x+ + = + ++ + = + ++ + = + ++ + = + + Then we write

4 2 2 2( 2)

(1)3 2 1 2

x x x Ax B Cx D

x x x x

+ + + + ++ + + + ++ + + + ++ + + + += += += += ++ + + ++ + + ++ + + ++ + + +

and 3 2 2 2

4 2 2 2

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

( ) ( ) (2 ) (2 )

Hence 1, 1, 2 1, 2 2

Solving simultaneously 0, 1, 1, 0 thus

( 2) 1

3 2 1 2

x x x Ax B x Cx D x

A C x B D x A C x B D

A C B D A C and B D

A B C D

x x x x

+ + + = + + + + ++ + + = + + + + ++ + + = + + + + ++ + + = + + + + +

= + + + + + + += + + + + + + += + + + + + + += + + + + + + ++ = + = + = + ++ = + = + = + ++ = + = + = + ++ = + = + = + +

= = = == = = == = = == = = =

+ + ++ + ++ + ++ + + = += += += ++ + + ++ + + ++ + + ++ + + +

3 21 2

4 2 2 2( 2) 1

tan ln 223 2 1 2

x x x dx x dxdx x x c

x x x x−−−−+ + ++ + ++ + ++ + + = + = + + += + = + + += + = + + += + = + + +

+ + + ++ + + ++ + + ++ + + +∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫

(6) Find 2

2 2(2 3)

++++++++

∫∫∫∫

Solution: we write 2

2 2 2 2 2

2 2 3 2

2 2 2 2 2

(2 3)then

( 1) ( 1) ( 1)

2 3 ( )( 1) ( ) ( )

0, 2, 0, 3

0, 2, 0, 1

(2 3) 2 1

( 1) ( 1) ( 1)

x Ax B Cx D

x Ax B x Cx D Ax Bx A C x B D

A B A C B D

A B C D

+ + ++ + ++ + ++ + += += += += ++ + ++ + ++ + ++ + +

+ = + + + + = + + + + ++ = + + + + = + + + + ++ = + + + + = + + + + ++ = + + + + = + + + + +∴ = = + = + =∴ = = + = + =∴ = = + = + =∴ = = + = + =

= = = == = = == = = == = = =

++++ = += += += ++ + ++ + ++ + ++ + +

2 2 2 2 2(2 3) 2 1

( 1) ( 1) ( 1)

xdx dx dx

++++ = += += += ++ + ++ + ++ + ++ + +

∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫

1To solve tan sec ,

1 sec 1 1cos sin2

2 4( 1) sec

dx put x u dx u dux

u dudx udu u u

==== ⇒⇒⇒⇒ ====++++

= = = += = = += = = += = = +++++

∫∫∫∫

∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫

1 12 2

2 2 2 2 2 2

1 1 1 1 1tan 2sin cos tan

2 4 2 2 1 11 1

tan2 2 1

(2 3) 2 1 1 12tan tan

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

5 1tan

2 2 ( 1)

xx u u x

x xdx dx dx x x

x x x x

− −− −− −− −

−−−−

− −− −− −− −

−−−−

= + = += + = += + = += + = ++ ++ ++ ++ +

= += += += +++++

++++∴ = + = + +∴ = + = + +∴ = + = + +∴ = + = + ++ + + ++ + + ++ + + ++ + + +

= + += + += + += + +++++

∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫

Exercise (7)

1 3: ln

1 1: ln

1 2: ln ( 1)( 4)

1 4 12 6: 3 ln(1 )

22 (1 ) 2(1 )

(2)7 6

(3)3 4

(4)(1 )

2 3 3(5)

xAns c

Ans x x c

Ans x x xx x c

x x x xdx

−−−−++++

++++++++

+ − ++ − ++ − ++ − +

−−−−− − − − +− − − − +− − − − +− − − − +

−−−− − +− +− +− +

−−−− + ++ ++ ++ + − +− +− +− + −−−−

− + − +− + − +− + − +− + − +

− +− +− +− +

∫∫∫∫

∫∫∫∫1 2

: ln3 22 2 3

12: ln( 1)

1 42 1: ln( 4) tan

22 2 4

1 12 1: ln ( 1) tan

22 2 1

2 4(7)

Ans x cx

xAns x c

xAns x x c

x x xdx

+ ++ ++ ++ +− +− +− +− +

+ + ++ + ++ + ++ + +++++

−−−−+ + + ++ + + ++ + + ++ + + +++++

−−−−+ − − ++ − − ++ − − ++ − − +++++

+ ++ ++ ++ +

+ −+ −+ −+ −

∫∫∫∫

3 23 2 2 11

: ln tan2 1 3 3( 1)

1 cosh 2: ln

( )( 1)

sin(10)

cos (1 cos )

x x x xAns c

xAns c

− + −− + −− + −− + −−−−−− + +− + +− + +− + +++++++++

++++++++

+ ++ ++ ++ + + ++ ++ ++ +

∫∫∫∫

Miscellaneous substitution:

ve integer put( )

(i)Form

dx m ax b uax b

+ + =+ + =+ + =+ + =++++∫∫∫∫

Example(1): Find 2

3 / 2(5 2 )

x−−−−∫∫∫∫

Solution: Put 5 2x u− =− =− =− = 2dx du∴ − =∴ − =∴ − =∴ − =

(((( ))))

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

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

( 3 / 2) ( 1 / 2) ( 1 / 2)

( 1 / 2) (1 / 2) (3 / 2)

1 (5 ) 1 25 108 8(5 2 )

1 25 108

125 10

25 208 3

1 50 220 5 2 5 2

8 35 2

x u u udx du du

u u u du

− − −− − −− − −− − −

−−−−

− − − − +− − − − +− − − − +− − − − += == == == =−−−−

−−−−= − += − += − += − +

−−−− = − += − += − += − +

= + − − −= + − − −= + − − −= + − − − −−−−

∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫

∫∫∫∫

ve integer put( )

(ii)Fom n

dx ax bdx m u

xx ax br +++++ =+ =+ =+ =

++++∫∫∫∫

Example(2): Find 3 / 2 5 / 2(3 5)

x x ++++∫∫∫∫

Solution:

23 5 5 5 5

, 3 ,5 3

x xy y dx dy dx dy x

x x yx

+ −+ −+ −+ −= + == + == + == + = ⇒⇒⇒⇒ ==== ⇒⇒⇒⇒ = ∴ == ∴ == ∴ == ∴ =− −− −− −− −

3 / 2 5 / 2 4 5 / 2(3 / 2) (5 / 2) 5 / 2

13 5 5(3 5) 5 ( )

dx x dy x dyxx x x yx xx

−−−−= == == == =++++++++ −−−−∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫

2 5 / 2 2 5 / 2 5 / 2

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

1 1 ( 3) 1 ( 6 9)5 5 1255

1 1( 6 9 ) 2 12 6

125 125

2 (3 5)1 12 6125 (3 5) (3 5)

dy y dy y y dy

x y y y

y y y dy y y y

x x xc

− − − − −− − − − −− − − − −− − − − −

− − − − − +− − − − − +− − − − − +− − − − − += = == = == = == = =

− −− −− −− − = − + = + += − + = + += − + = + += − + = + +

++++−−−− = + − += + − += + − += + − + + ++ ++ ++ +

∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫

∫∫∫∫

ve integer( )

1put ln ln

(iii)Forn

dxdx m

x ax b

dx dux n x u n

++++++++

= → = − → = −= → = − → = −= → = − → = −= → = − → = −

∫∫∫∫

Example(3): Find 3( 2)

x x ++++∫∫∫∫

Solution: 3

13ln ln ,

31 1 1 2

ln(1 2 ) ln(1 )1 3 1 2 6 6( 2) 3 ( 2)

dx duput x x u

u x udx du du

dx u cux x xu

= ∴ = − = −= ∴ = − = −= ∴ = − = −= ∴ = − = −

−−−−= − = − + = − + += − = − + = − + += − = − + = − + += − = − + = − + +++++++++ ++++

∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫

(iv)For

Thequadratic Form( canbechangedto a sum or difference of two squars

dx dxax bx cax bx c

ax bx c

+ ++ ++ ++ ++ ++ ++ ++ +

+ ++ ++ ++ +

∫ ∫∫ ∫∫ ∫∫ ∫

( )(v)For

weshalldivide the numerator into two parts , one is derivative of the function under the root and the other is constant

mx ndx

ax bx c

+ ++ ++ ++ +∫∫∫∫

2( )(vi)For

weshall divide the numerator into two parts , one is derivative of the function in denominator and the other part is constant

mx ndx

ax bx c

+++++ ++ ++ ++ +∫∫∫∫

Example(4): Find 2

x x+ ++ ++ ++ +∫∫∫∫

Solution:

3 3 32 22 4 3 2 (3 / 2) ( 2 1) 1 (3 / 2)

3 3. 2 tan 2( 1)

2 2( 1) (1/ 2)

dx dx dx

x x x x x x

x−−−−

= == == == =+ + + + + + − ++ + + + + + − ++ + + + + + − ++ + + + + + − +

= = + += = + += = + += = + ++ ++ ++ ++ +

∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫

∫∫∫∫

Note: To Complete the square of 2x ax++++ we add 2

by positive

sign and add the same value by negative sign 2 2

2 2 2 2 2 2 2( ) ( ) ( ) ( ) ( )2 2 4 2 2 2a a a a a a

x ax x ax x ax x+ = + + − = + + − = + −+ = + + − = + + − = + −+ = + + − = + + − = + −+ = + + − = + + − = + −

x x− −− −− −− −∫∫∫∫

Solution:

2 2 27 6 7 (6 ) 7 (6 9 9)

dx dx dx

x x x x x x= == == == =

− − − + − + + −− − − + − + + −− − − + − + + −− − − + − + + −∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫

12 2 2

( 3)sin

416 (6 9) 4 ( 3)

dx dx xc

−−−− ++++= = = += = = += = = += = = +− + + − +− + + − +− + + − +− + + − +

∫ ∫∫ ∫∫ ∫∫ ∫

+++++ ++ ++ ++ +

∫∫∫∫

Solution: 2(3 3) 6 1

4 5 (6 1) 6 4, 5

d x x x

Then put x A x B A A B

+ + = ++ + = ++ + = ++ + = ++ = + ++ = + ++ = + ++ = + + ⇒⇒⇒⇒ = + == + == + == + =

2 13, 4 5 (2 / 3)(6 1) (13 / 3)

3 34 5 (2 / 3)(6 1) (13 / 3)

3 3 3 32 (6 1) 13 1

(1)3 33 3 3 3

2 (6 1) 2ln 3 3 (2)

3 33 313 1 13 1

1 1 1 353 33( 1) 3( )3 3 36 36

A B x x

x xdx dx

x x x xx

dx dxx x x x

xNow dx x x

And dx dxx x x x

∴ = =∴ = =∴ = =∴ = = ⇒⇒⇒⇒ + = + ++ = + ++ = + ++ = + +

+ + ++ + ++ + ++ + +====+ + + ++ + + ++ + + ++ + + +

++++= += += += ++ + + ++ + + ++ + + ++ + + +

++++ = + += + += + += + ++ ++ ++ ++ +

====+ + + + ++ + + + ++ + + + ++ + + + +

∫ ∫∫ ∫∫ ∫∫ ∫

∫∫∫∫

∫ ∫∫ ∫∫ ∫∫ ∫

3 11 359 ( ) ( )6 36

16( )13 6 6. tan

9 35 3526 (6 1)

tan (3)3 35 35

(2),(3) (1)

4 5 2 26 (6 1)ln 3 3 tan

3 3 35 353 3

From in

x xdx x x c

−−−−

+ ++ ++ ++ +

++++====

+ ++ ++ ++ +∴ = + + + +∴ = + + + +∴ = + + + +∴ = + + + ++ ++ ++ ++ +

∫∫∫∫

Example(7): Find 2

3 6 10

−−−−

+ ++ ++ ++ +∫∫∫∫

Solution:

(3 6 10) 6 6

2 3 (6 6) 6 3, 6 2

1 1, 2 6( ) 5 2 3 ( 1/ 2)(6 6) 5

2 2(2 3 ) ( 1/ 2)(6 6) 5

3 6 10 3 6 10

d x x x

Then put x A x B A A B

A B x x

x xdx dx

x x x x

+ + = ++ + = ++ + = ++ + = +− = + +− = + +− = + +− = + + ⇒⇒⇒⇒ = − + == − + == − + == − + =

−−−−∴ = = − − =∴ = = − − =∴ = = − − =∴ = = − − = ⇒⇒⇒⇒ − = − + +− = − + +− = − + +− = − + +

− − + +− − + +− − + +− − + +====+ + + ++ + + ++ + + ++ + + +

∫ ∫∫ ∫∫ ∫∫ ∫

1 (6 6)5 (1)

2 3 6 10 3 6 101 (6 6)

3 6 10 (2)2 3 6 10

3 10 3 73 6 10 2 2 13 3

5 5 3( 1)sinh (3)

3 7 3 7( 1)

(4 5)(2),(3) (1) 3 6

x dx dx

x x x xx dx

Now x xx xdx dx dx

Andx x x x x x

xFrom in dx x

−−−−

− +− +− +− += += += += ++ + + ++ + + ++ + + ++ + + +

− +− +− +− + = − + += − + += − + += − + ++ ++ ++ ++ +

= == == == =+ ++ ++ ++ + + + + + ++ + + + ++ + + + ++ + + + +

++++= = += = += = += = ++ ++ ++ ++ +

++++∴ = − +∴ = − +∴ = − +∴ = − ++ ++ ++ ++ +

∫ ∫∫ ∫∫ ∫∫ ∫

∫∫∫∫

∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫

∫∫∫∫

∫∫∫∫15 3( 1)

10 sinh3 7

xx c−−−− +++++ + ++ + ++ + ++ + +

( ) ( )

y =G(x)2 y =G(x)

1=F(x)

(vii)For

(a)whenF(x)and G(x) are both linear put

(b)whenF(x)isquadratic and G(x) is linear put

(c)whenF(x)is linearand G(x) isquadratic put

(d)whenF(x)and G(x) are both q

dxF x G X∫∫∫∫

2 G(x)y =

F(x)uadratic put

Example(8): Find (4 1) 3 1

x x+ ++ ++ ++ +∫∫∫∫ case ( )a

Solution: 2 21

3 1 4 2 , ( 1)3

put x y dx ydy x y+ = ∴ = = −+ = ∴ = = −+ = ∴ = = −+ = ∴ = = −

(2 / 3) 12

12(4 1) 3 1 4 1(4 / 3)( 1) 1 ( )4

1 2 3 1 1coth (2 ) coth (2 3 1) ln

2 2 3 1 1

dx ydy dy dy

x x yy y y

xy x c

x− −− −− −− −

= = == = == = == = = + ++ ++ ++ + −−−−− +− +− +− + −−−−

− + −− + −− + −− + −= − = − + = += − = − + = += − = − + = += − = − + = ++ ++ ++ ++ +

∫ ∫ ∫ ∫∫ ∫ ∫ ∫∫ ∫ ∫ ∫∫ ∫ ∫ ∫

(2 / 3) 12

12(4 1) 3 1 4 1(4 / 3)( 1) 1 ( )4

1 2 3 1 1coth (2 ) coth (2 3 1) ln

2 2 3 1 1

dx ydy dy dy

x x yy y y

xy x c

x− −− −− −− −

= = == = == = == = = + ++ ++ ++ + −−−−− +− +− +− + −−−−

− + −− + −− + −− + −= − = − + = += − = − + = += − = − + = += − = − + = ++ ++ ++ ++ +

∫ ∫ ∫ ∫∫ ∫ ∫ ∫∫ ∫ ∫ ∫∫ ∫ ∫ ∫

Example(9): Find 2(4 9) 2 5

x x− −− −− −− −∫∫∫∫ case ( )b

Solution: 2 2

2 2 4 2

12 5 2 2 , ( 5)

(4 9) 2 5 ( 5) 9 10 16

put x y dx ydy x y

dx ydy dy

x x y y y

− = ∴ = = +− = ∴ = = +− = ∴ = = +− = ∴ = = +

∴ = =∴ = =∴ = =∴ = =− − + − + +− − + − + +− − + − + +− − + − + +∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫

2 2 2 2

1 16 6( 2)( 8) 2 8

1 1 1 1. tan . tan

6 62 2 2 2 2 2

1 2 5 1 2 5tan tan

6 2 2 12 2 2 2

dy dy dy

y y y y

− −− −− −− −

= = −= = −= = −= = −+ + + ++ + + ++ + + ++ + + +

= − += − += − += − +

− −− −− −− −= − += − += − += − +

∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫

Example(10): Find 2( 1) 4 2

x x x+ + ++ + ++ + ++ + +∫∫∫∫ (case ( )c )

Solution:

(((( )))) (((( ))))

2 21 1 1

11 , (1 )

( 1) 4 2 (1 ) 4 (1 ) 2

x y dx y dy x y yy

dx y dy

x x x y y y y y

− − −− − −− − −− − −

−−−−

− − −− − −− − −− − −

+ = = ∴ = − = −+ = = ∴ = − = −+ = = ∴ = − = −+ = = ∴ = − = −

−−−−∴ =∴ =∴ =∴ =+ + ++ + ++ + ++ + + − + − +− + − +− + − +− + − +

∫ ∫∫ ∫∫ ∫∫ ∫

1 12 2

1sin sin

2 2( 1)1 2 2 ( 1)

dy dy y xc c

xy y y

− −− −− −− − − − − −− − − −− − − −− − − − = = = − + = − += = = − + = − += = = − + = − += = = − + = − + ++++ + − − −+ − − −+ − − −+ − − −∫ ∫∫ ∫∫ ∫∫ ∫

Example(11): Find 2 2(1 ) 1

x x+ −+ −+ −+ −∫∫∫∫ case (d)

Solution: 2

22 2 2

1 4 1,

11 (1 )

x xdx yput y dy x

− − −− − −− − −− − −= ∴ = == ∴ = == ∴ = == ∴ = =+++++ ++ ++ ++ +

2 2 2 2 2

(1 ) (1 )

(1 ) 1 ( 4 )(1 ) 1 ( 4 ) 1

11 1 14 21 1 1 2 2 2

1 1sin (2 1)

2 2 1 1 2 2( )

1 1 3sin

dx x dy x dy

x x x x x x x

y dy dy

y y y y y yy y

−−−−

+ ++ ++ ++ +∴ = = =∴ = = =∴ = = =∴ = = =+ − − + − − −+ − − + − − −+ − − + − − −+ − − + − − −

−−−−++++ ++++− − −− − −− − −− − − = = == = == = == = =− − −− − −− − −− − − −−−−−−−−+ ++ ++ ++ +

− −− −− −− −= = − += = − += = − += = − +− −− −− −− −

− −− −− −− −= += += += + ++++

∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫

∫∫∫∫

(vii)Forn

x ax b++++∫∫∫∫ put

ln 2ln ,n dx dyx n x y n

−−−−= ∴ = − == ∴ = − == ∴ = − == ∴ = − =

x x ++++∫∫∫∫

Solution: 4 2

14 2 2

14ln 2 ln , 4

1 1 1sinh 3

2 2 2 33 3 1 3

1 3sinh

dx dyput x y x y

dx dy dyy

x x y y y

−−−−

−−−−−−−−

−−−−

−−−−= = ∴ = − == = ∴ = − == = ∴ = − == = ∴ = − =

− − −− − −− − −− − −∴ = = =∴ = = =∴ = = =∴ = = =+ + ++ + ++ + ++ + +

−−−−= += += += +

∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫

Exercise(7) Find

2 2 3/ 2

2 3 3/ 2 2 5/ 2 3/ 2

3 4(1) (2) (3)

(3 4) (5 3) (2 3)

1 1 1(4) (5) (6)

(3 4) (2 3) (3 1)

1 1 1(7) (8) (9)

(3 1) (1 2 ) (4 1)

1 1 1(10) (11) (12)

2 5 2 5 4 31

x x xdx dx dx

dx dx dxx x x x x x

− + −− + −− + −− + −

− + +− + +− + +− + +

+ − −+ − −+ − −+ − −

+ − − + + −+ − − + + −+ − − + + −+ − − + + −

−−−−

∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫

1 1(14) (15)

2 12 4 7 61 2 3 3 1

(16) (17) (18)2 3 2 5 4 3

3 3 5 3(19) (20) (21)

8 2 12 4 7 61 1 1

(22) (23) (24)2 3 2 5 4 3

dx dx dxx x x x x x

x x xdx dx dx

x x x x x xx x x

dx dx dxx x x x x x

− − − − −− − − − −− − − − −− − − − −+ + −+ + −+ + −+ + −

+ − − + + −+ − − + + −+ − − + + −+ − − + + −+ − −+ − −+ − −+ − −

− − − − − −− − − − − −− − − − − −− − − − − −

+ − − + + −+ − − + + −+ − − + + −+ − − + + −

− −− −− −− −

∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫

1 1(26) (27)

12 4 7 61 2 3 3 1

(28) (29) (30)2 3 2 5 4 33 3 5 3

(31) (32) (33)8 2 12 4 7 6

(34) (35) (36)( 1) 3 5 (2 1) 2 5 ( 3) 2 3

(37)(2

dx dx dxx x x x x

x x xdx dx dx

x x x x x xx x x

dx dx dxx x x x x xdx dx dx

x x x x x x

− − − −− − − −− − − −− − − −+ + −+ + −+ + −+ + −

+ − − + + −+ − − + + −+ − − + + −+ − − + + −+ − −+ − −+ − −+ − −

− − − − − −− − − − − −− − − − − −− − − − − −

+ − − ++ − − ++ − − ++ − − + + −+ −+ −+ −

∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫

2 2 2 2

(38) (39)1) 5 (2 3) 3 1 (3 1) 3

(40) (41)(3 1) 2 1 ( 2) 3 1

x x x x x x

x x x x

+ ++ ++ ++ + + − + ++ − + ++ − + ++ − + +

+ + − −+ + − −+ + − −+ + − −

∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫

∫ ∫∫ ∫∫ ∫∫ ∫

Definite integration Area under the curve: Given a continuous function ( )f x on the interval [ , ]a b such that

( ) 0f x >>>> we can approximate the area enclosed by the curve of ( )f x , x −−−− axis and the two lines ,x a x b= == == == = by dividing the interval [ , ]a b into subintervals by the set of points 0 1 2{ , , ,..., }np x x x x==== such that

0 1 2 ... na x x x x b= < < < < == < < < < == < < < < == < < < < = then the area given by:

1 0 1 2 1 2 1

( ) ( ) ( ) ( ) ... ( ) ( )

( ) ( )

n n n nn

k k kk

S x x f x x x f x x x f x

x x f x

−−−−

−−−−====

= − + − + + −= − + − + + −= − + − + + −= − + − + + −

= −= −= −= −∑∑∑∑

if we divide the interval into n equal subintervals with length 1( ), 1,2,...,k k kx x x k n−−−−∆ = − =∆ = − =∆ = − =∆ = − =

Then 1 1 2 2

( ) ( ) ... ( )

n n nn

S x f x x f x x f x

x f x====

= ∆ + ∆ + + ∆= ∆ + ∆ + + ∆= ∆ + ∆ + + ∆= ∆ + ∆ + + ∆

= ∆= ∆= ∆= ∆∑∑∑∑

Upper and lower sum:

x x b==== x a====

( )y f x====

To discuss the concept of integral of the function ( )f x , we must first introduce some notation. If [ , ]I a b==== is closed , bounded interval , then a partition of I is finite order d set 0 1: { , , .... }nP x x x==== of point of I such that

0 1 2 ... na x x x x b= < < < < == < < < < == < < < < == < < < < = The points of the partition P can be used to divide I into non-overlapping subintervals 0 1 1 2 1[ , ],[ , ], ...,[ , ]n nx x x x x x− . Let ( )f x continuous function ( )f x defined on the interval [ , ]a b and let P a partition of I we let

inf { ( ) : [ , ]}

sup{ ( ) : [ , ]}k k k

m f x x x x

M f x x x x−−−−

−−−−

= ∈= ∈= ∈= ∈= ∈= ∈= ∈= ∈

The lower sum of ( )f x corresponding to the partition P is defined to be

( ; ) ( )n

k k kk

L P f m x x −−−−====

= −= −= −= −∑∑∑∑

The upper sum of ( )f x corresponding to the partition P is defined to be

( ; ) ( )n

k k kk

U P f M x x −−−−====

= −= −= −= −∑∑∑∑

if ( )f x is positive function then the lower sum ( ; )L P f can be interpreted as the area of union of rectangles with base 1[ , ]k kx x−−−− and

0a x==== 1x 1nx −−−− x

height km ,Similarly the Upper sum ( ; )U P f can be interpreted as the area of union of rectangles with base 1[ , ]k kx x−−−− and height kM Riemann’s sum: Let ( )f x is real valued function defined on the interval [ , ]a b , and let

0 1 2{ , , , ..., }np x x x x==== is a partition on [ , ]a b . into n subintervals

0 1 1 2 1[ , ],[ , ], ...,[ , ]n nx x x x x x− , choose a points 1 2 3, , , ..., nξ ξ ξ ξξ ξ ξ ξξ ξ ξ ξξ ξ ξ ξ such that

1 1 2 3[ , ], , , , ...,k k kx x k nξξξξ −−−−∈ =∈ =∈ =∈ = We define Riemann’s sum in the form

1 0 1 2 1 2 1

( ; ) ( ) ,

( ) ( ) ( ) ( ) ... ( ) ( )

n k k k k kk

S P f f x x x x

x x f x x f x x f

ξξξξ

ξ ξ ξξ ξ ξξ ξ ξξ ξ ξ

−−−−====

−−−−

= ∆ ∆ = −= ∆ ∆ = −= ∆ ∆ = −= ∆ ∆ = −

= − + − + + −= − + − + + −= − + − + + −= − + − + + −

∑∑∑∑

Since ( )k k km f x M< << << << < then

1 1 11 1 1

( ) ( )( ) ( )

( ; ) ( ; ) ( ; )

k k k k k k k k kk k k

m x x f x x M x x

L P f S P f U P f

ξξξξ− − −− − −− − −− − −= = == = == = == = =

− ≤ − ≤ −− ≤ − ≤ −− ≤ − ≤ −− ≤ − ≤ −

≤ ≤≤ ≤≤ ≤≤ ≤

∑ ∑ ∑∑ ∑ ∑∑ ∑ ∑∑ ∑ ∑

Upper and lower Integrals: The lower integral of ( )f x on I is the number

lim ( ; ) lim ( )n

l k k kn n k

I L P f m x x −−−−→∞ →∞→∞ →∞→∞ →∞→∞ →∞ ====

= = −= = −= = −= = −

∑∑∑∑

The Upper integral of ( )f x on I is the number

lim ( ; ) lim ( )n

u k k kn n k

I U P f M x x −−−−→∞ →∞→∞ →∞→∞ →∞→∞ →∞ ====

= = −= = −= = −= = −

∑∑∑∑

Riemann’s Integral: (Riemann’s Criterion for integral) Let ( )f x is a continuous function on the interval [ , ]a b , and let p be a partition on [ , ]a b . ( )f x to be integrable on [ , ]a b if the limit

1lim ( ; ) lim ( )

R n k kn n k

I S P f f xξξξξ→∞ →∞→∞ →∞→∞ →∞→∞ →∞ ====

= = ∆= = ∆= = ∆= = ∆∑∑∑∑ exist and independent on

choosing the points 1 2 3, , , ..., nξ ξ ξ ξξ ξ ξ ξξ ξ ξ ξξ ξ ξ ξ . And we write

1( ) lim ( )

R k kn ka

f x dx I f xξξξξ→∞→∞→∞→∞ ====

= = ∆= = ∆= = ∆= = ∆∑∑∑∑∫∫∫∫

Corollary: If ( )f x is continuous function on [ , ]a b the

1lim ( ; ) lim ( )

n k kn n k

S P f f xξξξξ→∞ →∞→∞ →∞→∞ →∞→∞ →∞ ====

= ∆= ∆= ∆= ∆∑∑∑∑ is exist and independent on choosing

the points 1 2 3, , , ..., nξ ξ ξ ξξ ξ ξ ξξ ξ ξ ξξ ξ ξ ξ ,and ( )f x is integrable on [ , ]a b . Example(1): Show that the function 4 1( )f x x= −= −= −= − is integrable on the interval 0 1[ , ] and find the value of the integral. Solution: Since the function is continuous function on the interval 0 1[ , ] thus it is integrable on the interval 0 1[ , ] Consider

0 1 2 1{ 0, , , ..., , , ..., 1}r r np x x x x x x−−−−= = == = == = == = = be a partition to the interval v

into n subinterval with length 1b an n−−−− ==== and

1x 2x 3x 4x 5x 6x 7x 8x 9x 10x 0x

( )0 , 1k

b a k kx k n

n n−−−−= + = ≤ ≤= + = ≤ ≤= + = ≤ ≤= + = ≤ ≤ and 1

1( ) , 1,2, ...,k kx x k n

n−−−−− = =− = =− = =− = =

0 1 21 2

0, , , ..., 1nn

x x x xn n n

= = = = == = = = == = = = == = = = =

Choose rξξξξ such that

(((( ))))

2 2 21 1 1 1

11 2 12 2 2

2 1 4 2( ) 4 1 4 1 1

1 4 2( ; ) ( ) 1

1 1 4 1 14 2 2 1

4 ( 1) 1 2 ( 1. 2 1

n k k kr k

n n n n

k k k k

kx x k kn n n

kS P f x x f

k kn nn n n

n n n nn

ξξξξ

ξ ξξ ξξ ξξ ξ

ξξξξ

−−−−

−−−−= == == == =

= = = == = = == = = == = = =

−−−−++++ −−−−= = + == = + == = + == = + =

− −− −− −− − = − = − = −= − = − = −= − = − = −= − = − = −

−−−− ∴ = − = −∴ = − = −∴ = − = −∴ = − = −

= − − = − −= − − = − −= − − = − −= − − = − −

+ ++ ++ ++ + = − − == − − == − − == − − =

∑ ∑∑ ∑∑ ∑∑ ∑

∑ ∑ ∑ ∑∑ ∑ ∑ ∑∑ ∑ ∑ ∑∑ ∑ ∑ ∑

2 ( 1) 2lim ( ; ) lim 1 2 1 1n

n nS P f

nn→∞ →∞→∞ →∞→∞ →∞→∞ →∞

− −− −− −− −

++++ ∴ = − − = − =∴ = − − = − =∴ = − − = − =∴ = − − = − =

Example(2): Show that the function ( ) 4 1f x x= −= −= −= − is integrable on the interval [1, 7] and find the value of the integral. Solution: Since the function is continuous function on the interval [1, 7] thus it is integrable on the interval 1 7[ , ] Consider 0 1 2 1{ 1, , , ..., , , ..., 7 }r r np x x x x x x−−−−= = == = == = == = = be a partition to the interval [1, 7] Then

( ) 61 , 1k

b a k kx a k n

n n−−−−= + = + ≤ ≤= + = + ≤ ≤= + = + ≤ ≤= + = + ≤ ≤

( ) , 1,2, ...,k kx x k nn−−−−− = =− = =− = =− = =

0 1 26 6 6

1, 1 , 1 2,..., 1 7nx x x x nn n n

= = + = + ⋅ = + ⋅ == = + = + ⋅ = + ⋅ == = + = + ⋅ = + ⋅ == = + = + ⋅ = + ⋅ =

Choose rξξξξ such that

(((( ))))1 6 11 6 6 31 1 1

2 2k k

rkx x k kn n n

ξξξξ −−−− −−−−++++ −−−−= = + + + = += = + + + = += = + + + = += = + + + = +

(((( ))))

2 21 1 1

6 3 24 12( ) 4 1 4 1 1 3

6 24 12( ; ) ( ) 3

18 144 72

18 144 721 1

144 72 1 7218 1 18 72 1

n k k kr k

kS P f x x f

kn n n

n nn n

ξ ξξ ξξ ξξ ξ

ξξξξ−−−−= == == == =

= = == = == = == = =

−−−− = − = + − = + −= − = + − = + −= − = + − = + −= − = + − = + −

∴ = − = + −∴ = − = + −∴ = − = + −∴ = − = + −

= + −= + −= + −= + −

= + ⋅ + − ⋅ = + + −= + ⋅ + − ⋅ = + + −= + ⋅ + − ⋅ = + + −= + ⋅ + − ⋅ = + + −

∑ ∑∑ ∑∑ ∑∑ ∑

∑∑∑∑

∑ ∑ ∑∑ ∑ ∑∑ ∑ ∑∑ ∑ ∑

1 72lim ( ; ) lim 18 72 1 18 72 90n

n nS P f

n n→∞ →∞→∞ →∞→∞ →∞→∞ →∞

∴ = + + − = + =∴ = + + − = + =∴ = + + − = + =∴ = + + − = + =

Example(3):

Show that the function 2( ) 4 3f x x= += += += + is integrable on the interval

2 10[ , ] and find the value of 10

(4 3)x dx++++∫∫∫∫ .

Solution: Since the function is continuous function on the interval 2 10[ , ] thus it is integrable on the interval 2 10[ , ] Let

0 1 2 1{ 2, , , ..., , , ..., 20}r r np x x x x x x−−−−= = == = == = == = = be a partition on the interval 2 10[ , ] into n subintervals with length is

110 2 8

( )k k kb a

x x xn n n−−−−− −− −− −− −∆ = − = = =∆ = − = = =∆ = − = = =∆ = − = = =

(((( ))))0 1 2

8 162, 2 , 2 , ...,

8 1 8 82 , 2 , ..., 2 10k k n

x x xn n

k k nx x x

n n n−−−−

= = + = += = + = += = + = += = + = +

−−−−= + = + = + == + = + = + == + = + = + == + = + = + =

22 2 2

8 4( ) 4 2 3

64 32 16 16 644 4 3

256 64 128 64 25619

x x kn n

k k kn nn n n

ξξξξ

−−−− ++++= = + −= = + −= = + −= = + −

= + − += + − += + − += + − +

= + + + − − += + + + − − += + + + − − += + + + − − +

= + + + − −= + + + − −= + + + − −= + + + − −

23 3 2 2 3

8( ; ) ( ) ( ) ( )

152 2048 512 1024 512 2048

n k k k kk k

S p f x x f fn

k k kn n n n n n

ξ ξξ ξξ ξξ ξ−−−−= == == == =

= − == − == − == − =

= + + + − −= + + + − −= + + + − −= + + + − −

∑ ∑∑ ∑∑ ∑∑ ∑

∑∑∑∑

2152 2048 1 1 512

1024 1 512 2048 11 1

nn n n n

n n n n

= + + + += + + + += + + + += + + + +

+ + − − ++ + − − ++ + − − ++ + − − +

1x 2x 3x 4x 5x 6x 7x 8x 9x 10x 0x

(((( )))) (((( )))) (((( ))))

(((( ))))

1 1(1) 1 , (2) 1 2 1

1(3) 1

r n n r n n n

= + = + += + = + += + = + += + = + +

= += += += +

∑ ∑∑ ∑∑ ∑∑ ∑

∑∑∑∑

0 1 2{ , , , ..., }np x x x x====

(((( )))) (((( )))) (((( ))))0

1024 4040lim ; lim ; 152 2 512

n nx n

x x as nn

S p f S p f∆ → →∞∆ → →∞∆ → →∞∆ → →∞

∆ = ∆ → → ∞∆ = ∆ → → ∞∆ = ∆ → → ∞∆ = ∆ → → ∞

= = + + == = + + == = + + == = + + =

Darboux’s integral : (Darboux’s Criterion for integral) Let ( )f x is a continuous and bounded function on the interval [ , ]a b , and let p be a partition on [ , ]a b . ( )f x to be integrable (by Darboux’s Criterion) on [ , ]a b if the limit

1 11 1

lim ( ) lim ( )

lim ( ; ) lim ( ; )

k k k k k kn nk k

m x x M x x

L P f U P f

− −− −− −− −→∞ →∞→∞ →∞→∞ →∞→∞ →∞= == == == =

→∞ →∞→∞ →∞→∞ →∞→∞ →∞

− −− −− −− −

= == == == =

∑ ∑∑ ∑∑ ∑∑ ∑

f x dx I I I= = == = == = == = =∫∫∫∫

Example(4):

By using Riemann’s Criterion show that 2( ) 1, [0, 8]f x x x= − ∈= − ∈= − ∈= − ∈ is integrable on the interval 0 8[ , ] . Solution: Divide the interval 0 8[ , ] into n subinterval by the points

0 1 28 2.8 8

0, , , ..., , ..., 8k nk

x x x x xn n n

= = = == = = == = = == = = =

since the function is increasing then for the interval 1[ , ]k kx x−−−− we have

1( ), ( )k k k km f x M f x−−−−= == == == = thus

(((( ))))

1 21 1

22 2 2

23 3 3

8( 1) 8( 1) 64( 1)1 1

64 2 18( ; ) ( )

8 64 128 641

512 1016 512 8

k k kk k

k k km f

n n nk

M f kn n

k k nL P f x x m

k kn n n n

k knn n n

−−−−= == == == =

− − −− − −− − −− − − = = − = −= = − = −= = − = −= = − = −

= = −= = −= = −= = −

− + −− + −− + −− + − = − == − == − == − =

= − + −= − + −= − + −= − + −

∑ ∑∑ ∑∑ ∑∑ ∑

∑∑∑∑

(((( )))) (((( )))) (((( ))))3 3

11016 512 81 2 1

512 512 24 488lim ( ; ) 8 (1)

3 3 3n

n nnn n n n

L P f→∞→∞→∞→∞

++++⋅ + + − ⋅ + ⋅ − ⋅⋅ + + − ⋅ + ⋅ − ⋅⋅ + + − ⋅ + ⋅ − ⋅⋅ + + − ⋅ + ⋅ − ⋅

−−−−= − = == − = == − = == − = =

Similarly 2

8 64( ; ) 1

512 8 512( 1)(2 1) 8

512 488lim ( ; ) 8 (2)

U P f kn n

nk n n

−−−−

→∞→∞→∞→∞

= −= −= −= −

= − = ⋅ + + −= − = ⋅ + + −= − = ⋅ + + −= − = ⋅ + + −

∴ = − =∴ = − =∴ = − =∴ = − =

∑∑∑∑

From (1), (2) the function is integrable on v and 8

488( )

3f x dx ====∫∫∫∫

:)1(Theorem onsintegrability of continuous functi

Let ( ) : [ , ]f x a b �→ be a continuous function on the interval [ , ]a b .Then ( )f x is integrable on [ , ]a b . Proof: by the uniform continuity , since ( )f x is uniformly continuous function then if 0εεεε >>>> there exist ( ) 0δ δ εδ δ εδ δ εδ δ ε= >= >= >= > such that for any

, [ , ]x y a b∈∈∈∈ and ( )x y δ εδ εδ εδ ε− <− <− <− < then ( ) ( )f x f yb a

δδδδ− <− <− <− <−−−−

Now let n N>>>> be such that ( )

δ εδ εδ εδ ε−−−−>>>> and let

1 2 1{ ... ... }o r r np a x x x x x x b−−−−= = < < < < < < < == = < < < < < < < == = < < < < < < < == = < < < < < < < = be a portion to the interval [ , ]a b into n equal subintervals so that

1 ( )k kb a

δ εδ εδ εδ ε−−−−−−−−− = <− = <− = <− = <

Therefore we have

sup{ ( ) ( ) : , }

1,2,3, ...,

k k kw M m f x f y x yb a

for k n

εεεε= − = − ∈= − = − ∈= − = − ∈= − = − ∈−−−−

we have

( ; ) ( ; ) ( )( )

k k k kk

k k kk

U f p L f p x x M m

b ax x w n

b ab a

εεεε εεεε

−−−−====

− = − −− = − −− = − −− = − −

−−−−< − < =< − < =< − < =< − < =−−−−

= ⋅ − == ⋅ − == ⋅ − == ⋅ − =−−−−

∑∑∑∑

Then for every 0εεεε >>>> there exist a partition p such that

( ) ( )n nU p L p εεεε− <− <− <− < and since εεεε is arbitrary we can chose εεεε small

such that lim ( ) lim ( )n nn n

U p L p→∞ →∞→∞ →∞→∞ →∞→∞ →∞

==== Then ( )f x is integrable on [ , ]a b .

:)2(Theorem Integrability of monotone functions: Let ( ) : [ , ]f x a b �→ be a monotone function on the interval [ , ]a b .Then ( )f x is integrable on [ , ]a b . Proof: Let ( )f x increasing function and let 0εεεε >>>> is arbitrary positive number and let 0 1 2{ .... .... }r np a x x x x x b= = < < < < < < == = < < < < < < == = < < < < < < == = < < < < < < = is an a partition to [ , ]a b into n equal subintervals so that

1k kb a

x xn−−−−−−−−− =− =− =− = for 1,2,3, ...,k n====

since ( )f x is increasing function

on 1[ , ]k kx x−−−− then

1( ) ( )k k k km f x and M f x−−−−= == == == =

Therefore

(((( )))) (((( )))) (((( ))))

[[[[ ]]]]

1 11 1

1 0 2 1 1

( ; ) ( ; ) ( )( )

( ) ( ) ( ) ( )

( ) ( ) ( ) ( ) ... ( ) ( )

( ) ( )

k k k kk

k k k kk k

U f p L f p x x M m

b a b af x f x f x f x

b af x f x f x f x f x f x

f b f an

−−−−====

− −− −− −− −= == == == =

−−−−

− = − −− = − −− = − −− = − −

− −− −− −− −= ⋅ − = −= ⋅ − = −= ⋅ − = −= ⋅ − = −

−−−− = − + − + + −= − + − + + −= − + − + + −= − + − + + −

−−−−= −= −= −= −

∑∑∑∑

∑ ∑∑ ∑∑ ∑∑ ∑

Now if 0εεεε >>>> is given , we choose n N∈∈∈∈ such that

(((( ))))( ) ( ) ( )b a f b f an

εεεε− −− −− −− −

>>>> For corresponding partition p we have

[[[[ ]]]]( ) ( ) ( ) ( )n nb a

U p L p f b f an

εεεε−−−−− = − <− = − <− = − <− = − < and since εεεε is arbitrary we can

1( )f a m==== ( )nM f b==== 1 2M m====

chose εεεε small such that lim ( ) lim ( )n nn n

U p L p→∞ →∞→∞ →∞→∞ →∞→∞ →∞

==== Then ( )f x is

integrable on [ , ]a b .

Exercises

By mathematical induction prove that

(((( )))) (((( )))) (((( ))))

(((( ))))

1 11 , 1 2 1

(1) (2)

r n n r n n n

= + = + += + = + += + = + += + = + +

= += += += +

∑ ∑∑ ∑∑ ∑∑ ∑

∑∑∑∑

find the following sums

1 1102

1 120 152 2

( 5) ( 3)

( 3 5) (2 5)

(2 13) ( 2 4)

(4) (5)

(6) (7)

(8) (9)

= == == == =

+ ++ ++ ++ +∑ ∑∑ ∑∑ ∑∑ ∑

+ − −+ − −+ − −+ − −∑ ∑∑ ∑∑ ∑∑ ∑

− − +− − +− − +− − +∑ ∑∑ ∑∑ ∑∑ ∑

By using Darboux’s sums find

1 12 2

0 12 1

( ) ( 1)

(15) ( 1)

(10) (11)

(12) (13)

x x dx x dx

x dx x dx

−−−−− +− +− +− +

−−−−

∫ ∫∫ ∫∫ ∫∫ ∫

By using Riemann’s sums find the integrals (10)-(15)

:Properties of definite integral

Let ( ), ( )f x g x are integrable on the interval [ , ]a b and let [ , ]c a b∈∈∈∈

is an odd function

is an even function0

[ ( ) ( )] ( ) ( )

( ) ( ) ,

( ) ( )

( ) ( ) ( )

( )2 ( ) ( )

a a ab b

a ab a

a bb c b

f x g x dx f x dx g x dx

f x dx f x dx

f x dx f x dx f x dx

if f x

f x dxf x dx if f x

λ λ λλ λ λλ λ λλ λ λ

−−−−

± = ±± = ±± = ±± = ±

= ∈= ∈= ∈= ∈

= −= −= −= −

= += += += +

∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫

∫ ∫∫ ∫∫ ∫∫ ∫

∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫

∫∫∫∫∫∫∫∫

( ) ( )(8)

x dx f a b x dx

f x dx f a x dx

= + −= + −= + −= + −

= −= −= −= −

∫ ∫∫ ∫∫ ∫∫ ∫

(9) If ( )f x continuous function on [ , ]a b Then there exist [ , ]c a b∈∈∈∈

such that ( ) ( )( )b

af x dx f c b a= −= −= −= −∫∫∫∫

Mean value theorem: Theorem(3): If ( )f x continuous function on [ , ]a b then there exist [ , ]c a b∈∈∈∈ such

that ( ) ( )( )b

af x dx f c b a= −= −= −= −∫∫∫∫ .

Proof: Since ( )f x is continuous function on [ , ]a b then there exist a maximum M and minimum m values i.e. there exist , [ , ]u v a b∈∈∈∈ such that ( ) ( ) ( ) [ , ]m f u f x f v M x a b= ≤ ≤ = ∀ ∈= ≤ ≤ = ∀ ∈= ≤ ≤ = ∀ ∈= ≤ ≤ = ∀ ∈ then

( ) ( ) ( )

a a ab

m dx f x dx M dx

m b a f x dx M b a

f u f x dx f vb a

≤ ≤≤ ≤≤ ≤≤ ≤

− ≤ ≤ −− ≤ ≤ −− ≤ ≤ −− ≤ ≤ −

≤ ≤≤ ≤≤ ≤≤ ≤−−−−

∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫

∫∫∫∫

If ( ) ( )f u f v= then ( )f x is constant function and we can choose c such that [ , ]c a b∈∈∈∈ .

If ( ) ( )f u f v≠ then we can find c such that 1( ) ( )

af c f x dx

b a====

−−−− ∫∫∫∫

The fundamental theorem of calculus:

Theorem(4): Let ( )f x continuous function on the interval [ , ]a b then there exist a

point [ , ]c a b∈∈∈∈ such that ( ) ( ) ( )b

af x dx b a f c= −= −= −= −∫∫∫∫ .

Theorem(5): Differentiable theorem Let :[ , ]f a b →→→→ � be integrable on the interval [ , ]a b and let

:[ , ]aF a b →→→→ � be defined by ( ) [ , ]x

F f x x a b= ∀ ∈= ∀ ∈= ∀ ∈= ∀ ∈∫∫∫∫ Then aF

differentiable at any point [ , ]c a b∈∈∈∈ at which f continuous ,and ( ) ( )aF c f c′′′′ ====

Proof: Suppose that ( )f x is continuous at [ , ]c a b∈∈∈∈ . Let 0εεεε >>>> be given and let

( )δ δ εδ δ εδ δ εδ δ ε==== such that ( ) ( )f c h f c εεεε+ − <+ − <+ − <+ − < whenever [ , ]c h a b+ ∈+ ∈+ ∈+ ∈ and h δδδδ<<<< .

For any such h we use the observation 11

++++====∫∫∫∫

(((( ))))

( ) ( ) 1 1( ) ( ) ( )

1 1( ) ( )

c h c ha a

F c h F cf c f x dx f c dx

f x f c dx hh h

ε εε εε εε ε

+ ++ ++ ++ +

+ −+ −+ −+ −− = −− = −− = −− = −

= − ≤ == − ≤ == − ≤ == − ≤ =

∫ ∫∫ ∫∫ ∫∫ ∫

∫∫∫∫

Since 0εεεε >>>> is arbitrary it follows that

( ) ( )( ) lim ( )a a

F c h F cF c f c

h→→→→

+ −+ −+ −+ −′′′′ = == == == =

The function F is called antiderivative of a function f on the interval [ , ]a b if ( ) ( ) [ , ]F x f x x a b′′′′ = ∀ ∈= ∀ ∈= ∀ ∈= ∀ ∈ . fundamental theorem of calculus Let f be a continuous function on an interval [ , ]a b Then the function

: [ , ]F a b →→→→ � satisfies

( ) ( )x

aF x F a f− =− =− =− = ∫∫∫∫ if and only if ( ) ( ) [ , ]F x f x x a b′′′′ = ∀ ∈= ∀ ∈= ∀ ∈= ∀ ∈ .

Proof:

If ( ) ( )x

aF x F a f− =− =− =− = ∫∫∫∫ holds for all [ , ]x a b∈∈∈∈ then by differentiation

theorem we have ( ) ( ) ( ) [ , ]

( ) ( ) ( ) [ , ]a

F x F a F x x a b

F x F x f x x a b

− = ∀ ∈− = ∀ ∈− = ∀ ∈− = ∀ ∈′ ′′ ′′ ′′ ′∴ = = ∀ ∈∴ = = ∀ ∈∴ = = ∀ ∈∴ = = ∀ ∈

Conversely, if : [ , ]F a b →→→→ � is such that ( ) ( ) [ , ]F x f x x a b′′′′ = ∀ ∈= ∀ ∈= ∀ ∈= ∀ ∈ then ( ) ( ) ( ) [ , ]aF x F x f x x a b′ ′′ ′′ ′′ ′= = ∀ ∈= = ∀ ∈= = ∀ ∈= = ∀ ∈ There exist C such that ( ) ( ) [ , ]aF x F x C x a b= + ∀ ∈= + ∀ ∈= + ∀ ∈= + ∀ ∈ Since ( ) 0aF x ==== we se that ( )aF x C==== and hence

( ) ( )x

aF x F a f− =− =− =− = ∫∫∫∫

Corollary: Let f be a continuous function on an interval [ , ]a b and if

( ) ( ) [ , ]F x f x x a b′′′′ = ∀ ∈= ∀ ∈= ∀ ∈= ∀ ∈ .Then ( ) ( )b

aF b F a f− =− =− =− = ∫∫∫∫

:Examples

sin sin

( 3 4) 3 4

dx dx y

dx x dx t t

= −= −= −= −

+ + = + ++ + = + ++ + = + ++ + = + +

∫∫∫∫

:Example By using fundamental theorem of calculus find .

( 3 5)

x x dx+ ++ ++ ++ +

∫∫∫∫

Solution:

3 2 3 2

cos [ sin ]

cos sin sin

3( 3 5) ( 5 )

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

dx dx x dx

x dx b a

d xx x dx x x dx

b b a ab a

= −= −= −= −

+ + = + ++ + = + ++ + = + ++ + = + +

= + + − + += + + − + += + + − + += + + − + +

∫ ∫∫ ∫∫ ∫∫ ∫

∫∫∫∫

∫ ∫∫ ∫∫ ∫∫ ∫

Exercises

Using fundamental theorem of calculus find 10 2

2 02 2

(4 2) sin

(cos ) (sin )

(1) (2)

(3) (4)

(5) (6)

x dx e dx

x dx x dx

x e dx x x dx

ππππ

π ππ ππ ππ π

ππππ−−−−

+ ++ ++ ++ +

∫ ∫∫ ∫∫ ∫∫ ∫

2 2( ) , ( )

[ ( )] [ ( )]

(7) x x xf x g x

d df x dx g x dx

++++= == == == =+ ++ ++ ++ +

++++∫ ∫∫ ∫∫ ∫∫ ∫

(8) using fundamental theorem of calculus find the area under the curve of the following functions:

35 7( )f x x= += += += + bounded by 0 2 8, ,y x x= = == = == = == = = . Improper Integrals:

af x∫∫∫∫ is called an improper integral if

(a) the integrand ( )f x has one or more points of discontinuity on the interval a x b≤ ≤≤ ≤≤ ≤≤ ≤ , or (b) at least of the limits of integration is infinite. Discontinuous integrand: (1)If ( )f x is continuous on the interval a x b≤ <≤ <≤ <≤ < but is discontinuous

at x b==== we define

( ) lim ( )b b

a af x dx f x dx

εεεε

εεεε ++++

−−−−

→→→→====∫ ∫∫ ∫∫ ∫∫ ∫ provided the limit exist

(2)If ( )f x is continuous on the interval a x b< ≤< ≤< ≤< ≤ but is discontinuous at x a==== we define

( ) lim ( )b b

a af x dx f x dx

εεεε ++++→→→→ ++++====∫ ∫∫ ∫∫ ∫∫ ∫ provided the limit exist

(3)If ( )f x is continuous on the interval a x b≤ ≤≤ ≤≤ ≤≤ ≤ but is discontinuous at [ , ]x c a b= ∈= ∈= ∈= ∈ we define

( ) lim ( ) lim ( )b c b

a a cf x dx f x dx f x dx

εεεε

ε εε εε εε ε εεεε+ ++ ++ ++ +

−−−−

→ →→ →→ →→ → ++++= += += += +∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫ provided the limit

exist. Infinite limits of integration: (1)If ( )f x is continuous on the interval a x b≤ ≤≤ ≤≤ ≤≤ ≤ we define

( ) lim ( )b

ba af x dx f x dx

∞∞∞∞

→∞→∞→∞→∞====∫ ∫∫ ∫∫ ∫∫ ∫ provided the limit exist

(2)If ( )f x is continuous on the interval a x b≤ ≤≤ ≤≤ ≤≤ ≤ we define

( ) lim ( )b b

a af x dx f x dx

→−∞→−∞→−∞→−∞−∞−∞−∞−∞====∫ ∫∫ ∫∫ ∫∫ ∫ provided the limit exist

(3)If ( )f x is continuous on the interval a x b≤ ≤≤ ≤≤ ≤≤ ≤ we define

( ) lim ( )a

a af x dx f x dx

∞∞∞∞

→∞→∞→∞→∞−∞ −−∞ −−∞ −−∞ −====∫ ∫∫ ∫∫ ∫∫ ∫

provided the limit exist. Examples: Example(1):

Find 3

x−−−−∫∫∫∫

Solution: The integrand is discontinuous at 2x ==== we consider

33 31 1

2 20 0 00 0 0

3lim lim sin lim sin

3 39 9

sin 12

dx dx x

εεεεεεεε

ε ε εε ε εε ε εε ε ε

εεεε

ππππ

−−−−−−−−− −− −− −− −

→ → →→ → →→ → →→ → →

−−−−

−−−− = = == = == = == = = − −− −− −− −

= == == == =

∫ ∫∫ ∫∫ ∫∫ ∫

Example(2):

Find 2

x−−−−∫∫∫∫

Solution: The integrand is discontinuous at 3x ==== we consider

22 2 2

00 0 00 0 0

1lim lim ln 2 lim ln

1 1lim ln ln

dx dxx

εεεεεεεε εεεε

ε ε εε ε εε ε εε ε ε

εεεε εεεε

−−−−−−−− −−−−

→ → →→ → →→ → →→ → →

→→→→

= = − − == = − − == = − − == = − − = − − −− − −− − −− − −

= −= −= −= −

∫ ∫∫ ∫∫ ∫∫ ∫

the limit does not exist then the function doesn’t integrable Example(3):

Show that 2

( 1)x −−−− is not integrable on the interval [0,4]

Solution: The integrand is discontinuous at 1x ==== value between the limit 0 and 4 we consider 4 1 4

2 2 20 00 0 1

0 0 0 00 1

1 1 1lim lim

( 1) ( 1) ( 1)

1 1 1 1 1lim lim lim 1 lim

εεεε

ε εε εε εε ε εεεεεεεε

ε ε ε εε ε ε εε ε ε εε ε ε εεεεε ε εε εε εε ε

+ ++ ++ ++ +

+ + + ++ + + ++ + + ++ + + +

−−−−

→ →→ →→ →→ → ++++−−−−

→ → → →→ → → →→ → → →→ → → →++++

= += += += +− − −− − −− − −− − −

− − −− − −− − −− − − = + = − + += + = − + += + = − + += + = − + + − −− −− −− −

∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫

the limit does not exist then the function dos not integrable in [0,4] Example(4):

Find 2

∞∞∞∞

++++∫∫∫∫

Solution: The upper limit of integration is infinite. We consider

1 1 12 2

1 1lim lim tan lim tan tan 0

2 2 2 2 44 4

dx dx x b

ππππ∞∞∞∞− − −− − −− − −− − −

→∞ →∞ →∞→∞ →∞ →∞→∞ →∞ →∞→∞ →∞ →∞

= = = − == = = − == = = − == = = − = + ++ ++ ++ +

∫ ∫∫ ∫∫ ∫∫ ∫

Example(5):

Find 0

2xe dx−∞−∞−∞−∞∫∫∫∫

Solution: The lower limit of integration is infinite. We consider

[[[[ ]]]]00 0

2 2 2 21 1 1 1lim lim lim 1 1 0

2 2 2 2x x x a

a a ba ae dx e dx e e

→−∞ →−∞ →−∞→−∞ →−∞ →−∞→−∞ →−∞ →−∞→−∞ →−∞ →−∞−∞−∞−∞−∞

= = = − = − == = = − = − == = = − = − == = = − = − = ∫ ∫∫ ∫∫ ∫∫ ∫

Hence 0

xe dx−∞−∞−∞−∞

====∫∫∫∫

Example(6):

Find x x

∞∞∞∞

−−−−−∞−∞−∞−∞ ++++∫∫∫∫

Solution: Both limit of integration is infinite. We consider

lim lim tan1 1

lim tan tan 02 2

x xa axx x x x aa aa

dx e dx e dxe

e e e e

e eπ ππ ππ ππ π

∞ ∞∞ ∞∞ ∞∞ ∞−−−−

−−−− −−−−→∞ →∞→∞ →∞→∞ →∞→∞ →∞−∞ −∞ −−∞ −∞ −−∞ −∞ −−∞ −∞ −

− − −− − −− − −− − −→∞→∞→∞→∞

= = == = == = == = = + + ++ + ++ + ++ + +

= − = − == − = − == − = − == − = − =

∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫

Hence 2x x

ππππ∞∞∞∞

−−−−−∞−∞−∞−∞

====++++

∫∫∫∫ .

Exercise(9)

Evaluate the following integrals:

(((( ))))

2 22 0

(1) (2)

(3) (4)

(5) (6)

(7) (8)

(9) (10)

dx x dxx x

dx xdx

x dxdx

∞ ∞∞ ∞∞ ∞∞ ∞

−∞−∞−∞−∞

−−−−∞∞∞∞

−∞ −−∞ −−∞ −−∞ −∞ ∞∞ ∞∞ ∞∞ ∞

−∞−∞−∞−∞

++++ ++++

− −− −− −− −

−−−−++++

−−−−

∫ ∫∫ ∫∫ ∫∫ ∫

Evaluation of some definite integrals: Example(1):

Find 2

0sinm x dx

ππππ

∫∫∫∫

Solution:

0 0sin sin sinm mx dx x x dx

−−−−====∫ ∫∫ ∫∫ ∫∫ ∫ use integration by parts

21 2 220 0

sin sin

( 1)sin cos cos

cos sin ( 1)sin cos

u x dv x dx

du m x x dx v x

I x x m x x dx

ππππππππ

−−−−

− −− −− −− −

= == == == =

= − = −= − = −= − = −= − = −

∴ = − + −∴ = − + −∴ = − + −∴ = − + − ∫∫∫∫

if is even then

0 ( 1) sin (1 sin )

( 1) sin ( 1) sin

( 1) ( 1) sin

( 1).( 3)( 5)... 3.1.( 1).( 2)... 4.2

m x x dx

m xdx m x dx

I m I m xdx

mm m m

m I Im m m

I x dx

ππππ

−−−−

= + − −= + − −= + − −= + − −

= − − −= − − −= − − −= − − −

+ − = −+ − = −+ − = −+ − = −

−−−−∴ =∴ =∴ =∴ =

− − −− − −− − −− − −====− −− −− −− −

= == == == =

= == == == =∴∴∴∴

∫∫∫∫

∫ ∫∫ ∫∫ ∫∫ ∫

∫∫∫∫

[[[[ ]]]]since

is even positive integer

if is odd then 1

1).( 3)( 5)... 3.1,

.( 1).( 2)... 4.2 2

( 1).( 3)( 5)... 4.2.( 1).( 2)... 5.3

sin cos 1

m m mm m m

m m mm I I

I dx x

ππππ ππππ

ππππ− − −− − −− − −− − −− −− −− −− −

− − −− − −− − −− − −====− −− −− −− −

= = − == = − == = − == = − =∫∫∫∫

is odd positive integer

( 1).( 3)( 5)... 4.2sin

.( 1).( 2)... 5.3m

mm m m

I x dxm m m

ππππ− − −− − −− − −− − −= == == == =∴∴∴∴ − −− −− −− −∫∫∫∫

2 26 76 7

5.3.1 6.4.2sin . , sin

6.4.2 2 7.5.3I x dx I x dx

π ππ ππ ππ πππππ= = = == = = == = = == = = =∫ ∫∫ ∫∫ ∫∫ ∫

Example(2):

Find 2

0cosm x dx

ππππ

∫∫∫∫

Solution:

0 0cos cos cosm mx dx x x dx

−−−−====∫ ∫∫ ∫∫ ∫∫ ∫ use integration by parts

21 2 220 0

cos cos

( 1)cos sin sin

cos sin ( 1)cos sin

0 ( 1) cos (1 cos )

( 1) cos ( 1) cos

u x dv x dx

du m x x dx v x

I x x m x x dx

m x x dx

m x dx m x dx

ππππππππ

ππππ

−−−−

− −− −− −− −

−−−−

= == == == =

= − − == − − == − − == − − =

∴ = + −∴ = + −∴ = + −∴ = + −

= + − −= + − −= + − −= + − −

= − − −= − − −= − − −= − − −

∫∫∫∫

∫ ∫∫ ∫∫ ∫∫ ∫

if is even then

( 1) ( 1) cos

( 1).( 3)( 5)... 3.1.( 1).( 2)... 4.2

I m I m xdx

mm m m

m I Im m m

ππππ

−−−−

+ − = −+ − = −+ − = −+ − = −

−−−−∴ =∴ =∴ =∴ =

− − −− − −− − −− − −====− −− −− −− −

∫∫∫∫

since2

ππππππππ= == == == =∫∫∫∫

is even positive integer

( 1).( 3)( 5)... 3.1cos ,

.( 1).( 2)... 4.2 2m

mm m m

I x dxm m m

ππππππππ− − −− − −− − −− − −= == == == =∴∴∴∴ − −− −− −− −∫∫∫∫

[[[[ ]]]]since

if is odd then 1

( 1).( 3)( 5)... 4.2.( 1).( 2)... 5.3

cos sin 1

mm m m

m I Im m m

I dx x

ππππ ππππ

− − −− − −− − −− − −====− −− −− −− −

= = == = == = == = =∫∫∫∫

is odd positive integer

( 1).( 3)( 5)... 4.2cos

.( 1).( 2)... 5.3m

mm m m

I x dxm m m

ππππ− − −− − −− − −− − −= == == == =∴∴∴∴ − −− −− −− −∫∫∫∫

2 26 76 7

5.3.1 6.4.2cos . , cos

6.4.2 2 7.5.3I x dx I x dx

π ππ ππ ππ πππππ= = = == = = == = = == = = =∫ ∫∫ ∫∫ ∫∫ ∫

221 1 1 2

sin .cos

1 1sin .cos sin .cos

m n m n

x x dx

nx x x x dx

m n m n

ππππ

ππππππππ+ − + −+ − + −+ − + −+ − + −−−−− = += += += + + ++ ++ ++ +

∫∫∫∫

221 1 2

10 sin .cos (1)

sin .cos

1 1sin .cos sin .cos

10 sin .cos (2)

m n m n

nx x dx

and x x dx

mx x x x dx

m n m n

mx x dx

ππππ

ππππππππ

ππππ

+ −+ −+ −+ −

− + −− + −− + −− + −

−−−−

−−−−= += += += +++++

− −− −− −− − = += += += + + ++ ++ ++ +

−−−−= += += += +++++

∫∫∫∫

applying the formula (2) in the R.H.S. of (1) the formula (1) will reduce to

2 2 2 2

( 1)( 1)sin .cos sin .cos (1)

( )( 2)m n m nn m

x x dx x x dxm n m n

− −− −− −− −− −− −− −− −====+ + −+ + −+ + −+ + −∫ ∫∫ ∫∫ ∫∫ ∫

and by repeating of the formula we get

sin .cos

( 1)( 1) ( 3)( 3). sin .cos

( )( 2) ( 4)( 6)

x x dx

n m n mx x dx

m n m n m n m n

ππππ

− −− −− −− −− − − −− − − −− − − −− − − −====+ + − + − + −+ + − + − + −+ + − + − + −+ + − + − + −

∫∫∫∫

if m and n are both even

( 1)( 3)...1.( 1)( 3)...1sin .cos 1

( )( 2)( 4) ... 2

( 1)( 3)...1.( 1)( 3)...1.

( )( 2)( 4) ... 2 2

m n m m n nx x dx dx

m n m n m n

m m n nm n m n m n

ππππ

− − − −− − − −− − − −− − − −====+ + − + −+ + − + −+ + − + −+ + − + −

∫ ∫∫ ∫∫ ∫∫ ∫

Solved examples:

5.3.1.3.1sin .cos .

10.8.6.4.2 2 2x x dx

ππππππππ====∫∫∫∫

7.5.3.1 35sin . ( 8, 0)

8.6.4.2 2 256x dx m n

πππππ ππ ππ ππ π= = = == = = == = = == = = =∫∫∫∫

if m and n are both odd.

( 1)( 3)...2.( 1)( 3)...2sin .cos . sin cos

( )( 2)( 4) ... 4

( 1)( 3)...2.( 1)( 3)...2 1.

( )( 2)( 4) ... 4 2

m n m m m nx x dx x x dx

m n m n m n

m m n nm n m n m n

π ππ ππ ππ π− − − −− − − −− − − −− − − −====+ + − + −+ + − + −+ + − + −+ + − + −

− − − −− − − −− − − −− − − −====+ + − + −+ + − + −+ + − + −+ + − + −

∫ ∫∫ ∫∫ ∫∫ ∫

solved Examples:

4.2.2 1 1sin .cos .

8.6.4 2 24x x dx

ππππ

= == == == =∫∫∫∫

6.4.2.1 16cos ( 8, 0)

7.5.3.1 35x dx m n

ππππ

= = = == = = == = = == = = =∫∫∫∫

if m is odd and n is even.

( 1)( 3)...2.( 1)( 3)...1sin .cos . sin

( )( 2)( 4) ... 3

( 1)( 3)...2.( 1)( 3)...1.1

( )( 2)( 4) ... 3

m n m m n nx x dx x dx

m n m n m n

m m n nm n m n m n

π ππ ππ ππ π− − − −− − − −− − − −− − − −====+ + − + −+ + − + −+ + − + −+ + − + −− − − −− − − −− − − −− − − −====+ + − + −+ + − + −+ + − + −+ + − + −

∫ ∫∫ ∫∫ ∫∫ ∫

Example:

4.2.3.1 8sin .cos .1

9.7.5.3 315x x dx

ππππ

= == == == =∫∫∫∫

if m is even and n is odd.

( 1)( 3)...1.( 1)( 3)...2sin .cos . cos

( )( 2)( 4) ... 3

( 1)( 3)...1.( 1)( 3)...2.1

( )( 2)( 4) ... 3

m n m m n nx x dx x dx

m n m n m n

m m n nm n m n m n

π ππ ππ ππ π− − − −− − − −− − − −− − − −====+ + − + −+ + − + −+ + − + −+ + − + −

− − − −− − − −− − − −− − − −====+ + − + −+ + − + −+ + − + −+ + − + −

∫ ∫∫ ∫∫ ∫∫ ∫

Example:

3.1.6.4.2 16sin .cos .1

11.9.7.5.3 5511x x dx

ππππ

= == == == =∫∫∫∫

we can evaluate these integral through the following definition

with the following properties

0( ) (1)

(1) ( 1) ( ) !

1(2) ( )

n xx e dx n

n n n n

ππππ

∞∞∞∞−−−− = Γ= Γ= Γ= Γ

Γ + = Γ =Γ + = Γ =Γ + = Γ =Γ + = Γ =

Γ =Γ =Γ =Γ =

∫∫∫∫

we can prove that

1 1( ) ( )

2 2sin .cos2

2 ( )2

x x dxm n

ππππ + ++ ++ ++ +Γ ΓΓ ΓΓ ΓΓ Γ==== + ++ ++ ++ +ΓΓΓΓ

∫∫∫∫

Solved examples:

5 5 3 37 1 12 6 4 2 2 2 2 2 2 22

9 5 31 7 12 8 2 2 2 2 2 2

( ) ( ) . . . . 3sin .cos

1.5.4.3.2.1 5122 ( )

( ) ( ) . . . . 35sin

2 (5) 2.4.3.2.1 256

m nx x dx

ππππ

π ππ ππ ππ π ππππ

+ ++ ++ ++ +

Γ ΓΓ ΓΓ ΓΓ Γ= = == = == = == = =

ΓΓΓΓ

Γ ΓΓ ΓΓ ΓΓ Γ= = == = == = == = =

ΓΓΓΓ

∫∫∫∫

12 7 29 5 37 1

0 2 2 2 2 2

5 3 12 5 4 2 2 211 9 5 37 1

0 2 2 2 2 2 2

(3) (2) 2.1.1 1sin .cos

2 (5) 2.4.3.2.1 24

( ) (4) .3.2.1. 16cos

352 ( ) 2. . . .

(3) ( ) 2.1. . 8sin .cos

3152 ( ) 2. . . . .

sin .cos

x x dx

ππππ

ππππππππ

Γ ΓΓ ΓΓ ΓΓ Γ= = == = == = == = =ΓΓΓΓ

Γ ΓΓ ΓΓ ΓΓ Γ= = == = == = == = =

ΓΓΓΓ

Γ ΓΓ ΓΓ ΓΓ Γ= = == = == = == = =

ΓΓΓΓ

∫∫∫∫

∫∫∫∫5 3 12 2 2

13 9 5 311 7 12 2 2 2 2 2 2

( ) (4) . .3.2.1 1655112 ( ) 2. . . . . .

xππππ

ππππΓ ΓΓ ΓΓ ΓΓ Γ

= = == = == = == = =ΓΓΓΓ

Numerical integrals Exercise(10)

Application of definite integration 1- Plane Areas by integration: If ( )f x is continuous and non-negative on the interval a x b≤ ≤≤ ≤≤ ≤≤ ≤ , the

definite integral 1

( ) lim ( )b n

k kn ka

f x dx f x x→∞→∞→∞→∞ ====

= ∆= ∆= ∆= ∆∑∑∑∑∫∫∫∫ where 1( )k k kx x x −−−−∆ = −∆ = −∆ = −∆ = −

can be give a geometric interpretation. Let the interval [ , ]a b be subdivided into n subintervals by the points

1 2{ , , , ..., , ..., }o k nx a x x x x b= == == == = then the perpendicular to x-axis at these points are 0 1 2{ ( ), ( ), ( ), ..., ( ), ..., ( )}k nf x a f x f x f x f x b= == == == = respectively divided the area under the curve of ( )f x into n strips. Approximate each strip by a rectangle whose base is lower base of the strip and whose altitude is the ordinate erected at the point kx of the

subinterval. Hence 1

f x x====

∆∆∆∆∑∑∑∑ is simply the sum of the areas of n

approximating rectangles and the limit of the sum is ( )b

af x dx∫∫∫∫ which

represent to the area under the curve of ( )f x enclosed by x-axis and ,x a x b= == == == = .

Then we have the following cases :

(1) Area bounded by the curve ( )y f x==== and thex axis−−−− and the two

ordinates ,x a x b= == == == = is b

ay dx∫∫∫∫ .

x b==== x a====

( )y f x====

1kx x −−−−==== kx x====

1( )kf x −−−−

(2) Area bounded by the curve ( )y f x==== and they axis−−−− and the two

ordinates ,y a y b= == == == = is b

ax dy∫∫∫∫ .

(3) To find the area enclosed between two curves

1 2( ), ( )y f x y g x= == == == = we solve the two equation simultaneously to obtain the points of intersection say ( , ), ( , )a b c d then

1 2( )c

aA y y dx= −= −= −= −∫∫∫∫

(4) Area of closed curve Let the equation of the curve be ( )y f x= ±= ±= ±= ± the equation has two values of y say 1 2,y y and let ,x a x b= == == == = is the two lines at

which 1 2y y==== then 1 2( )b

aA y y dx= −= −= −= −∫∫∫∫

(5) Area of parametric curve ( ), ( )x f t y g t= == == == = given by

aA xdy ydx= −= −= −= −∫∫∫∫

(6) Area in polar coordinates The area bounded by the curve ( )r f θθθθ==== and two radii vectors

drawn from the origin direction given by 2

2A r dθθθθ

θθθθθθθθ==== ∫∫∫∫

Example(1): Find the area bounded by the curve 2y x==== ,x-axis and the two lines

1, 3x x= == == == = . Solution:

3A f x dx x dx= = == = == = == = =∫ ∫∫ ∫∫ ∫∫ ∫

Example(2): Find the area above the x-axis and under the parabola 24y x x= −= −= −= −

Solution: 3 3

32( ) (4 )

3A f x dx x x dx= = − == = − == = − == = − =∫ ∫∫ ∫∫ ∫∫ ∫

Example(3): Find the area bounded by the parabola 26y x x= −= −= −= − and 2 2y x x= −= −= −= − Solution: The parabolas intersect at the points (0,0) ) and(4,8) The area =the area under 26y x x= −= −= −= − - the area under 2 2y x x= −= −= −= − from

0x ==== to 4x ====

Area = (((( )))) (((( ))))4

3x x x x dx − − − =− − − =− − − =− − − =

∫∫∫∫

Example(4): Find the area bounded by the cardoid (1 cos )r a θθθθ= += += += + Solution:

2 22 2 22 2 2 2

1 1 3(1 cos ) (1 2cos cos )

2 2 2 2a a

A r d a d dπ π ππ π ππ π ππ π π ππππθ θ θ θ θ θθ θ θ θ θ θθ θ θ θ θ θθ θ θ θ θ θ= = + = + + == = + = + + == = + = + + == = + = + + =∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫

Exercise(11)

(1) Find the area bounded as follows: 2

, 0, 2, 6 [ . : 39]

224 , 0, 1, 3 [ . : ]

1 , 10 [ . : 36]

1259 , 3 [ . : ]

6cos , sin [ . : ]

2cos cos 2 1, 2sin sin 2 [ . : 6 ]

9 (3 )

(5)(6)

y x y x x Ans

y x x y x x Ans

x y x Ans

y x y x Ans

x a t y b t Ans ab

x t t y t t Ans

The loop ay x a x

ππππππππ

= = = == = = == = = == = = =

= − = = == − = = == − = = == − = = =

= + == + == + == + =

= − = += − = += − = += − = +

= == == == == − − = −= − − = −= − − = −= − − = −

= −= −= −= −2

32 2 2 2

8 3[ . : ]

2( ) [ . : ]

4 2 2 [ . : ]3

3( sin ), (1 cos ) [ . : ]

ay x a x Ans

y x from x to x Ans

ax a t t y a t Ans

= −= −= −= −

= − = − == − = − == − = − == − = − =

= − = −= − = −= − = −= − = −

+ =+ =+ =+ =

(2) volume and solid revolution: A solid of revaluation is generated by revolving a plane area about a line, called the axis of revaluation. In the plane. The volume of a solid of revaluation may be found by using one of the following procedures. Disc method: make a sketch showing the area involved, a reprehensive strip perpendicular to the axis of rotation ,Write the volume of the cylindrical shell generated by rotation (= mean circumference height thickness) × ×× ×× ×× × then integrate this unit of volume fromx a to x b= == == == = x=a

If ( )f x is continuous function and we want to find the volume generated by the are under the curve from x a to x b= == == == =

[[[[ ]]]]22 ( )b b b

a a aV dV y dx f x dxπ ππ ππ ππ π= = == = == = == = =∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫

Shell method: Let 1 2 1{ ... ... }o r r np a x x x x x x b−−−−= = < < < < < < < == = < < < < < < < == = < < < < < < < == = < < < < < < < = be a portion to the interval [ , ]a b into n equal subintervals make a sketch showing the area involved on the interval 1[ , ]k kx x−−−− , a reprehensive strip parallel to the axis of rotation when the strip rotate around the axis of rotation we have a cylindrical shell with volume 2 2

12 ( ) 2 ( )k kx f c x f cπ ππ ππ ππ π −−−−−−−− where 1[ , ]k kc x x−−−−∈∈∈∈ then

2 21 1 1

( ) ( ) ( )( ) ( )

( )( ) ( ) 2 ( )

lim lim 2 ( ) 2 ( )

k k k k k k k k k k

k kk k k k k k k k

k k kkn nk k a

V x f x f x x x x f

x xx x x f f x

V V f x xf x dx

π ξ π ξ π ξπ ξ π ξ π ξπ ξ π ξ π ξπ ξ π ξ π ξ

π ξ π ξ ξ ξπ ξ π ξ ξ ξπ ξ π ξ ξ ξπ ξ π ξ ξ ξ

π ξ ξ ππ ξ ξ ππ ξ ξ ππ ξ ξ π

− − −− − −− − −− − −

−−−−−−−−

→∞ →∞→∞ →∞→∞ →∞→∞ →∞= == == == =

∆ = − = − +∆ = − = − +∆ = − = − +∆ = − = − +++++= ∆ + = ∆ == ∆ + = ∆ == ∆ + = ∆ == ∆ + = ∆ =

∴ = ∆ = ∆ =∴ = ∆ = ∆ =∴ = ∆ = ∆ =∴ = ∆ = ∆ =∑ ∑∑ ∑∑ ∑∑ ∑ ∫∫∫∫

2 ( )b

aV xf x dxππππ==== ∫∫∫∫

Example(1): Find the volume generated by revolving the area bounded by the parabola 2 8y x==== about its latus rectum. Solution:

24 42 2

256(2 ) 2 (2 )

V x dy dyπ ππ ππ ππ π− −− −− −− −

= − = − == − = − == − = − == − = − =∫ ∫∫ ∫∫ ∫∫ ∫

Example(2): Find the volume generated by revolving about x axis−−−− the area between the first arch of the cycloid sin , 1 cosx t t t t= − = −= − = −= − = −= − = − and the x axis−−−− . Solution:

2 2 22 2 3

0 0 02

(1 cos ) (1 cos ) (1 cos )

(1 3cos 3cos cos )

5 33cos cos 2 (1 sin )cos

5 3 13sin sin 2 (sin sin ) 5

t t tt

V y dx t t dt t dt

t t t dt

t t t t dt

t t t t t

π π ππ π ππ π ππ π π

ππππ

= = == = == = == = =

= = == = == = == = =====

========

= = − − = −= = − − = −= = − − = −= = − − = −

= − + − == − + − == − + − == − + − =

= − + − −= − + − −= − + − −= − + − −

= − + − − == − + − − == − + − − == − + − − =

∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫

∫∫∫∫

Example(3): Find the volume generated by revolving about y axis−−−− the area between the first arch of the cycloid sin , 1 cosx t t t t= − = −= − = −= − = −= − = − and the x axis−−−− . Solution:

2 2 3 3

2 2 ( sin )(1 cos )(1 cos )

2 ( 2 cos cos sin 2sin cos cosh 2 sin )

3 1 1 1 12 2( sin cos ) ( sin 2 cos 2 ) cos sin cos 6

4 2 2 4 3

V xydx t t t t dt

t t t t t t t t t t dt

t t t t t t t t t t

ππππ

= == == == =

= == == == =====

= = − − −= = − − −= = − − −= = − − −

= − + − + −= − + − + −= − + − + −= − + − + −

= − + + + + + + == − + + + + + + == − + + + + + + == − + + + + + + =

∫ ∫∫ ∫∫ ∫∫ ∫

∫∫∫∫

Exercise(12)

Find the volume generated by revolving the the given plane area about the given line, using the disc method

22 , 0, 0, 5; [ . 2500 ](1) y x y x x x axis Ans ππππ= = = = −= = = = −= = = = −= = = = −

25616, 0, 8; [ . ]

(1 ); [ . ]35

4 9 36; [ . 16 ]

x y y x x axis Ans

y x x x axis Ans

x y x axis Ans

ππππ

− = = = −− = = = −− = = = −− = = = −

= − −= − −= − −= − −

+ = −+ = −+ = −+ = −

Find the volume generated by revolving the the given plane area about the given line, using the shell method.

2 , 0, 0, 5; [ .625 ]

320, 0, 2; 8 [ . ]

5 6, 0; [ . ]6

y x y x x y axis Ans

y x y x y Ans

y x x y y axis Ans

ππππππππ

ππππ

= = = = −= = = = −= = = = −= = = = −

= = = == = = == = = == = = =

= − + = −= − + = −= − + = −= − + = −

Find the volume generated by revolving the the given plane area about the given line, using any appropriate method.

1sin 2 ; [ . ]

sin , 1 cos ; [ .5 ]

642cos cos 2 1, 2sin sin 2 ; [ . ]

y x x axis Ans

y y x axis Ans

x y x axis Ans

ππππ

θ θ θ πθ θ θ πθ θ θ πθ θ θ πππππθ θ θ θθ θ θ θθ θ θ θθ θ θ θ

= −= −= −= −

= − = − −= − = − −= − = − −= − = − −

= − − = − −= − − = − −= − − = − −= − − = − −

(3) Length of Arc: To find the length of an arc AB of the curve of the function ( )f x in the interval [ , ]a b let the interval be divided into subintervals by points

0 1 1, , ..., , , ...,k k nx a x x x x b−−−−= == == == = and erect perpendiculars to determine the points 1 1, , ..., , , ...,o k k nP A P P P P B−−−−= == == == = on the arc. For the length

22 2 2 2

1 1 1( ) ( ) ( ) ( ) 1 kk k k k k k k k k

yP P x x y y x y x

x− − −− − −− − −− − − ∆∆∆∆= − + − = ∆ + ∆ = + ∆= − + − = ∆ + ∆ = + ∆= − + − = ∆ + ∆ = + ∆= − + − = ∆ + ∆ = + ∆ ∆∆∆∆

1lim 1 1

kn k k a

y dyAB x dx

x dx→∞→∞→∞→∞ ====

∆∆∆∆ ∴ = + ∆ = +∴ = + ∆ = +∴ = + ∆ = +∴ = + ∆ = + ∆∆∆∆ ∑∑∑∑ ∫∫∫∫

if the equation of the curve given in the parametric form ( ), ( )x x t y y t= == == == = then

dx dyAB L dt

dt dt ∴ = = +∴ = = +∴ = = +∴ = = +

∫∫∫∫

if the equation of the curve given in the polar form ( )r f θθθθ==== then

AB L r dd

θθθθ

θθθθθθθθ

θθθθ ∴ = = +∴ = = +∴ = = +∴ = = +

∫∫∫∫

Example(1):

Find the length of the arc of the curve 32y x==== from 0x ==== to 5x ==== .

Solution:

(((( ))))

9 11 1 4 9

1 1 2 1 3354 9 . . 4 9

2 2 3 9 27

L x dx x

′′′′+ = + = ++ = + = ++ = + = ++ = + = +

= + = + == + = + == + = + == + = + =

∫∫∫∫

Example(2):

Find the length of the arc of the curve 3

123x y

−−−−==== from 0y ==== to 4y ==== .

Solution:

(((( ))))

81 11 1 4 81

1 1 2 1 84 81 . . 4 81 (82 82 1)

2 2 3 81 234

L y dy x

+ = + = ++ = + = ++ = + = ++ = + = +

= + = + = −= + = + = −= + = + = −= + = + = −

∫∫∫∫

Example(3): Find the length of the arc of the curve 424 48xy x= += += += + from 2x ==== to

4x ==== . Solution:

2 4 2 4 4 2

4 4 2 4 8 42 2

( 16) 64 ( 16)1 1

64 64 641 1

64 ( 16) 64 32 2568 8

dy xdx x

dy x x xdx x x x

x x x x xx x

−−−−====

− −− −− −− − ∴ + = + = +∴ + = + = +∴ + = + = +∴ + = + = +

= + − = + − += + − = + − += + − = + − += + − = + − +

8 4 42 2

44 44 2 2 3

22 2 2

1 132 256 ( 16)

1 1 1 1 16 17( 16) ( 16 )

8 8 3 68

x x xx x

L x dx x x dx x unitsxx

−−−−

= + + = += + + = += + + = += + + = +

= + = + = − == + = + = − == + = + = − == + = + = − =

∫ ∫∫ ∫∫ ∫∫ ∫

Example(4): Find the length of the arc of the curve 2 3,x t y t= == == == = from 0t ==== to 4t ==== . Solution:

2 22 4 2

4 4 42 2 2 3/ 2

4 9 4 9

1 2 14 9 18 4 9 . (4 9 )

18 3 18

8(37 37 1)

dx dyt t

dx dyt t t t

L t t dt t t dt t

= == == == =

∴ + = + = +∴ + = + = +∴ + = + = +∴ + = + = +

= + = + = += + = + = += + = + = += + = + = +

= −= −= −= −

∫ ∫∫ ∫∫ ∫∫ ∫

Example(5): Find the length of the arc of the cycloid sin , 1 cosx yθ θ θθ θ θθ θ θθ θ θ= − = −= − = −= − = −= − = − from

0θθθθ ==== to 2θ πθ πθ πθ π==== . Solution:

2 22 2 2 2

1 cos , sin

(1 cos ) sin 1 2cos cos sin

2(1 cos 4sin 2sin2 2

2sin 2 2cos 82 2

dx dyd d

L d unitsππππππππ

θ θθ θθ θθ θθ θθ θθ θθ θ

θ θ θ θθ θ θ θθ θ θ θθ θ θ θθ θθ θθ θθ θ

θ θθ θθ θθ θθθθθ

= − == − == − == − =

∴ + = − + = − + +∴ + = − + = − + +∴ + = − + = − + +∴ + = − + = − + +

= − = == − = == − = == − = =

= = − == = − == = − == = − =

∫∫∫∫

Example(6): Find the circumference of cardoid (1 cos )r a θθθθ= −= −= −= − Solution:

22 2 2 2 2 2 2

2 22 2 22 2 2

0 0 0 0

(1 cos )

1(1 cos ) sin (2 2cos ) 2 (1 cos )

4 sin2

4 sin 2 sin 4 cos 82 2 2

drr a a a a

drL r a d a d a a

πππππ π ππ π ππ π ππ π π

θθθθ

θθθθθθθθ

θ θ θ θθ θ θ θθ θ θ θθ θ θ θθθθθ

θθθθ

θ θ θθ θ θθ θ θθ θ θθ θθ θθ θθ θθθθθ

= −= −= −= −

∴ =∴ =∴ =∴ =

+ = − + = − = −+ = − + = − = −+ = − + = − = −+ = − + = − = −

= + = = = − == + = = = − == + = = = − == + = = = − =

∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫

Exercise

In the following find the length of the entire curve: units]

units]

[ .(104 13 125) / 27

[ .17 / 12

1[ .3 2 ln (2 2)

6 3 1 2

ln 1 2 2

27 4( 2) (2,0) (11,6 3)

ln(1 ) 1/ 4

y x x x

xy x x x

y x x x

−−−−

− + +− + +− + +− + +

= = == = == = == = =

= + = == + = == + = == + = =

= = == = == = == = =

= −= −= −= −

= − == − == − == − = units]

units]

tofrom

1[ .ln(21 / 5)

24[ . 2( 1)

cos , sin 0 4

2cos cos 2 1, 2sin sin 2

(6)(7)

x e t y e t t t

x yθ θ θ θθ θ θ θθ θ θ θθ θ θ θ

−−−−

= = = == = = == = = == = = == + + = += + + = += + + = += + + = +

(4) Area of Surface of Revolution:

The area of the surface generated by revolving the arc AB of the continuous curve about a line in its plane If ( , ), ( , )A a c B b d are two points of the curve ( )y f x==== where

( ) ( )f x and f x′′′′ are continuous functions and ( )y f x==== dos not change in its sign on the interval [ , ]a b The area of the surface generated by revolving the arc AB about x-axis is given by:

dyS ydL y dx

dxy dy

ππππ

= = += = += = += = +

= += += += +

∫ ∫∫ ∫∫ ∫∫ ∫

∫∫∫∫

If ( , ), ( , )A a c B b d are two points of the curve ( )x g y==== where ( ) ( )g y and g y′′′′ are continuous functions and ( )y f x==== dos not change in

its sign on the interval [ , ]a b The area of the surface generated by revolving the arc AB about y-axis is given by:

dyS xdL x dx

dxx dy

ππππ

= = += = += = += = +

= += += += +

∫ ∫∫ ∫∫ ∫∫ ∫

∫∫∫∫

If 1 2( ), ( )A t t B t t= == == == = are two points of the curve defined by the parametric equation ( ), ( )x f t y y t= == == == = The area of the surface generated by revolving the arc AB about x-axis is given by:

dx dyS ydL y dt

dt dtπ ππ ππ ππ π = = += = += = += = +

∫ ∫∫ ∫∫ ∫∫ ∫

and the area of the surface generated by revolving the arc AB about y-axis is given by:

dx dyS xdL x dt

dt dtπ ππ ππ ππ π = = += = += = += = +

∫ ∫∫ ∫∫ ∫∫ ∫

Example(1): Find the area of the surface of revaluation generated by revolving about the x-axis the arc of the parabola 2 12y x==== from 0x ==== to 3x ==== . Solution:

We use 22 1b

S y y dxππππ ′′′′= += += += +∫∫∫∫ Now

[[[[ ]]]]

6 612 2 12

1236 3

3 31 12 1 12 12 36

y x yy yy x

xy y x x x

′ ′′ ′′ ′′ ′==== ⇒⇒⇒⇒ ==== ⇒⇒⇒⇒ = == == == =

′′′′∴ = =∴ = =∴ = =∴ = =

′′′′+ = + = + = ++ = + = + = ++ = + = + = ++ = + = + = +

22 1 2 12 36 24(2 2 1)b b

S y y dx x dxπ π ππ π ππ π ππ π π′′′′∴ = + = + = −∴ = + = + = −∴ = + = + = −∴ = + = + = −∫ ∫∫ ∫∫ ∫∫ ∫

Example(2): Find the area of the surface of revaluation generated by revolving about the y-axis the arc of the parabola 3x y==== from 0y ==== to 1y ==== . Solution:

We shall use 2

dxS x dy

dyππππ

= += += += +

∫∫∫∫ Now

dxx y y

==== ⇒⇒⇒⇒ ====

+ = ++ = ++ = ++ = +

13 4 3 4

02 1 9 2 1 9 (10 10 1)

S y y dy y y dyπππππ ππ ππ ππ π∴ = + = + = −∴ = + = + = −∴ = + = + = −∴ = + = + = −∫ ∫∫ ∫∫ ∫∫ ∫

Example(3): Find the area of the surface of revaluation generated by revolving about the x-axis the arc of 2 4 2lny x y+ =+ =+ =+ = from 1y ==== to 3y ==== . Solution:

We shall use 2

dxS y dy

dyππππ

= += += += +

∫∫∫∫ Now

2 2 22 2 2

2 2 4 4 2

1 2 1 12

1 (1 ) 11 1 4 (1 )

1 14 1 2 2 1

dy y y

dx yx y y

y y y y yy y

+ =+ =+ =+ =

= −= −= −= −

−−−−= − == − == − == − =

−−−−+ = + = + −+ = + = + −+ = + = + −+ = + = + −

= + − + = + += + − + = + += + − + = + += + − + = + +

1 ( 1) 1 1( 1)

2 12 1

dxS y dy y y dy

y dy y y

ππππππππ

πππππ ππ ππ ππ π

++++= + = = += + = = += + = = += + = = +

∴ = + = +∴ = + = +∴ = + = +∴ = + = +

= + = + == + = + == + = + == + = + =

∫ ∫∫ ∫∫ ∫∫ ∫

∫∫∫∫

Example(4): Find the area of the surface of revaluation generated by revolving a loop of the curve 2 2 2 2 48a y a x x= −= −= −= − about the x-axis . Solution:

We shall use 2

dyS y dx

dxππππ = += += += +

∫∫∫∫ Now

2 2 4 2 2 22 2 2 2 4 2

2 3 2 2 3 2 2 3 22

4 2 2 2 2 42 2 44

2 3 2 2 2 22

2 2 2 4

16 2 4

(2 4 ) (2 4 ) (2 4 )

256 32 ( )256

(2 4 ) (3 2 )1 1

32 ( ) 8

a x x x a xa y a x x y

a yy a x x

a x xy

a x x a x x a x xy

a y a a x xa x xa

a x x a xy

a a x x

− −− −− −− −= −= −= −= − ⇒⇒⇒⇒ = == == == =

′′′′∴ = −∴ = −∴ = −∴ = −

−−−−′′′′ ====

− − −− − −− − −− − −′′′′ = = == = == = == = = −−−−−−−−

− −− −− −− −′′′′+ = + =+ = + =+ = + =+ = + =−−−−

2 2 2 2 2 22

2 2 2 20 0

( ) (3 2 )2 1 2 .

8 8 ( )

x a x a xS y y dx dx

a a a xπ ππ ππ ππ π

−−−−

− −− −− −− −′′′′∴ = + =∴ = + =∴ = + =∴ = + =−−−−

∫ ∫∫ ∫∫ ∫∫ ∫

2 2 2 22 2

2 2 2 24 4

1 12 (3 2 ) . (3 2 )( 4 )

1 (3 2 ). 9

4 2 44 32

x a x dx a x x dxa a

a x aa a

ππππππππ

−−−−= − = − −= − = − −= − = − −= − = − −

− − −− − −− − −− − − = = − == = − == = − == = − =

∫ ∫∫ ∫∫ ∫∫ ∫

Example(5): Find the area of the surface of revaluation generated by revolving the

ellipse 2 2

116 4x y+ =+ =+ =+ = about the x-axis .

Solution:

We shall use 2

dyS y dx

dxππππ = += += += +

∫∫∫∫ Now

11, 16

16 4 21 24 16 2 16

x yy x

dy x xdx x x

+ = = −+ = = −+ = = −+ = = −

− −− −− −− −∴ = =∴ = =∴ = =∴ = =− −− −− −− −

squareunits

2 2 2 2

2 2 22 2

4(16 )1 1

4(16 ) 4(16 )

1 4(16 ) 11 16 64 3

2 44(16 )

2 1 64 32

3 3 4 364 3 32sin 8 1

2 8 92 3

dy x x xdx x x

dy x xy x x

dyS y dx x dx

ππππππππ

ππππ π ππ ππ ππ π

− −− −− −− −

−−−−

− +− +− +− + + = + =+ = + =+ = + =+ = + = − −− −− −− −

− +− +− +− + + = − = −+ = − = −+ = − = −+ = − = − −−−−

∴ = + = −∴ = + = −∴ = + = −∴ = + = −

= − + = += − + = += − + = += − + = +

∫ ∫∫ ∫∫ ∫∫ ∫

Example(6): Find the area of the surface of revaluation generated by revolving about the x-axis the hypocycloid 3 3cos , sinx a y aθ θθ θθ θθ θ= == == == = . Solution:

We shall use 2 2

dx dyS y d

d dπ θπ θπ θπ θ

θ θθ θθ θθ θ = += += += +

∫∫∫∫ Now

2 22 4 2 2 4 2 2 2 2

2 23 2 2 2 2 4

cos , sin

3 cos sin , 3 sin cos

9 cos sin 9 sin cos 9 sin cos

sin 9 sin cos 3 sin cos

x a y a

dx dya a

dx dya a a

dx dyy a a a

dx dyS y

θ θθ θθ θθ θ

θ θ θ θθ θ θ θθ θ θ θθ θ θ θθ θθ θθ θθ θ

θ θ θ θ θ θθ θ θ θ θ θθ θ θ θ θ θθ θ θ θ θ θθ θθ θθ θθ θ

θ θ θ θ θθ θ θ θ θθ θ θ θ θθ θ θ θ θθ θθ θθ θθ θ

ππππθ θθ θθ θθ θ

= == == == =

= − == − == − == − =

+ = + =+ = + =+ = + =+ = + =

+ = =+ = =+ = =+ = =

∴ = +∴ = +∴ = +∴ = +

∫∫∫∫

[[[[ ]]]]

2 / 22 4

2(2 ) 3 sin cos

12 12sin5

ππππ

θ π θ θ θθ π θ θ θθ π θ θ θθ π θ θ θ

π ππ ππ ππ πθθθθ

= == == == =

∫∫∫∫

Example(7): Find the volume generated by revolving about x axis−−−− the area between the first arch of the cycloid ( sin ), (1 cos )x a t t t a t= − = −= − = −= − = −= − = − and the x axis−−−− . Solution:

2 22 2 2 2

22 3 2

( sin ), (1 cos )

(1 cos ), sin

(1 cos ) sin 2 sin2

2 4 (1 cos )sin2

648 sin

x a t t y a t

dx dya t a t

dx dy tdL dt a t a t a dt

dx dy tS y dt a a t dt

ta dt a

ππππ

= − = −= − = −= − = −= − = −

= − == − == − == − =

= + = − + == + = − + == + = − + == + = − + =

∴ = + = −∴ = + = −∴ = + = −∴ = + = −

= == == == =

∫ ∫∫ ∫∫ ∫∫ ∫

∫∫∫∫

Exercise(15)

Find the area of the surface of revaluation generated by revolving the given arc about the given axis.

from to

20 2; [ .4 1 . .]

10 3; [ . 982 82 1) / 9 . .]

0 2; [ . 9 82 ln( 82 9) . .]2

18 (1 ), ; [ . . .]

y mx x x x axis Ans m m s u

y x x x x axis Ans s u

y mx x x y axis Ans s u

y x x loop x axis Ans s u

Anarch of x

ππππ

= = = − += = = − += = = − += = = − +

= = = − −= = = − −= = = − −= = = − −

= = = − + += = = − + += = = − + += = = − + +

= − −= − −= − −= − −

====2 2 2

( sin ), (1 cos );

[ .64 ( 4) . .]

1282cos cos 2 , 2sin sin 2 [ . . .]

a y a x axis

Ans a e e s u

x y Ans s u

ππππππππθ θ θ θθ θ θ θθ θ θ θθ θ θ θ

−−−−

− = − −− = − −− = − −− = − −

− +− +− +− +

= − = −= − = −= − = −= − = −

integral calculus

repeated linear factors

proper rational fraction

odd positive integer

dx du du

dv dx du

du dx dx

dx dx dx

dx ax bx

Education

unit 3. integral calculus (2...

unit 3. integral calculus (2 credits)vlasta/mathematics/unit...

chapter-2 : integral calculus

integral calculus by arihant b016

integral calculus lecture

calculus ii video library (integral...

text book of integral calculus

integral calculus[1]

integral calculus

unit 4 applications of integral integral calculus calculus...

integral calculus differentiation under the integral … ·...

chapter 3 integral calculus

integral calculus made easy

integral calculus - exercisesvlasta/mathematics/unit...

notes on calculus ii integral calculus - math department,...

differential and integral calculus 1000021985

integral calculus -...

integral calculus of vector - free

integral transforms and operational calculus

a1 calculus 3 integral