Integral calculus
description
Transcript of Integral calculus
Calculus
of
Integration
Prof. Dr. M. Abul-Ez
Mathematics Department
Faculty of Science
Sohag University
Mathematics For Engineering
2
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
(1) ( ) ( )
(2) ( )
(3)
(4) 11
(5) ln
(6)ln
(7)
mm
xx
x x
df x f x c
dxu v dx udx v dx
udx udx
xx dx c m
mdx
x cx
aa dx c
a
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
x dx x c
x dx x c
x dx x c
= − += − += − += − += += += += +
= += += += +
= += += += +
∫∫∫∫
∫∫∫∫
∫∫∫∫
∫∫∫∫
Indefinite Integration
3
(12) sec ln sec tan
(13) cosec ln csc cot
x dx x x c
dx x x c
= + += + += + += + +
= − += − += − += − +∫∫∫∫
∫∫∫∫
2
2
(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
x x dx x c
= += += += +
= − += − += − += − += += += += +
= − += − += − += − +
∫∫∫∫
∫∫∫∫
∫∫∫∫
∫∫∫∫
Integration tends to inverse of Trigonometric functions: 1
2 2 2 1
1
2 2 21
1
2 2 2 1
1
2 2 2
1sin
(18)1
cos
1tan
(19)1
cot
1sec
(20)1
cosec
1coth
(21)1
ln2
bxc
dx b abxa b x c
b a
bxc
dx ab abxa b x c
ab a
bxc
dx a adxbxx b x a c
a a
bxc
ab adxdx
bx ab x a cab bx a
−−−−
−−−−
−−−−
−−−−
−−−−
−−−−
−−−−
++++==== −−−−−−−− ++++
++++==== −−−−++++ ++++
++++==== −−−−−−−− ++++
++++==== −−−−−−−− ++++++++
∫∫∫∫
∫∫∫∫
∫∫∫∫
∫∫∫∫
1
2 2 2
2 2
12 2
1tanh
(22)1
ln2
ln( )(23)
sinh
bxc
ab adxa bxa b x c
ab a bx
x x a cdxx
cx aa
−−−−
−−−−
++++==== ++++−−−− ++++ −−−−
+ + ++ + ++ + ++ + +==== ++++++++
∫∫∫∫
∫∫∫∫
Mathematics For Engineering
4
2 2
12 2
22 2 2 2 1
22 2 2 2 1
22 2 2 2 1
ln( )(24)
cosh
1(25) sin
2 2
1(26) sinh
2 2
1(27) cosh
2 2
x x a cdxx
cx aa
a xa x x a x c
a
a xx a x x a c
a
a xx a x x a c
a
−−−−
−−−−
−−−−
−−−−
+ − ++ − ++ − ++ − +==== ++++−−−−
− = − + +− = − + +− = − + +− = − + +
+ = + + ++ = + + ++ = + + ++ = + + +
− = − − +− = − − +− = − − +− = − − +
∫∫∫∫
∫∫∫∫
∫∫∫∫
∫∫∫∫
In the following some lows witch we use to integrate the square of trigonometric functions
[[[[ ]]]]
[[[[ ]]]]
[[[[ ]]]]
2 2
2 2
2 2
2
2
(1)cos sin 1,
(2)1 tan sec ,
(3)cot 1 csc
1(4) sin (1 cos 2 )
21
(5)cos (1 cos 2 )2
1(6)cos cos cos( ) cos( )
21
(7)sin sin cos( ) cos( )21
(8)sin cos sin( ) sin( )2
x x
x x
x x
x x
x x
x y x y x y
x y x y x y
x y x y x y
+ =+ =+ =+ =
+ =+ =+ =+ =
+ =+ =+ =+ =
= −= −= −= −
= += += += +
= + + −= + + −= + + −= + + −
= − − += − − += − − += − − +
= + + −= + + −= + + −= + + −
Integration of square of trigonometric functions:
2
2
2
1 1 1(1) sin (1 cos 2 ) cos 2
2 2 2
1 1 1(2) cos (1 cos 2 ) cos 2
2 2 2
(3) sec tan ,
x dx x dx x x c
x dx x dx x x c
x dx x c
= − = − += − = − += − = − += − = − +
= + = + += + = + += + = + += + = + +
= += += += +
∫ ∫∫ ∫∫ ∫∫ ∫
∫ ∫∫ ∫∫ ∫∫ ∫
∫∫∫∫
Indefinite Integration
5
2
2 2
2 2
(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( )
2
1 sin( ) sin( )2 ( )
1(9) sin cos sin( ) sin( )
2
ax bx dx a b x a b x dx
a b x a b xc
a b a b
ax bx dx a b x a b x dx
a b x a b xc
a b a b
ax bx dx a b x a b x dx
= + + −= + + −= + + −= + + −
+ −+ −+ −+ −= + += + += + += + + + −+ −+ −+ −
= − − += − − += − − += − − +
− +− +− +− += + += + += + += + + − +− +− +− +
= + + −= + + −= + + −= + + −
====
∫ ∫∫ ∫∫ ∫∫ ∫
∫ ∫∫ ∫∫ ∫∫ ∫
∫ ∫∫ ∫∫ ∫∫ ∫
1 cos( ) cos( )2 ( ) ( )
a b x a b xc
a b a b
− + −− + −− + −− + −+ ++ ++ ++ + + −+ −+ −+ −
Solved Examples:
5 6
1 43 3 3
1
1
1:
6
34
1 ( )( )
( 1)
1 1ln ( )
( )
1 1 ( )( ) , 1
( 1)( )
Exampl(1)
Exampl(2) :
Exampl(3) :
Exampl(4) :
Exampl(5) :
nn
nn
n
x dx x c
x dx x dx x c
ax bax b dx c
a n
dx ax b cax b a
ax bdx ax b dx c n
a nax b
++++
− +− +− +− +−−−−
= += += += +
= = += = += = += = +
+++++ = ++ = ++ = ++ = +++++
= + += + += + += + +++++
++++= + = + ≠= + = + ≠= + = + ≠= + = + ≠− +− +− +− +++++
∫∫∫∫
∫ ∫∫ ∫∫ ∫∫ ∫
∫∫∫∫
∫∫∫∫
∫ ∫∫ ∫∫ ∫∫ ∫
Mathematics For Engineering
6
1 3 1 3 3 52 2 2 2 2 2
23 5 2 3 2
4 3
1 2
2 2(1 ) ( )
3 5
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
x
= + += + += + += + +++++
− = − = − = − +− = − = − = − +− = − = − = − +− = − = − = − +
+ ++ ++ ++ +++++
∫∫∫∫
∫ ∫ ∫ ∫∫ ∫ ∫ ∫∫ ∫ ∫ ∫∫ ∫ ∫ ∫
∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫
3 2
3 5 2 5 6 3 6
1 3 3234 4 4
44 3
3 33 2 32 2
2 3
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)
9 9 9
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
ux
iii x x dx u du u c x c
−−−−
= + ∴ == + ∴ == + ∴ == + ∴ =
+ = = + = + ++ = = + = + ++ = = + = + ++ = = + = + +
= = = = + += = = = + += = = = + += = = = + +++++
+ = = + = + ++ = = + = + ++ = = + = + ++ = = + = + +
∫ ∫∫ ∫∫ ∫∫ ∫
∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫
∫ ∫∫ ∫∫ ∫∫ ∫
3 2 3 2
3 2 2
2 2 2
3 5 32 2 2 22 2 2
1 ( ) 1
( ) 1 1
( 1) 1 1
2 2( 1) 1 ( 1) ( 1)
10 6
Exampl (9) :
x x dx x x x x dx
x x x dx x x dx
x x x dx x x dx
x x dx x x dx x x c
+ = + − ++ = + − ++ = + − ++ = + − +
= + + − += + + − += + + − += + + − +
= + + − += + + − += + + − += + + − +
= + − + = + − + += + − + = + − + += + − + = + − + += + − + = + − + +
∫ ∫∫ ∫∫ ∫∫ ∫
∫ ∫∫ ∫∫ ∫∫ ∫
∫ ∫∫ ∫∫ ∫∫ ∫
∫ ∫∫ ∫∫ ∫∫ ∫
2 23
3 3
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
( 1)
Exampl (10) :
Exampl (11) :
Exampl (12) :
dx dxx c
x x
x dx x dxx c
x xx x x
dx dx dx dxx x x x
dxx x c
x
= = − += = − += = − += = − +− −− −− −− −
− −− −− −− −= − = − += − = − += − = − += − = − +
− −− −− −− −+ + + ++ + + ++ + + ++ + + +
= = += = += = += = ++ + + ++ + + ++ + + ++ + + +
= + = + + += + = + + += + = + + += + = + + + ++++
∫ ∫∫ ∫∫ ∫∫ ∫
∫ ∫∫ ∫∫ ∫∫ ∫
∫ ∫ ∫ ∫∫ ∫ ∫ ∫∫ ∫ ∫ ∫∫ ∫ ∫ ∫
∫∫∫∫
Indefinite Integration
7
2 2 2 2 2
2 2 3
1 1tan 2ln sec
2 2
1 1(sec ) sec ( ) ln sec tan
2 21
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
2 4 6
3 7 112 2 2
5 9 132 2 2
5 5 6
:
(1 ) (1 ) (1 3 3 )
1 3 3( )
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 x
x x xdx
x x x x
dxx dx x dx x dx
x
x x x x c
e e dx e d e e c
+ + + + ++ + + + ++ + + + ++ + + + += == == == =
= + + += + + += + + += + + +
= + + += + + += + + += + + +
= + + + += + + + += + + + += + + + +
+ = + + = + ++ = + + = + ++ = + + = + ++ = + + = + +
∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫
∫∫∫∫
∫ ∫ ∫ ∫∫ ∫ ∫ ∫∫ ∫ ∫ ∫∫ ∫ ∫ ∫
∫ ∫∫ ∫∫ ∫∫ ∫
3 3 3
2 22
2 2
2 22 5 2
2 5 2 5
2 4
1 1(3 )
3 3ln
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 x x
x xx
x x
x xx x
x x
x
a dx a dx a ca
e d edx dx e c
e e
e dx d ee d e
e e
ec
−−−−
−−−−
= = += = += = += = +
−−−−= = − += = − += = − += = − +− −− −− −− −
−−−−= = − −= = − −= = − −= = − −− −− −− −− −
−−−−= += += += +−−−−
∫ ∫∫ ∫∫ ∫∫ ∫
∫ ∫∫ ∫∫ ∫∫ ∫
∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫
4 4
1 1sin4 cos4 , (19) cos 3 sin 3
4 3
cos sin cos ( sin )
Example(21):
Example(22):
x dx x c x dx x c
x x dx x x dx
−−−−= + = += + = += + = += + = +
= − −= − −= − −= − −
∫ ∫∫ ∫∫ ∫∫ ∫
∫ ∫∫ ∫∫ ∫∫ ∫
Mathematics For Engineering
8
3 2 3 4
5 2 5 2
5 6
5 2 4 2
2 2 2
1tan sec tan (tan ) tan
4
cot cosec cot ( cosec )
1cot (cot ) cot
6
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
2 4 6 3 5 7
3 5 7
4 3 4 2
4 2
n sin )sin (cos )
(sin 2sin sin )(cos )
sin cos
1 2 1( 2 )
3 5 7
1 2 1sin sin sin
3 5 7
cos sin cos sin (sin )
cos (1 cos )(sin )
Example(26):
x x x xdx
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
4 6 5 7
5 7
3 5
4 4
(cos cos )(sin )
cos sin
1 1( )
5 7
1 1cos cos
5 7
sin cos
sin cos
x x xdx
let y x dy xdx
I y y dy y y c
x x c
Try to solve x x dx
x x dx
= −= −= −= −
==== ⇒⇒⇒⇒ = −= −= −= −
∴ = − = − +∴ = − = − +∴ = − = − +∴ = − = − +
= − += − += − += − +
∫∫∫∫
∫∫∫∫
∫∫∫∫
∫∫∫∫
Indefinite Integration
9
4 3
3 4
2
2
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
x xx
x xdx dx x x dx x c
x xx
====
= = += = += = += = +
= = = += = = += = = += = = +
= = = − += = = − += = = − += = = − +
∫ ∫∫ ∫∫ ∫∫ ∫
∫∫∫∫
∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫
∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫
2 2
2 2
2
2
(1 cos ) (1 cos )1 cos 1 cos sin
cos
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
x x
xdx x x dx
x x c
x x x x xx dx x dx dx
x x x xx x c
− −− −− −− −= == == == =++++ −−−−
= −= −= −= −
= −= −= −= −= − + += − + += − + += − + +
+ ++ ++ ++ += == == == =+ ++ ++ ++ +
= + += + += + += + +
∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫
∫ ∫∫ ∫∫ ∫∫ ∫
∫ ∫∫ ∫∫ ∫∫ ∫
∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫
2
4
cosec cotcosec cosec
cosec cot
cosec cosec cotln cosec cot
cosec cot
int
( ) sin (2 3cos ) ,
sin sin( ) , ( )
(2 3cos ) (2 3cos )
e(32):
Example(33):
x xx dx x dx
x x
x x xdx x x c
x x
FindThefolowing egrals
i I x x dx
x dx x dxii J iii K
x x
−−−−====−−−−
−−−−= = − += = − += = − += = − +−−−−
= += += += +
= == == == =++++ ++++
∫ ∫∫ ∫∫ ∫∫ ∫
∫∫∫∫
∫∫∫∫
∫∫∫∫
32
32
2 3cos 3sin
1 1 2( ) sin (2 3cos ) .
3 3 3
2(2 3cos )
9
let u x du x dx
i I x x dx u du u
x c
= += += += + ⇒⇒⇒⇒ = −= −= −= −
− −− −− −− − ∴ = + = =∴ = + = =∴ = + = =∴ = + = =
−−−− = + += + += + += + +
∫∫∫∫
∫ ∫∫ ∫∫ ∫∫ ∫
Mathematics For Engineering
10
(((( ))))
4 34 4
3
2 2
2
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 )
9
(1 tan ) (1 2tan tan )
1 2tan (sec 1) 2tan
Example(34):
xdx duii J u c x c
x u
xdx duiii K u du u
x u
x c
x dx x x dx
x x dx x
− −− −− −− −
−−−−
− − −− − −− − −− − −= = = + = + += = = + = + += = = + = + += = = + = + +++++
− − − −− − − −− − − −− − − − = = = == = = == = = == = = = ++++
= + += + += + += + +
+ = + ++ = + ++ = + ++ = + +
= + + − == + + − == + + − == + + − =
∫ ∫∫ ∫∫ ∫∫ ∫
∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫
∫ ∫∫ ∫∫ ∫∫ ∫
∫∫∫∫ (((( ))))
(((( ))))(((( ))))
2
2
2 2
2
2
2 2 2
sec
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):
x dx
x x c
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
++++
= + += + += + += + +
++++
+ = + ++ = + ++ = + ++ = + +
= + + −= + + −= + + −= + + −
= + = − += + = − += + = − += + = − +
+ = + ++ = + ++ = + ++ = + +
∫∫∫∫
∫∫∫∫
∫ ∫∫ ∫∫ ∫∫ ∫
∫∫∫∫
∫∫∫∫
∫ ∫∫ ∫∫ ∫∫ ∫
(((( ))))(((( ))))
2 2
2
sin sin sin
tan 2 tan tan
cos
(sec 3 1) 2tan 3 sec3 sec 3
2sec 3 2tan 3 sec3 1
2 2tan 3 sec3
3 3
cos (sin )
1sec (tan )
ln
sin
Example(37):
Example(38):
Example(39):
x x x
x x x
x
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
22
(cos )
lnsin )cot lnsin cot
1 1lnsin
2 2
Example(40): (
x xe d x e c
x x dx let y x dy xdx
I ydy y c x c
− = − +− = − +− = − +− = − +
==== ⇒⇒⇒⇒ ====
∴ = = + = +∴ = = + = +∴ = = + = +∴ = = + = +
∫∫∫∫
∫∫∫∫
∫∫∫∫
Indefinite Integration
11
(((( ))))3 2 3
2 2 2
2 2
2
2
(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
21
ln(tan 1 tan )2
co
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
x x c
dx
+ = + + ++ = + + ++ = + + ++ = + + +
= + + + += + + + += + + + += + + + +
= + + + += + + + += + + + += + + + +
= + + + + += + + + + += + + + + += + + + + +
+ + + ++ + + ++ + + ++ + + +
∫ ∫∫ ∫∫ ∫∫ ∫
∫∫∫∫
∫∫∫∫
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
x u
= == == == =− −− −− −− −−−−−
= −= −= −= − ⇒⇒⇒⇒ ====
∴ = = = − +∴ = = = − +∴ = = = − +∴ = = = − +−−−−
∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫
∫ ∫∫ ∫∫ ∫∫ ∫
(((( ))))(((( ))))(((( ))))(((( ))))
3
3 2 3
2 2
2 2
2 2
2 2
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
2
Example(43): x dx
x dx x x x dx
x x x x dx
x x x x dx
x x x x x dx
x x x x dx
d
++++
+ = + + ++ = + + ++ = + + ++ = + + +
= + + − += + + − += + + − += + + − +
= + + − + −= + + − + −= + + − + −= + + − + −
= + + − + −= + + − + −= + + − + −= + + − + −
= − + + += − + + += − + + += − + + +
= −= −= −= −
∫∫∫∫
∫ ∫∫ ∫∫ ∫∫ ∫
∫∫∫∫
∫∫∫∫
∫∫∫∫
∫∫∫∫
2 2
2
12
12
2tan 3sec tan sec
12 2ln sec 3tan tan
2
sin39
1tan
5 55
Example(44):
Example(45):
x x dx x dx x x dx
x x x x c
dx xdx c
xdx x
cx
−−−−
−−−−
+ + ++ + ++ + ++ + +
= − + + + += − + + + += − + + + += − + + + +
= += += += +−−−−
= += += += +++++
∫ ∫ ∫ ∫∫ ∫ ∫ ∫∫ ∫ ∫ ∫∫ ∫ ∫ ∫
∫∫∫∫
∫∫∫∫
Mathematics For Engineering
12
3 2 3
2
2
2
3
(1 cos ) (1 3cos 3cos cos )
31 3cos (1 cos 2 ) cos cos
2
3 31 3cos cos 2 cos (1 sin )
2 2
5 34cos cos 2 sin cos
2 2
5 3 14sin sin 2 sin
2 4 3
Example(46):
x dx x x x dx
x x x x dx
x x x x dx
x x x x dx
x x x x c
+ = + + ++ = + + ++ = + + ++ = + + +
= + + + += + + + += + + + += + + + +
= + + + + −= + + + + −= + + + + −= + + + + −
= + + −= + + −= + + −= + + −
= + + − += + + − += + + − += + + − +
∫ ∫∫ ∫∫ ∫∫ ∫
∫∫∫∫
∫∫∫∫
∫∫∫∫
3 2 3
2
2
2
3
(1 sin ) (1 3sin 3sin sin )
31 3sin (1 cos 2 ) sin sin
2
3 31 3sin cos 2 sin (1 cos )
2 2
5 34sin cos 2 sin cos
2 2
5 3 14cos sin 2 cos
2 4 3
Example(47):
x dx x x x dx
x x x x dx
x x x x dx
x x x x dx
x x x x c
+ = + + ++ = + + ++ = + + ++ = + + +
= + + − += + + − += + + − += + + − +
= + + − + −= + + − + −= + + − + −= + + − + −
= + + −= + + −= + + −= + + −
= − + + += − + + += − + + += − + + +
∫ ∫∫ ∫∫ ∫∫ ∫
∫∫∫∫
∫∫∫∫
∫∫∫∫
(((( ))))
2
1 12 22 3
2
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
x x x
dx dx xc
x x
− −− −− −− −
−−−−
= = = = += = = = += = = = += = = = ++ ++ ++ ++ + ++++
= = += = += = += = ++ ++ ++ ++ +
∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫
∫ ∫∫ ∫∫ ∫∫ ∫
Indefinite Integration
13
(((( )))) (((( ))))
12 2 2
12 2
2 2 31 3
6 3 2 3 2
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 ( )
sin
cos
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
x x x
xdx
−−−−
−−−−
−−−−
= = += = += = += = +− −− −− −− −′′′′
= += += += +−−−−
= = = += = = += = = += = = +− − −− − −− − −− − −
++++
∫ ∫∫ ∫∫ ∫∫ ∫
∫∫∫∫
∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫
12
2
4 2 2 2 2
1 2
2
4 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
x x
x c
let u x du x xdx
x xdx du
u
−−−−
−−−−
= − = − += − = − += − = − += − = − +++++
−−−−= == == == =+ + ++ + ++ + ++ + +
= += += += +
==== ⇒⇒⇒⇒ ====−−−−∴ = =∴ = =∴ = =∴ = =
+ ++ ++ ++ +
∫ ∫∫ ∫∫ ∫∫ ∫
∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫
∫ ∫∫ ∫∫ ∫∫ ∫1 1 2
12 2 2
12 2 2
3 2
2 2
21
1n tan (cos )
2
1 4 1 4sin
4 4 525 16 5 (4 )
( 2)sin
24 ( 2) 2 ( 2)
3 4 3 4(3 4)
1 1
34 4tan
2
Example(53):
Example(54):
Example(55):
Example
u c x c
dx dx xc
x x
dx dx xc
x x
x x xdx x dx
x x
xx x c
− −− −− −− −
−−−−
−−−−
−−−−
−−−−+ = ++ = ++ = ++ = +
= = += = += = += = +− −− −− −− −
++++= = += = += = += = +− + − +− + − +− + − +− + − +
− +− +− +− + = − += − += − += − + + ++ ++ ++ +
= − + += − + += − + += − + +
∫ ∫∫ ∫∫ ∫∫ ∫
∫ ∫∫ ∫∫ ∫∫ ∫
∫ ∫∫ ∫∫ ∫∫ ∫
2 2 2 2
1
4 13 ( 4 4) 9 ( 2) 3
1 ( 2)tan
3 3
(56): dx dx dx
x x x x x
xc−−−−
= == == == =+ + + + + + ++ + + + + + ++ + + + + + ++ + + + + + +
++++= += += += +
∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫
Mathematics For Engineering
14
2 2 2
12 2
2 2 2
2 2
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
2
Example(57):
Example(58):
dx dx dx
x x x x x x
dx xc
x
x x xdx dx dx
x x x x x xx
dx dxx x x x
x
−−−−
= == == == =+ − − − − − + −+ − − − − − + −+ − − − − − + −+ − − − − − + −
−−−−= = += = += = += = +− −− −− −− −
+ − − − − − − −+ − − − − − − −+ − − − − − − −+ − − − − − − −= == == == =− − − − − −− − − − − −− − − − − −− − − − − −
− − − − −− − − − −− − − − −− − − − −= += += += +− − − −− − − −− − − −− − − −
− − −− − −− − −− − −====
∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫
∫∫∫∫
∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫
∫ ∫∫ ∫∫ ∫∫ ∫
2 2
2 1
2
2
) 1
5 4 9 ( 2)
1 ( 2).2 5 4 sin
2 3
Another solution (5 4 ) 2 4
put 3 ( (5 4 )) ( 2 4)
1 1, 1, 3 ( 2 4) 1,
2 2
dx dxx x x
xx x c
dx x x
dxd
x A x x B A x Bdx
A B x x
−−−−
+ =+ =+ =+ =− − − +− − − +− − − +− − − +
− +− +− +− += − − + += − − + += − − + += − − + +
− − = − −− − = − −− − = − −− − = − −
+ = − − + = − − ++ = − − + = − − ++ = − − + = − − ++ = − − + = − − +
− −− −− −− −∴ = = + = − − +∴ = = + = − − +∴ = = + = − − +∴ = = + = − − +
∫ ∫∫ ∫∫ ∫∫ ∫
12
2 2
12
2 2 2
2 12
( 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
−−−−
−−−−
−−−−
− − +− − +− − +− − +++++ ====− − − −− − − −− − − −− − − −
− −− −− −− − − − −− − −− − −− − −= + == + == + == + =
− − − − − −− − − − − −− − − − − −− − − − − −− +− +− +− +
+ = − − + ++ = − − + ++ = − − + ++ = − − + +− +− +− +− +
∫ ∫∫ ∫∫ ∫∫ ∫
∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫
∫∫∫∫
22 3
9 12 81 39
2 3 (18 12) , 3 129 9
Example(59): xdx
x x
put x A x B A B A
++++− +− +− +− +
+ = − ++ = − ++ = − ++ = − + ⇒⇒⇒⇒ = = + == = + == = + == = + =
∫∫∫∫
Indefinite Integration
15
2 2
2 2
2 2
2 2
2 1
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
x x x
xx x −−−−
+ − ++ − ++ − ++ − +∴ =∴ =∴ =∴ =− + − +− + − +− + − +− + − +
−−−−= += += += +− + − +− + − +− + − +− + − +
−−−−= += += += +− + − + +− + − + +− + − + +− + − + +
−−−−= += += += +− + − +− + − +− + − +− + − +
−−−−= − + += − + += − + += − + +
∫ ∫∫ ∫∫ ∫∫ ∫
∫ ∫∫ ∫∫ ∫∫ ∫
∫ ∫∫ ∫∫ ∫∫ ∫
∫ ∫∫ ∫∫ ∫∫ ∫
2)2
c++++
[[[[ ]]]][[[[ ]]]]
2
2 2 2 2
2 2
2 2
22 (4 2 )
41 1
, 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
x x
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
x x x
++++ + = − ++ = − ++ = − ++ = − +−−−−
− −− −− −− −∴ = = + = − +∴ = = + = − +∴ = = + = − +∴ = = + = − +
− +− +− +− ++ − − −+ − − −+ − − −+ − − −= = −= = −= = −= = −− − − −− − − −− − − −− − − −
− −− −− −− −= −= −= −= −− − − +− − − +− − − +− − − +
− −− −− −− −= − = − −= − = − −= − = − −= − = − −− − −− − −− − −− − −
∫∫∫∫
∫ ∫ ∫ ∫∫ ∫ ∫ ∫∫ ∫ ∫ ∫∫ ∫ ∫ ∫
∫ ∫∫ ∫∫ ∫∫ ∫
∫ ∫∫ ∫∫ ∫∫ ∫2 11 ( 2)
sin2 2
xx c−−−− −−−−− +− +− +− +
Exercise(1)
Integrate the following functions with respect to x :
3
4 3
2 2 3 72
1 2(1) (3 2 4 ) (2)( 3)( 4) (3)( )
2
1(4)( 2) (5)(3 1) (6)
4 10)
( 1)(7) (2 3) (8) (3 4) (9)
2 4
x x x x x xx
x xx
xx x x x
x x
− + − + + +− + − + + +− + − + + +− + − + + +
+ −+ −+ −+ −−−−−+++++ −+ −+ −+ −
+ ++ ++ ++ +
Mathematics For Engineering
16
2 2
2
23 4
2
3 sin tan 2
2 25
2
2 3 2
2
(1 ) 1(10) (11) (12)
13 1
2 2 2(13) (14) (15) ( 2)
2 2 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)
( 1)
x x x
x
x
x x
x x xxx x
x x xx x
x x x
a x x
x ex x
x e
x x x e exx
−−−−
+ −+ −+ −+ −++++++++
+ + ++ + ++ + ++ + + ++++++++ + ++ ++ ++ +
−−−−++++−−−− ++++
+ + ++ + ++ + ++ + +++++
2 3 4
5 2 4 22
2
2
(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)
xe
x x x x x x
xx x x
xx x x x
x x xx
xx xx
−−−−
+++++ ++ ++ ++ +++++++++
− −− −− −− −++++3 5
2
2 2
22 4
cot ) (1 tan) 1(38) (39)
1 sin 2 1 cos 2 5
sec sin cos(40) (41) (42)
1 cos 24 tan 1
x
x
xx x x
x e x x
xx e
+ ++ ++ ++ +− −− −− −− − ++++
++++− −− −− −− −
22 2
2 22
2 2
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 x
x xx xx x
x x x xx xx x
x x x x
+ −+ −+ −+ −− −− −− −− −− −− −− −− −
+ + − ++ + − ++ + − ++ + − ++ −+ −+ −+ −−−−−
+ − + −+ − + −+ − + −+ − + −
Indefinite Integration
17
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 xx x
x x x x
x x x x
x e e xx e e
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
2 2
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 x
x x
x x
x y x y x y
x y x y x y
x y x y x y
x y x y x y
− =
− =
− =+ = +− = −+ = +− = −
[ ]
[ ]
2 2
2
2
2
2sinh cosh
(9) cosh 2 cosh sinh
1(10) sinh cosh 2 1
21
(11) cosh cosh 2 12
(12) cosh sinh
(13) cosh sinh
2tanh(14)tanh 2
1 tanh
x
x
x x x
x x x
x x
x x
x x e
x x e
xx
x
−
=
= +
= −
= +
+ =
− =
=+
we can proof this lows by using the definition in the following we stat integration formula for hyperbolic functions
Mathematics For Engineering
18
2
2
2 2 2
(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 dx x c
x dx x c
x dx x c
x dx x c
x dx x c
x x dx x c
x x dx x c
dx
bb x a
−−−−
= += += += += += += += += += += += +
= += += += +
= += += += +
= − += − += − += − += − += − += − += − +
= − += − += − += − +
====++++
∫∫∫∫
∫∫∫∫
∫∫∫∫
∫∫∫∫
∫∫∫∫
∫∫∫∫
∫∫∫∫
∫∫∫∫
∫∫∫∫1
12 2 2
1
2 2 2
1
2 2 2
2
1(10) cosh
1tanh
(11)1
ln2
1coth
(12)1
ln2
Solved Examples:
1 cosh cosh(1) sech
cosh cosh 1 si
bxc
a
dx bxc
b ab x a
bxc
dx ab aa bxa b x c
ab a bx
bxc
dx ab abx ab x a c
ab bx a
x xx dx dx dx
x x
−−−−
−−−−
−−−−
++++
= += += += +−−−−
++++==== ++++−−−− ++++ −−−−
++++==== −−−−−−−− ++++ ++++
= = == = == = == = =++++
∫∫∫∫
∫∫∫∫
∫∫∫∫
∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫1
2
2
3 2 2
3
tan (sinh )nh
1 1 1(2) sinh (cosh2 1) sinh2
2 4 2
(3) cosh 2 (1 sinh 2 )cosh2 cosh2 sinh 2 cosh2
1 1 sinh 2sinh2
2 2 3
dx x cx
x dx x dx x x c
x dx x x dx x dx x x dx
xx c
−−−−= += += += +
= − = − += − = − += − = − += − = − +
= + = += + = += + = += + = +
= + += + += + += + +
∫∫∫∫
∫ ∫∫ ∫∫ ∫∫ ∫
∫ ∫ ∫ ∫∫ ∫ ∫ ∫∫ ∫ ∫ ∫∫ ∫ ∫ ∫
Indefinite Integration
19
2 2
3 3
5 53 3
8 2 8 2
(4) cosh2 .2
1 1 1( ) ( )
2 2 3
(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
[[[[ ]]]]
2 2
12
2
2
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 cx
x dx x dx x x c
x d
−−−−
= += += += +
= = == = == = == = =++++
= = += = += = += = +++++
= − = − += − = − += − = − += − = − +
∫∫∫∫
∫ ∫ ∫ ∫∫ ∫ ∫ ∫∫ ∫ ∫ ∫∫ ∫ ∫ ∫
∫∫∫∫
∫ ∫∫ ∫∫ ∫∫ ∫
∫∫∫∫ [[[[ ]]]]
2 2
4 2 2
2 2 2 3
1 1 1cosh6 1 sinh6
2 2 6
1: tanh 5 1 sech 5 tanh5
5
: sech 1 tanh sech
1sech tanh sech tanh tanh
3
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
= + = + += + = + += + = + += + = + +
= − = − += − = − += − = − += − = − +
= −= −= −= −
= − = − += − = − += − = − += − = − +
∫∫∫∫
∫ ∫∫ ∫∫ ∫∫ ∫
∫ ∫∫ ∫∫ ∫∫ ∫
∫ ∫∫ ∫∫ ∫∫ ∫
3 2
2
2
3
: sinh 4 sinh 4 (sinh4 )
(cosh 4 1)(sinh4 )
(cosh 4 (sinh4 ) (sinh4 )
1 1cosh 4 cosh4
12 4
Example(7) x dx x x dx
x x dx
x x dx x dx
x x c
====
= −= −= −= −
= −= −= −= −
= − += − += − += − +
∫ ∫∫ ∫∫ ∫∫ ∫
∫∫∫∫
∫ ∫∫ ∫∫ ∫∫ ∫
Mathematics For Engineering
20
3 33 3
4 24 2
3 32 2 2 3 3
55
( ) 1: cosh3 ( )
2 2
1 1( )
2 2 4 2
( ) 1: sinh3 ( )
2 2
1 1( )
2 2 5
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
−−−−−−−−
−−−−−−−−
−−−−−−−−
− −− −− −− −
++++= = += = += = += = +
= + = + += + = + += + = + += + = + + −−−−
−−−−= = −= = −= = −= = −
= + = + += + = + += + = + += + = + +
∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫
∫∫∫∫
∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫
∫∫∫∫
(((( ))))
12
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
x x x x
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
dx x
x
− −− −− −− −
−−−−
− −− −− −− −
− −− −− −− −
−−−−
= − = −= − = −= − = −= − = −
= −= −= −= −
= − − − − += − − − − += − − − − += − − − − +
= − − − += − − − += − − − += − − − +
= − += − += − += − +
====++++
∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫
∫ ∫∫ ∫∫ ∫∫ ∫
∫∫∫∫
∫∫∫∫
12
3 2
2
2
3
4
: cosh525
: sinh 4 sinh 4 (sinh4 )
(cosh 4 1)(sinh4 )
(cosh 4 (sinh4 ) (sinh4 )
1 1cosh 4 cosh4
12 4
Example(12)
Example(13)
c
dx xc
x
x dx x x dx
x x dx
x x dx x dx
x x c
−−−−
++++
= += += += +−−−−
====
= −= −= −= −
= −= −= −= −
= − += − += − += − +
∫∫∫∫
∫ ∫∫ ∫∫ ∫∫ ∫
∫∫∫∫
∫ ∫∫ ∫∫ ∫∫ ∫
Indefinite Integration
21
Exercise(2)
Integrate the following functions: 2 2
2 2 3
25 2
3 4 5 2
4 22
(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 x e x e x
xx x x x
x
x x x x x x
xx
+ ++ ++ ++ +−−−−
++++ 2
2
2
3 5 2
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 x
x xx x x x
x x xx
xx xx
x xx x x
+++++ ++ ++ ++ +++++++++
− +− +− +− +++++
+ ++ ++ ++ +− −− −− −− − −−−−
Mathematics For Engineering
22
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
let
then
by substituting in the rule then
x
x
x x x
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= − += − += − += − + .
Indefinite Integration
23
Example (2): Find: 2 lnx x dx∫∫∫∫ Solution: Consider the partition 2lnu x dv x dx= == == == =
Then 31
3x
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= = += = += = += = +
322
(1 )3
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
1
( ) ,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∴ = = − + = − + +∴ = = − + = − + +∴ = = − + = − + +∴ = = − + = − + +∫ ∫∫ ∫∫ ∫∫ ∫
Mathematics For Engineering
24
(try with thepartation what hapen)
try to solve is positive integer
sin
sinnn
u x dv xdx
I x x dx n
= == == == =
==== ∫∫∫∫
Example (5): Find cosx x dx∫∫∫∫ Solution: let cos
sin
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:
2 sin
2 cos
let u x dv xdx
du xdx v x
= == == == =∴ = = −∴ = = −∴ = = −∴ = = −
by substituting in the rule
2 cos 2 cos (1)
cos (5)
cos cos
sin
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∴ = − = −∴ = − = −∴ = − = −∴ = − = −∫∫∫∫
2 2
2
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:
2 cos
2 sin
u x dv x dx
du xdx v x
= == == == =∴ = =∴ = =∴ = =∴ = =
Indefinite Integration
25
2 2
2
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 2
2
2 2( )
x
x
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:
1
2
sin
1
u x dv dx
dxdu let v x
x
−−−−= == == == =
∴ = =∴ = =∴ = =∴ = =−−−−
1 12 2
1 2sin sin
21 1
x dx x dxI udv uv v du x x x x
x x
− −− −− −− −= = − − = −= = − − = −= = − − = −= = − − = −− −− −− −− −
∫ ∫ ∫ ∫∫ ∫ ∫ ∫∫ ∫ ∫ ∫∫ ∫ ∫ ∫
1 1 22
1 2 1sin sin .2 1
2 21
x dxx x x x x c
x
− −− −− −− −−−−−= + = + − += + = + − += + = + − += + = + − +−−−−
∫∫∫∫
try tosolve where is a positive integer1sinx x dx m−−−−∫∫∫∫
Example (10): Find 1tan x dx−−−−
∫∫∫∫ Solution
1
2
tan
1
u x dv dx
dxdu let v x
x
−−−−= == == == =
∴ = =∴ = =∴ = =∴ = =++++
1 1 12 2
1 2tan tan tan
21 1
x dx x dxudv x dx x x x x
x x− − −− − −− − −− − −∴ = = − = −∴ = = − = −∴ = = − = −∴ = = − = −
+ ++ ++ ++ +∫ ∫ ∫ ∫∫ ∫ ∫ ∫∫ ∫ ∫ ∫∫ ∫ ∫ ∫
Mathematics For Engineering
26
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.
sin
1cos
ax
ax
let u e dv bx dx
du a e dx v bxb
= == == == =
−−−−∴ = =∴ = =∴ = =∴ = =
(((( ))))
where
1 1sin cos cos
cos cos
cos (1)
cos
ax ax ax
axax
ax
ax
I e bx dx e bx bx a e dxb b
e abx e bx dx
b b
e aI bx J
b b
J e bx dx
− −− −− −− − = = −= = −= = −= = −
−−−−= += += += +
−−−−∴ = +∴ = +∴ = +∴ = +
====
∫ ∫∫ ∫∫ ∫∫ ∫
∫∫∫∫
∫∫∫∫
Similarly taking the second integral J let cos
1sin
1 1 1sin sin . sin sin
1sin (2)
ax
ax
ax 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
b b
= == == == =
∴ = =∴ = =∴ = =∴ = =
∴ = − = −∴ = − = −∴ = − = −∴ = − = −
= −= −= −= −
∫ ∫∫ ∫∫ ∫∫ ∫
from (1),(2) we get 2
2 2
2
2 2
1cos sin cos sin
cos sin
ax axax ax
axax
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
bb b
− −− −− −− − ∴ = + − = + −∴ = + − = + −∴ = + − = + −∴ = + − = + −
−−−−+ = ++ = ++ = ++ = +
Indefinite Integration
27
2 2 2
2 2 2
2
2 2 2
2 2 2 2
(1 ) ( ) cos sin
cos sin
cos sin
axax
axax
ax ax
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 2
2 2 2 2
2 2 2 2
sin cos sin
1 1sin sin cos
1sin cos
11 sin cos
co
axax
axax ax
axax
axax
ax
eI e bx dx b bx a bx c
b a
a a e aJ e bx I e bx bx J
b b b b b b
ae ae bx bx J
b b b
a a a b aeJ J J J e bx bx
bb b b b
J e
= = − + += = − + += = − + += = − + +++++
−−−−= − = − += − = − += − = − += − = − +
= + −= + −= + −= + −
+++++ = + = = ++ = + = = ++ = + = = ++ = + = = +
∴ =∴ =∴ =∴ =
∫∫∫∫
[[[[ ]]]]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
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 )
dxI
x====
++++∫∫∫∫
Mathematics For Engineering
28
Solution:
consider
5 25 2
3 23 2
22
22
(1 )(1 )
(1 )(1 )
dxI x dx
x
dxJ x dx
x
−−−−
−−−−
= = += = += = += = +++++
= = += = += = += = +++++
∫ ∫∫ ∫∫ ∫∫ ∫
∫ ∫∫ ∫∫ ∫∫ ∫
3 2
5 2
3 2 5 2
3 2 5 2
3 2 3 2 5 2
3 2 3 2 5 2 3 2
2
2
2 2 2
2 2 2
2 2 2
2 2 2 2
(1 )
3 (1 )
(1 ) 3 (1 )
(1 ) 3 (1 1)(1 )
(1 ) 3 (1 ) (1 )
(1 ) 3 (1 ) (1 ) (1 ) 3 3
3 (1
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 x x dx x x J I
I x
−−−−
−−−−
− −− −− −− −
− −− −− −− −
− − −− − −− − −− − −
− − − −− − − −− − − −− − − −
= + == + == + == + =
∴ = − + =∴ = − + =∴ = − + =∴ = − + =
= + + += + + += + + += + + +
= + + + − += + + + − += + + + − += + + + − +
= + + + − += + + + − += + + + − += + + + − +
= + + + − + = + + −= + + + − + = + + −= + + + − + = + + −= + + + − + = + + −
∴ =∴ =∴ =∴ =
∫∫∫∫
∫∫∫∫
∫∫∫∫
∫∫∫∫
3 22) 2x J−−−−+ ++ ++ ++ +
To solve consider with the partation
3 2
3 2 1 2
1 2
1 2 1 2
1 2
2
2 2
2
2 3 2
2 2 2 3 2 2 2 2 3 2
2 2 1 2 2
(1 ) 2(1)
3 3
(1 ) (1 )
(1 )
(1 )
(1 ) (1 ) (1 ) (1 1)(1 )
(1 ) (1 ) (1 )
x xI J
dx dxJ K
x x
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 2
1 21 2
3 2
1 2
3 2 2
22
2
2
(1 )
(1 )(1 )
(1 ) 23 3(1 )
dx x x K J
xJ x x c
x
x x xI c
x
−−−−
−−−−
−−−−
−−−− = + + −= + + −= + + −= + + −
= + = += + = += + = += + = +++++
++++∴ = + +∴ = + +∴ = + +∴ = + +++++
Indefinite Integration
29
Example (13): Find 3sec x dx∫∫∫∫ Solution:
3 2
2
3 2 2
3 3
3
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
====
= == == == =
= == == == =
∴ = − = − −∴ = − = − −∴ = − = − −∴ = − = − −
= − − = − += − − = − += − − = − += − − = − +
= + == + == + == + =
∫ ∫∫ ∫∫ ∫∫ ∫
∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫
∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫
∫ ∫∫ ∫∫ ∫∫ ∫
3
ln sec tan
1 1sec sec tan ln sec tan
2 2
x x c
x dx x x x x c
+ + ++ + ++ + ++ + +
∴ = + + +∴ = + + +∴ = + + +∴ = + + +∫∫∫∫
Example (14): Find 2 2I x a dx= += += += +∫∫∫∫ Solution:
2 2 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 2 2 2 2 1
2 2 2
(14)
( )
( )
sinh
2 sinh
2
I x a dx let u x a dv dx
x dxdu v x
x a
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
a
xx a dx x x a a
a
xx a dx x
−−−−
−−−−
= + = + == + = + == + = + == + = + =
= == == == =++++
+ −+ −+ −+ −= + = + − = + −= + = + − = + −= + = + − = + −= + = + − = + −+ ++ ++ ++ +
++++= + − += + − += + − += + − ++ ++ ++ ++ +
= + − + += + − + += + − + += + − + +
∴ + = + +∴ + = + +∴ + = + +∴ + = + +
∴ + =∴ + =∴ + =∴ + =
∫∫∫∫
∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫
∫ ∫∫ ∫∫ ∫∫ ∫
∫∫∫∫
∫∫∫∫
∫∫∫∫2
2 1sinh2
a xa c
a−−−−+ + ++ + ++ + ++ + +
Mathematics For Engineering
30
Example (15): Find 2 2I a x 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 2 2 2 2 1
22 2 2 2 1
( )
( )
sin
2 sin
sin2 2
let u a x dv dx
x dxdu v x
a 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
a
xa x dx x a x a
a
x a xa x dx a x
a
−−−−
−−−−
−−−−
= − == − == − == − =
−−−−= == == == =−−−−
− − −− − −− − −− − −= + = − − = − −= + = − − = − −= + = − − = − −= + = − − = − −− −− −− −− −
−−−−= − − += − − += − − += − − +− −− −− −− −
= − − − += − − − += − − − += − − − +
∴ − = − +∴ − = − +∴ − = − +∴ − = − +
∴ − = − +∴ − = − +∴ − = − +∴ − = − +
∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫
∫ ∫∫ ∫∫ ∫∫ ∫
∫∫∫∫
∫∫∫∫
∫∫∫∫ c++++
Exercise(3)
Integrate the following function with respect to x : 2 3
2 3
4 2 3 2
1 1 1 1
2
(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 (
n
x x x x
x x x x 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
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 x
−−−−
− − −− − −− − −− − −
++++
Indefinite Integration
31
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
m
1
2
1 2 2m
1 2 2
1
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
m m m
m
m
m m
m m
m m
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
x x m x x dx
x x m
− −− −− −− −
−−−−
−−−−
− −− −− −− −
− −− −− −− −
−−−−
= = = −= = = −= = = −= = = −
= = −= = −= = −= = −
∴ = − = −∴ = − = −∴ = − = −∴ = − = −
= − + −= − + −= − + −= − + −
= − + − −= − + − −= − + − −= − + − −
= − + −= − + −= − + −= − + −
∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫
∫∫∫∫
∫∫∫∫
∫∫∫∫
(((( ))))
2
1m 2 m
1m 2
1m 2
( 1) sin
I sin cos ( 1) ( 1)I
1 ( 1) I sin cos ( 1)
1 ( 1)I sin cos
m
mm
mm
mm
xdx m xdx
x x m I m
m x x m I
mx x I
m m
−−−−
−−−−−−−−
−−−−−−−−
−−−−−−−−
− −− −− −− −
= − + − − −= − + − − −= − + − − −= − + − − −
∴ + − = − + −∴ + − = − + −∴ + − = − + −∴ + − = − + −
− −− −− −− −= += += += +
∫∫∫∫
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
m m m-1
m m m-1
1If I I I
1If I I I
ln ln
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
m m
mx e dx x e
a a
mx a dx x a
a a
x dx x x m
= = −= = −= = −= = −
= = −= = −= = −= = −
= = −= = −= = −= = −
∫∫∫∫
∫∫∫∫
∫∫∫∫
Mathematics For Engineering
32
m
1m m-22
1
m m m-2
2
m m m-2
If I sin
( 1)I cos sin I
tanIf I tan I I
1
sec tan 2If I sec I I
1 1
(6)
then show that
(7) then show that
(8) then show that
m
mm
mm
mm
x ax dx
x m m max x ax
a aa
xdx
m
x x mdx
m m
−−−−
−−−−
−−−−
====
− −− −− −− −= + −= + −= + −= + −
= = −= = −= = −= = −−−−−
−−−−= = += = += = += = +− −− −− −− −
∫∫∫∫
∫∫∫∫
∫∫∫∫
m
2
m m-2
n
n
If I cosec
cosec cot 2I I
1 1
If I cos , sin
I sin . cos
(9) then show that
(10)
show that
m
m
n nn
n nn n n
x dx
x x mm m
x bx dx J x bx dx
x bx nJ J x bx n I
−−−−
====
− −− −− −− −= += += += +− −− −− −− −
= == == == =
= − = −= − = −= − = −= − = −
∫∫∫∫
∫ ∫∫ ∫∫ ∫∫ ∫
Exercise(4)
If
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
(1)
(2)
(3)
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
n
− −− −− −− −
− +− +− +− +
= == == == =
= − = −= − = −= − = −= − = −
====
= == == == =
∫ ∫∫ ∫∫ ∫∫ ∫
∫∫∫∫
∫ ∫∫ ∫∫ ∫∫ ∫
prove that
Find thereducion formula for
are positive integer.
12
12
2 2
sin cos ( 1) .
cos sin ( 1) .
1-1sin sin( 1) sin sin
1 ,
,
(4)
(5)
nn n
nn n
nm
I x x n I
nJ x x n J
n nx n x dx x nxn
x x dx
m n
−−−−−−−−
−−−−−−−−
= + −= + −= + −= + −
= − + −= − + −= − + −= − + −
+ =+ =+ =+ =∫∫∫∫
++++ ∫∫∫∫
Indefinite Integration
33
find
find
find
find
Find thereducion formula for
Find thereducion formula for
4cos cos
3sin sin
3sin sin
3cos cos
sinh cosh tanh
(6)
(7)
(8)
(9)
(10) (11) (12)
(13
ax n axxdx xdx
ax n axxdx e xdx
n x dx x x dx
n x dx x x dx
n n nx dx x dx x dx
e e
e
x
x
∫∫∫∫
∫∫∫∫
∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫
∫ ∫∫ ∫∫ ∫∫ ∫
∫∫∫∫
∫∫∫∫
∫ ∫∫ ∫∫ ∫∫ ∫
cosech sech
Show thatIf
coth
2(2 3) ( 2)sin 1 21 2 2 2 2( cos ) ( 1)( ) ( 1)( )
( cos )
) (14) (15)
(16)
n n n x dx
n n
a n I n Im x n nIn na b x n a b n a b
x dx x dx
dx
a b xI
∫∫∫∫
− −− −− −− −− −− −− −− −= + −= + −= + −= + −−−−−+ − − − −+ − − − −+ − − − −+ − − − −
∫ ∫∫ ∫∫ ∫∫ ∫
==== ∫∫∫∫++++
Mathematics For Engineering
34
Trigonometric Integrals The following identities are employed to find the Trigonometric Integrals
[[[[ ]]]]
[[[[ ]]]]
[[[[ ]]]]
[[[[ ]]]]
2 2
2 2
2 2
2
2
2
2
(1) sin cos 1
(2) tan 1 sec
(3)1 cot csc
1(4) sin 1 cos 2
2
1(5) cos 1 cos 2
2
(6) sin 2 2sin cos
(7) 1 cos 2 2sin
(8)1 cos 2 2cos
1(9) sin cos sin( ) sin( )
2
1(10) sin sin cos( ) cos( )
2
(
x x
x x
x x
x x
x x
x x x
x x
x x
x y x y x y
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= − = += − = += − = += − = +
Indefinite Integration
35
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 2
3 2 3 3
sin sin (sin ) (1 cos )(sin )
cos sin
1 1sin (1 ) cos cos
3 3
x dx x x dx x x dx
let y x dy x dx
x dx y dy y y x x c
= = −= = −= = −= = −
= = −= = −= = −= = −
∴ = − − = − − = − − +∴ = − − = − − = − − +∴ = − − = − − = − − +∴ = − − = − − = − − +
∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫
∫ ∫∫ ∫∫ ∫∫ ∫
Example (2): Find 5sin x dx∫∫∫∫ Solution:
5 4 2 2
5 2 2 2 2 3
2 3
sin sin (sin ) (1 cos ) (sin )
cos sin
1sin (1 ) (1 2 )
3
1cos cos cos
3
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
= = −= = −= = −= = −
= = −= = −= = −= = −
∴ = − − = − − + = − − +∴ = − − = − − + = − − +∴ = − − = − − + = − − +∴ = − − = − − + = − − +
= − − + += − − + += − − + += − − + +
∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫
∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫
Example (3): Find 4sin x dx∫∫∫∫ Solution:
4 2 2 2
2
1(3) sin sin sin (1 cos2 )
4
1 1 11 2cos2 cos 2 1 2cos2 (1 cos4 )
4 4 2
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
= = −= = −= = −= = −
= − + = − + −= − + = − + −= − + = − + −= − + = − + −
= − − = − − += − − = − − += − − = − − += − − = − − +
∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫
∫ ∫∫ ∫∫ ∫∫ ∫
∫∫∫∫
Mathematics For Engineering
36
(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
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 2
3 2 3 3
cos cos (cos ) (1 sin )(cos )
sin cos
1 1cos (1 ) sin sin
3 3
x dx x x dx x x dx
let y x dy x dx
x dx y d y y y c x c
= = −= = −= = −= = −
= == == == =
∴ = − = − + = − +∴ = − = − + = − +∴ = − = − + = − +∴ = − = − + = − +
∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫
∫ ∫∫ ∫∫ ∫∫ ∫
Example (5): Find 5cos x dx∫∫∫∫ Solution:
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
3 3
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
= = −= = −= = −= = −
= == == == =
∴ = − = − +∴ = − = − +∴ = − = − +∴ = − = − +
= − + = − + += − + = − + += − + = − + += − + = − + +
∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫
∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫
Example (6): Find 4cos x dx∫∫∫∫ Solution:
4 2 2 21cos cos cos (1 cos 2 )
4x dx x x dx x dx= = += = += = += = +∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫
Indefinite Integration
37
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
3 2 3
3 2 2
3 2 2
4 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
4 2
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= = −= = −= = −= = −∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫
Mathematics For Engineering
38
4 2 4
4 2 2 2
4 2 2 2 2
4 2 2 2 2
4 2 2 2 2
5 3
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
5 3
x x dx x dx
x x dx x dx
x x dx 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
1
m m m-2tan
If I tan Then I I( 1)
mm axdx
a m
−−−−= = −= = −= = −= = −
−−−−∫∫∫∫
(4)Integrals in the form: cotm x dx∫∫∫∫
(i) if m is odd positive integer then m-1 is even
1 2 1 2cot cot (cot ) (cosec 1) (cot )
cosec cosec cot
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 integration by parts
2 2
2 2 2
cot cot (cosec 1)
cot cosec cot
m m
m m
x dx x x dx
x x dx x dx
−−−−
− −− −− −− −
= −= −= −= −
= −= −= −= −
∫ ∫∫ ∫∫ ∫∫ ∫
∫ ∫∫ ∫∫ ∫∫ ∫
In general we can use the reduction formula
1
m m m-2cot
If I cot Then I I( 1)
mm ax dx
a m
−−−−= = − −= = − −= = − −= = − −
−−−−∫∫∫∫
Example (9): Find 4cot 3x dx∫∫∫∫ Solution:
4 2 2 2 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
= = −= = −= = −= = −
= −= −= −= −
∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫
∫ ∫∫ ∫∫ ∫∫ ∫
Indefinite Integration
39
2 2 2
2 2 2
3
cot 3 cosec 3 (cosec 3 1)
cot 3 cosec 3 cosec 3
1 1cot 3 cot 3
9 3
x x dx x dx
x x dx x dx dx
x x x c
= − −= − −= − −= − −
= − += − += − += − +
= + + += + + += + + += + + +
∫ ∫∫ ∫∫ ∫∫ ∫
∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫
(5)Integrals in the form: secm x dx∫∫∫∫
(i) if m is odd positive integer then m-1 is even
1 2 ( 1) 2sec sec (sec ) (tan 1) (sec )
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) / 2
sec sec (sec )
tan sec (tan 1)
m m
m
x dx x x dx
put y x and x x
−−−−
−−−−
====
= = += = += = += = +
∫ ∫∫ ∫∫ ∫∫ ∫
to get the reduction formula
2
m m-2sec tan 2
I sec I1 1
mm x m
dxm 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
2 6
x dx x x dx x x dx
x dx x x dx x x c
= = += = += = += = +
= + = + += + = + += + = + += + = + +
∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫
∫ ∫∫ ∫∫ ∫∫ ∫
(6)Integrals in the form: cosecm x dx∫∫∫∫
(i) if m is odd positive integer then m-1 is even
Mathematics For Engineering
40
1 2 ( 1) 2cosec cosec (cosec ) (cot 1) (cosec )
cosec cosec cot
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/ 2
cosec cosec (cosec )
cot cosec (cot 1)
m m
m
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 4 2
2 2 2 4 2
3 5
(11) cosec cosec (cosec ) (1 cot ) (cosec )
(1 2cot cot )(cosec )
(cosec 2cot cosec cot cosec )
1 2 1cot cot cot
3 5
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
= == == == =
∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫
Indefinite Integration
41
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
2 4
4 6
4 6
5 7
(13) sin cos sin cos (sin )
(1 cos )cos (sin )
(cos cos )(sin )
(cos sin cos sin )
1 1cos cos
5 7
ax ax dx ax 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:
[[[[ ]]]]
24 2
2
2
2 2
1 1(14) sin cos (1 cos2 ) (1 cos2 )
2 2
1(1 cos2 )(1 cos2 )(1 cos2 )
8
1(1 cos2 )(1 cos 2 )
8
1(1 cos2 )sin 2
8
1sin 2 cos2 sin 2
8
1 1(1 cos4
8 2
ax ax dx ax ax dx
ax ax ax dx
ax ax dx
ax ax dx
ax dx ax ax dx
= − += − += − += − +
= − − += − − += − − += − − +
= − −= − −= − −= − −
= −= −= −= −
= −= −= −= −
= −= −= −= −
∫ ∫∫ ∫∫ ∫∫ ∫
∫∫∫∫
∫∫∫∫
∫∫∫∫
∫ ∫∫ ∫∫ ∫∫ ∫
2
3
3
) 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
x ax ax ca a
−−−−
= − − += − − += − − += − − +
= − − += − − += − − += − − +
∫ ∫∫ ∫∫ ∫∫ ∫
Mathematics For Engineering
42
The reduction formula for cos sinm nx x dx∫∫∫∫
Case (1): if m and n are positive integer:
1 1
1 2 1
2 2
2 2
2 2
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
m n n m
m n n m
m n n m
m n n m
dx x
dx
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
x
n x x m n x x−−−−= − − + += − − + += − − + += − − + +
by integrating both sides w.r.tox we have
are positive integers
1 1
2
,( 2) ,
1 1, ,( 2)
sin .cos
( 1)sin .cos ( )cos sin
( 1) ( )
1 ( 1)cos sin sin .cos
( ) ( )
,
m n
m n n m
m n m n
m n m nm n m n
x x
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
m n
+ −+ −+ −+ −
−−−−
−−−−
+ −+ −+ −+ −−−−−
= − − + += − − + += − − + += − − + += − − + += − − + += − − + += − − + +
−−−−= = −= = −= = −= = −+ ++ ++ ++ +∴∴∴∴
∫ ∫∫ ∫∫ ∫∫ ∫
∫∫∫∫
Example(15):
4 6 5 54,6 4,4
5 5 5 34,2
5 5 5 3 54,0
44
1 1sin cos sin cos
10 2
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
= = −= = −= = −= = −
= − −= − −= − −= − −
= − − −= − − −= − − −= − − −
= = − += = − += = − += = − +
∫∫∫∫
∫∫∫∫
Indefinite Integration
43
5 54,6
5 3 5
5 5 5 3 5
1sin cos
10
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
I x x
x x x x x x x
x x x x x x x
x x
∴ =∴ =∴ =∴ =
− − − − +− − − − +− − − − +− − − − +
= − + −= − + −= − + −= − + −
+ −+ −+ −+ −
Case (2): if m and n are negative integers :
1 1
1 1
2 2
2 2
2
sin .cos
sin .( 1).cos .( sin ) cos .( 1)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
m n n m
m n m n
m n m n
m n m n
dx x
dx
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
2
sin cos
( 2)sin .cos ( 1)sin cos
m n
m n m n
x x
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), ,
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
m m
m n
+ − ++ − ++ − ++ − +
+ −+ −+ −+ −++++
+ ++ ++ ++ +++++
= − + + + += − + + + += − + + + += − + + + +
= − + + + += − + + + += − + + + += − + + + +
+ ++ ++ ++ += = += = += = += = ++ ++ ++ ++ +∴∴∴∴
∫ ∫∫ ∫∫ ∫∫ ∫
∫∫∫∫
Example(16):
Mathematics For Engineering
44
3 24, 3 ( 2, 3)4 3
( 2, 3)3 2
1 5sin cos
3 3sin cos
1 533sin cos
dxI x x I
x x
Ix x
− −− −− −− −− − − −− − − −− − − −− − − −
− −− −− −− −
−−−−= = += = += = += = +
−−−−= += += += +
∫∫∫∫
1 22, 3 0, 3 0, 32
3 2 20, 3 3
2 3
0, 3
2, 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
2 2
1
sin
I x x I Ix x
dxI x dx x x dx x x x x dx
x
x x x x dx x x xdx x dx
I x x x x
I
− −− −− −− −− − − −− − − −− − − −− − − −
−−−−
−−−−
− −− −− −− −
−−−−= − + = += − + = += − + = += − + = +
= = = = −= = = = −= = = = −= = = = −
= − + = − −= − + = − −= − + = − −= − + = − −
= −= −= −= −
−−−−∴ =∴ =∴ =∴ =
∫ ∫ ∫ ∫∫ ∫ ∫ ∫∫ ∫ ∫ ∫∫ ∫ ∫ ∫
∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫
2
( 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:
1 1
1 1 2
2 2
sin .cos
sin .( 1).cos .( sin ) cos .( 1)sin (cos )
( 1)sin .cos ( 1)sin cos
m n
m n n m
m n m n
dx x
dx
x n x x x m x x
n x x m x x
− +− +− +− +
− + −− + −− + −− + −
− +− +− +− +
= + − + −= + − + −= + − + −= + − + −
= − + + −= − + + −= − + + −= − + + −
by integrating both sides w.r.to x we have
is positive and is negative integars
1 1
2 2
, ( 2),( 2)
1 1, ( 1),( 2
sin .cos
( 1) sin .cos ( 1) sin cos
( 1) ( 1)
1 ( 1)cos sin 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
mI x x dx x x I
n n
m n
− +− +− +− +
− +− +− +− +
− +− +− +− +
− +− +− +− +− =− =− =− =
= − + + −= − + + −= − + + −= − + + −= − + + −= − + + −= − + + −= − + + −
− −− −− −− −= = += = += = += = ++ ++ ++ ++ +∴∴∴∴
∫ ∫∫ ∫∫ ∫∫ ∫
∫∫∫∫
Indefinite Integration
45
Example(17): 6 5
5 7(6, 8) (4, 6) (4, 6)8 7
33 5
(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
cos
x dx xI x x I I
x x
xI x x I I
x
xI x x I I
x
dxI
x
−−−−− − −− − −− − −− − −
−−−−− − −− − −− − −− − −
−−−−− − −− − −− − −− − −
−−−−
= = − + = −= = − + = −= = − + = −= = − + = −− −− −− −− −
−−−−= + = −= + = −= + = −= + = −− −− −− −− −
−−−−= + = −= + = −= + = −= + = −− −− −− −− −
====
∫∫∫∫
2
3
(2, 4) (4, 6)3 5 3
6 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 x 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:
1 1
1 2 1
2 2
sin .cos
sin .( 1).cos .( sin ) cos .( 1)sin cos
( 1)sin .cos ( 1)sin cos
m n
m n n m
m n m n
dx x
dx
x n x x x m x x
n x x m x x
+ −+ −+ −+ −
+ − −+ − −+ − −+ − −
+ −+ −+ −+ −
= − − + += − − + += − − + += − − + +
= − − + += − − + += − − + += − − + +
by integrating both sides w.r.to x we have
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)
1 ( 1)cos sin 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
m m
m n
+ − + −+ − + −+ − + −+ − + −
+ −+ −+ −+ −
+ −+ −+ −+ −+ −+ −+ −+ −
= − − + += − − + += − − + += − − + +
= − − + += − − + += − − + += − − + +
−−−−= = += = += = += = ++ ++ ++ ++ +∴∴∴∴
∫ ∫∫ ∫∫ ∫∫ ∫
∫∫∫∫
Mathematics For Engineering
46
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
4
( 6,46
cos 1 3 cos 3cos sin
5 5 5sin 5sin
1 1 cos 1cos sin
3 3 33sin
csec cotsin
cos 1cot
33sin
cos
sin
x dx xI x x I I
x x
xI x x I I
x
dxI x dx x
x
xI
x
x dxI
x
−−−−− − −− − −− − −− − −
−−−−− − −− − −− − −− − −
−−−−
−−−−
−−−−
−−−−= = − − = −= = − − = −= = − − = −= = − − = −
−−−−= + = −= + = −= + = −= + = −− −− −− −− −
= = = −= = = −= = = −= = = −
−−−−∴ = +∴ = +∴ = +∴ = +
====
∫∫∫∫
∫ ∫∫ ∫∫ ∫∫ ∫
3
) 5 3cos cos 1
cot55sin 5sin
x xc
x x
−−−−= + − += + − += + − += + − +∫∫∫∫
we can solve this example without reduction formulas
44 2 5
6cos 1
cot cosec cot5sin
x dxx x dx x c
x
−−−−= = += = += = += = +∫ ∫∫ ∫∫ ∫∫ ∫
(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
y x
− −− −− −− −====
====∫ ∫∫ ∫∫ ∫∫ ∫
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
3 2 2
5 3 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
x x x dx
====
= −= −= −= −
= −= −= −= −
∫ ∫∫ ∫∫ ∫∫ ∫
∫∫∫∫
∫∫∫∫
Indefinite Integration
47
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
5 3 2 35
3 3
3 3 2 3 3 2
3
(16) sec tan sec (sec 1) (sec sec )
(1)
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
I I I
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
x x
= − = −= − = −= − = −= − = −
∴ = −∴ = −∴ = −∴ = −
= = == = == = == = =
= −= −= −= −
= − = − −= − = − −= − = − −= − = − −
= −= −= −= −
∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫
∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫
∫∫∫∫
∫ ∫∫ ∫∫ ∫∫ ∫
5 3
35 3
3 23
2
2 3
sec sec )
1sec tan (2)
4
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
2
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
x x x
−−−−
∴ = +∴ = +∴ = +∴ = +
= = == = == = == = =
= − = −= − = −= − = −= − = −
= − − = − −= − − = − −= − − = − −= − − = − −
= + += + += + += + +
∫∫∫∫
∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫
∫ ∫∫ ∫∫ ∫∫ ∫
∫ ∫∫ ∫∫ ∫∫ ∫
(3)x
(((( ))))
(((( ))))
3 33 3 3
3
3
1 1(1),(2) (3) sec tan sec tan 3
4 4
1 3sec tan sec tan ln sec tan
4 2
1 3sec tan sec tan ln sec tan
4 8
from and I x x I I x x I
x x x x x x
x x x x x x c
= + − = −= + − = −= + − = −= + − = −
= − + += − + += − + += − + +
= − + + += − + + += − + + += − + + +
Mathematics For Engineering
48
Example (21): Find 3 3sec tanax ax dx∫∫∫∫
Solution: 3 3 2 2
2 2
4 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
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
y x
− −− −− −− −====
====∫ ∫∫ ∫∫ ∫∫ ∫
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
y x
−−−−====
====∫ ∫∫ ∫∫ ∫∫ ∫
Example (22): Find 3 3cot cosecax ax dx∫∫∫∫
Solution: 3 3 2 2
2 2
2 4
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 ( )
3 5
1
ax ax dx ax ax 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+ ++ ++ ++ +
Indefinite Integration
49
Example (23): Find 3 4cot cosecax ax dx∫∫∫∫
Solution: 3 4 3 2 2
3 2 2
3 5 2
2
3 4 3 5 4 6
4
(19) cot cosec cot cosec (cosec )
cot (1 cot )(cosec )
(cot cot )(cosec )
cot cosec
1 1 1 1cot cosec ( y ) y
4 6
1 1cot co
4 6
ax ax dx ax 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
axa a
====
= += += += +
= += += += +
= = −= = −= = −= = −
− −− −− −− − ∴ = + = +∴ = + = +∴ = + = +∴ = + = +
−−−−= −= −= −= −
∫ ∫∫ ∫∫ ∫∫ ∫
∫∫∫∫
∫∫∫∫
∫ ∫∫ ∫∫ ∫∫ ∫
6
4 4
t
try to solve cot cosec
ax c
ax ax dx
++++
∫∫∫∫
As similar we can integrate the following
sinh , cosh , tanh
cosech , sech , coth
sinh cosh
sech tanh
cosech coth
n n n
n n n
n m
n m
n m
x dx x dx x dx
x x dx x dx
x x dx
x x dx
x x dx
∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫
∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫
∫∫∫∫
∫∫∫∫
∫∫∫∫
Mathematics For Engineering
50
Exercise(5)
Find 3 2
4 4 2 5
3 2 3 3
4
10 2
5 5
(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
n
x dx x dx
x x dx x x dx
x x dx x x dx
x x dx x dx
x dx x x dx
x x dx x dx
x
∫ ∫∫ ∫∫ ∫∫ ∫
∫ ∫∫ ∫∫ ∫∫ ∫
∫ ∫∫ ∫∫ ∫∫ ∫
∫ ∫∫ ∫∫ ∫∫ ∫
∫ ∫∫ ∫∫ ∫∫ ∫
∫ ∫∫ ∫∫ ∫∫ ∫3 5
3 2
33
4
3 5 3 4
2 4 3 5
2 4 2 3
3
(14) sin cos
(15) sin cos (16) sin5 cos
cos(17) (18) tan
sin
(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 x dx x x dx
x x dx x x dx
x x dx x x dx
x
∫ ∫∫ ∫∫ ∫∫ ∫
∫ ∫∫ ∫∫ ∫∫ ∫
∫ ∫∫ ∫∫ ∫∫ ∫
∫ ∫∫ ∫∫ ∫∫ ∫
∫ ∫∫ ∫∫ ∫∫ ∫
∫ ∫∫ ∫∫ ∫∫ ∫4 3 3
5 5
6 5
5 5
4 4
5 3 2 4
3 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 dx x dx
x dx x dx
x dx x dx
x x dx x x dx
x x dx
∫ ∫∫ ∫∫ ∫∫ ∫
∫ ∫∫ ∫∫ ∫∫ ∫
∫ ∫∫ ∫∫ ∫∫ ∫
∫ ∫∫ ∫∫ ∫∫ ∫
∫ ∫∫ ∫∫ ∫∫ ∫
∫ ∫∫ ∫∫ ∫∫ ∫
∫∫∫∫4 3
2 4
8) coth cosech
(39) coth cosech
x x dx
x x dx
∫∫∫∫
∫∫∫∫
Indefinite Integration
51
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
2
22 2 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
b bb
aa b x a b a a
b
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
2
22 2 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
b bb
aa b x a b a a
b
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 sec , sec sec tan
( sec ) sec 1 tan
(iii) a a ab x a put x x dx d
b bb
ab x a b a a a
b
θ θ θ θ θθ θ θ θ θθ θ θ θ θθ θ θ θ θ
θ θ θθ θ θθ θ θθ θ θ
− = = =− = = =− = = =− = = =
∴ − = − = − =∴ − = − = − =∴ − = − = − =∴ − = − = − =
2
22 2
2
2 2 2
(iv)For , tansin cos 2
2sin 2sin cos ,
2 2 1
1cos cos sin ,
2 2 1
1 1 1sec (1 tan ) (1 )
2 2 2 2 2
dx dx xput u
a b x a b x
x x ux
u
x x ux
u
x xdu dx dx u dx
====± ±± ±± ±± ±
∴ = =∴ = =∴ = =∴ = =++++
−−−−= − == − == − == − =++++
= = + = += = + = += = + = += = + = +
∫ ∫∫ ∫∫ ∫∫ ∫
Mathematics For Engineering
52
2
2 2 22 2 1
, sin , cos1 1 1
du u udx x x
u u u
−−−−∴ = = =∴ = = =∴ = = =∴ = = =+ + ++ + ++ + ++ + +
2
22 2
2
2 2 2
2
2 2 2
(v)For , tanhsinh cosh 2
2sinh 2sinh cosh ,
2 2 1
1cosh cosh sinh ,
2 2 1
1 1 1sech (1 tanh ) (1 )
2 2 2 2 2
2 2 1, sinh , cosh
1 1 1
dx dx xput u
a b x a b x
x x ux
u
x x ux
u
x xdu dx dx u dx
du u udx x x
u u u
====± ±± ±± ±± ±
∴ = =∴ = =∴ = =∴ = =−−−−
++++= + == + == + == + =−−−−
= = − = −= = − = −= = − = −= = − = −
++++∴ = = =∴ = = =∴ = = =∴ = = =− − −− − −− − −− − −
∫ ∫∫ ∫∫ ∫∫ ∫
Example(1): Find 2
2 4
x dx
x −−−−∫∫∫∫
Solution:
(((( ))))
2 2 2 2
2 23
2
2 2 2 2
2sec 2sec tan ,
14 4sec 4 2tan , tan sec 1 4
2
4sec (2sec tan )4sec
2tan4
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 x
x dx dd
x
x x x xx x x x c
θ θ θ θθ θ θ θθ θ θ θθ θ θ θ
θ θ θ θθ θ θ θθ θ θ θθ θ θ θ
θ θ θ θθ θ θ θθ θ θ θθ θ θ θ θ θθ θθ θθ θθθθθ
θ θ θ θθ θ θ θθ θ θ θθ θ θ θ
= ∴ == ∴ == ∴ == ∴ =
− = − = = − = −− = − = = − = −− = − = = − = −− = − = = − = −
∴ = =∴ = =∴ = =∴ = =−−−−
= + += + += + += + +
= − + + − = − + + − += − + + − = − + + − += − + + − = − + + − += − + + − = − + + − +
∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫
Indefinite Integration
53
Example(2): Find 29
x dx
x−−−−∫∫∫∫
Solution:
[[[[ ]]]]
(((( ))))
2 2 2
2 22
2
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 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
dx
x x++++∫∫∫∫
Solution: 2
2 2
2
2
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)
3 3
2 3 9 42 3tan tan ,cot , cosec
3 2 2
1ln cosec
3
x dx d
x
dx d d d
x x
d
x xx
x x
I
==== ⇒⇒⇒⇒ ====
+ = + =+ = + =+ = + =+ = + =
∴ = = =∴ = = =∴ = = =∴ = = =++++
= = −= = −= = −= = −
++++==== ⇒⇒⇒⇒ = = == = == = == = =
∴ =∴ =∴ =∴ =
∫ ∫ ∫ ∫∫ ∫ ∫ ∫∫ ∫ ∫ ∫∫ ∫ ∫ ∫
∫∫∫∫
Q
θ θ θθ θ θθ θ θθ θ θ
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
θ θ θ θθ θ θ θθ θ θ θθ θ θ θ
θ θ θ θθ θ θ θθ θ θ θθ θ θ θ
θθθθ21 9 4 3
cot ln3 2 2
xc
x x++++− = − +− = − +− = − +− = − +θθθθ
Example(3): Find 5 4cos
dxx++++∫∫∫∫
Solution:
Mathematics For Engineering
54
(((( ))))
(((( ))))
22 2
2
2
2 2 22
2
1 12
tan2
1cos cos sin ,
2 2 1
2
1
2 25 4cos 1 5(1 ) 4(1 )
(1 ) 5 41
tan2 2 2 2tan tan3 3 3 39
xput u
x x ux
u
dudx
u
dx du dux u u u
uu
xdu u
cu
− −− −− −− −
====
−−−−∴ = − =∴ = − =∴ = − =∴ = − =++++
====++++
= == == == =++++ −−−− + + −+ + −+ + −+ + −+ ++ ++ ++ + ++++
= = = += = = += = = += = = + ++++
∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫
∫∫∫∫
Example(3): Find 12 13sin
dxx++++∫∫∫∫
Solution:
2 2
222
2 2
2 2tan sin 2sin cos ,
2 2 2 1 1
2 212 13sin 2 12(1 ) 13(2 )1 12 13
1
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
u u
dx du dux u u uu
u
du du duu uu u u u
du duu
u u
= ∴ = = == ∴ = = == ∴ = = == ∴ = = =+ ++ ++ ++ +
= == == == =++++ + ++ ++ ++ + + ++ ++ ++ + ++++
= = == = == = == = =+ ++ ++ ++ + + + + ++ + + ++ + + ++ + + +
− −− −− −− −= + = + += + = + += + = + += + = + ++ ++ ++ ++ +
∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫
∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫
∫ ∫∫ ∫∫ ∫∫ ∫ ln 3 25
1 1ln 2tan 3 ln 3tan 2
15 2 15 2
u
x xc
++++
−−−−= + + + += + + + += + + + += + + + +
Indefinite Integration
55
Exercise(6)
Prove that
2 3/ 2 2
2 22
2 2
22 2 2
2 2 2
21
2 2 3/ 2 2 2
2 2 2
2 22 2
1(1)
(4 ) 4 4
25 5 25(2) 5ln 25
1(3)
(4) 4 4 2ln( 4)2
(5) sin( ) ( )
(6) 4 4 2ln( 4)2
(7)
xdx c
x x
x xdx x c
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
x
−−−−
= += += += +−−−− −−−−
− − −− − −− − −− − −= + − += + − += + − += + − +
−−−−= − += − += − += − +−−−−
+ = + + + + ++ = + + + + ++ = + + + + ++ = + + + + +
= − += − += − += − +− −− −− −− −
− = − + + − +− = − + + − +− = − + + − +− = − + + − +
++++ = += += += +
∫∫∫∫
∫∫∫∫
∫∫∫∫
∫∫∫∫
∫∫∫∫
∫∫∫∫
∫∫∫∫2 2
2 2
2 2
2 5/ 2 2 3/ 2
2 2 3/ 2 2 2 2
2
2 2
ln2
(8)(4 ) 12(4 )
(9)( ) ( )
9(10)
99
a x a ac
x a a
x dx xc
x x
dx xc
a x a a x
dx xc
xx x
+ −+ −+ −+ −+ ++ ++ ++ ++ ++ ++ ++ +
= += += += +− −− −− −− −
= += += += +++++ ++++
−−−−= − += − += − += − +−−−−
∫∫∫∫
∫∫∫∫
∫∫∫∫
22 2
2
22 2 2 2 2 5/ 2 2 2 3/ 2
1(11) 16 8ln 16
216
1(12) ( ) ( )
5 3
x dxx x x x c
x
ax a x dx a x a x c
= − + + − += − + + − += − + + − += − + + − + −−−−
− = − − − +− = − − − +− = − − − +− = − − − +
∫∫∫∫
∫∫∫∫
Mathematics For Engineering
56
22
2 3/ 2 2
(13) ln( 2 4 13)4 13
2(14)
(4 ) 4 4
dxdx x x x c
x xdx x
cx x x x
= − + − + += − + − + += − + − + += − + − + +− +− +− +− +
−−−−= += += += +−−−− −−−−
∫∫∫∫
∫∫∫∫
12 2 2
1 2 1 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
x x dx x x x x c
x x dx x x x x c
−−−−
− −− −− −− −
− −− −− −− −
= + += + += + += + ++ ++ ++ ++ +
= − + − += − + − += − + − += − + − +
= − − − += − − − += − − − += − − − +
∫∫∫∫
∫∫∫∫
∫∫∫∫
sech 12 2 2
12 2 2
2
2
22
2
2 1 2
1(18)
1(19) tan
1 3 5 5(20) ln
5 33 5
1 2 3(21) 2 3 ln
22 3
2 1(22) 8 2 2 2 sin ( ) ( 2) 8 2
2 2
(23)2 sin
dx bxc
a ax a b xdx bx
cab aa b x
dx xc
xx x
xdx x x xx x c
x x
xx x dx x x x c
dxx
−−−−
−−−−
−−−−
−−−−= += += += +−−−−
= += += += +++++
+ −+ −+ −+ − = += += += + ++++
− + − −− + − −− + − −− + − − = − − + += − − + += − − + += − − + + − −− −− −− −
−−−−− = + − − +− = + − − +− = + − − +− = + − − +
====++++
∫∫∫∫
∫∫∫∫
∫∫∫∫
∫∫∫∫
∫∫∫∫
∫∫∫∫12 1
tan (1 2tan )23 3
x−−−− ++++
find
2
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
++++ −−−−
+ ++ ++ ++ +− −− −− −− −
∫ ∫∫ ∫∫ ∫∫ ∫
∫ ∫∫ ∫∫ ∫∫ ∫
Indefinite Integration
57
3
2 3/ 24 2(28) (29)
(9 4 )1(1 cos )
(30) (31)3 2tan (1 sin )
dx x dx
xx xdx x dx
dxx x
++++++++−−−−
− +− +− +− +
∫ ∫∫ ∫∫ ∫∫ ∫
∫ ∫∫ ∫∫ ∫∫ ∫
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
dx dxx x x x
dxx
+ −+ −+ −+ −++++
+ − ++ − ++ − ++ − +
+ −+ −+ −+ −
+ −+ −+ −+ −
+ −+ −+ −+ −
++++
∫ ∫∫ ∫∫ ∫∫ ∫
∫ ∫∫ ∫∫ ∫∫ ∫
∫ ∫∫ ∫∫ ∫∫ ∫
∫ ∫∫ ∫∫ ∫∫ ∫
∫ ∫∫ ∫∫ ∫∫ ∫
∫∫∫∫ )5 3cos
(44) (45)4 5sin 5 3sin
(46) (47)4 5cosh 5 3cosh
(48) (49)4 5sinh 5 3sinh
dxx
dx dxx x
dx dxx x
dx dxx x
−−−−
+ −+ −+ −+ −
+ −+ −+ −+ −
+ −+ −+ −+ −
∫∫∫∫
∫ ∫∫ ∫∫ ∫∫ ∫
∫ ∫∫ ∫∫ ∫∫ ∫
∫ ∫∫ ∫∫ ∫∫ ∫
Mathematics For Engineering
58
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
2 21 1
x xx
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
nAA A
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
Indefinite Integration
59
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
dx
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 ( )
1 1,
4 4
A B
A B
= += += += +
= = −= = −= = −= = −
Mathematics For Engineering
60
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
4B B
−−−−= − → == − → == − → == − → =
substitute in (1) the value of ,A B we have:
2
2
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
x
−−−− = + = −= + = −= + = −= + = − − + − +− + − +− + − +− + − + −−−− = − = − == − = − == − = − == − = − = − + − +− + − +− + − +− + − + −−−−
−−−− = − − + = += − − + = += − − + = += − − + = + ++++
∫ ∫ ∫ ∫∫ ∫ ∫ ∫∫ ∫ ∫ ∫∫ ∫ ∫ ∫
(2) Find 3 2( 1)
6
x dx
x x x
+++++ −+ −+ −+ −
∫∫∫∫
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
Indefinite Integration
61
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:
3 2
3/10
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
x x
+ −+ −+ −+ −= + −= + −= + −= + −− +− +− +− ++ −+ −+ −+ −
−−−−−−−−= + − − + = += + − − + = += + − − + = += + − − + = +++++
∫ ∫ ∫ ∫∫ ∫ ∫ ∫∫ ∫ ∫ ∫∫ ∫ ∫ ∫
(3) Find 3 2(3 5)
1
x dx
x x x
++++− − +− − +− − +− − +
∫∫∫∫
Solution: First factor the denominator :
3 2 3 2 2
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
− − + = − − − = − − −− − + = − − − = − − −− − + = − − − = − − −− − + = − − − = − − −
= − − = − − + = − += − − = − − + = − += − − = − − + = − += − − = − − + = − +
Then
3 2 2(3 5)
(1)1 11 ( 1)
x A B Cx xx x x x
++++ = + += + += + += + ++ −+ −+ −+ −− − + −− − + −− − + −− − + −
and 2
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= −= −= −= − ⇒⇒⇒⇒ ==== ⇒⇒⇒⇒ ====
Mathematics For Engineering
62
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
x x x
+ −+ −+ −+ −= + += + += + += + ++ −+ −+ −+ −− − + −− − + −− − + −− − + −
++++= + − − − = − += + − − − = − += + − − − = − += + − − − = − +− − −− − −− − −− − −
∫ ∫ ∫ ∫∫ ∫ ∫ ∫∫ ∫ ∫ ∫∫ ∫ ∫ ∫
(4) Find 4 3
3 2( 1)x x x dx
x x
− − −− − −− − −− − −−−−−
∫∫∫∫
Solution: The integrand is an improper fraction. By division
4 3
3 2 3 2 2
2 2
( 1) 1 1(1)
( 1)
1(2)
( 1)( 1)
x x x x xx x
x x x x x x
x A B Cx xx x x
− − − + +− − − + +− − − + +− − − + += − = −= − = −= − = −= − = −− − −− − −− − −− − −
++++ = + += + += + += + +−−−−−−−−
2
2 2
2
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
+ − −+ − −+ − −+ − −= + += + += + += + +−−−−−−−−
4 3
3 2 2 2
22
2
( 1) 1 2 1 2( 1)( 1)
2 1 2 1 12ln 2ln 1
( 1) 2
1 12ln
2 1
x x x xdx x dx x dx
x xx x x x x
x dx x x xx x xx
xx c
x x
− − − + − −− − − + − −− − − + − −− − − + − −= − = − + += − = − + += − = − + += − = − + + −−−−− −− −− −− −
−−−−+ + + = + − − −+ + + = + − − −+ + + = + − − −+ + + = + − − − −−−−
= − + += − + += − + += − + +−−−−
∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫
∫∫∫∫
Indefinite Integration
63
(5) Find 3 2
4 2( 2)
3 2
x x x dx
x x
+ + ++ + ++ + ++ + ++ ++ ++ ++ +
∫∫∫∫
Solution: First factor the denominator : 4 2 2 23 2 ( 1)( 2)x x x x+ + = + ++ + = + ++ + = + ++ + = + + Then we write
3 2
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
3 2
3 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
x x x x
+ + + = + + + + ++ + + = + + + + ++ + + = + + + + ++ + + = + + + + +
= + + + + + + += + + + + + + += + + + + + + += + + + + + + ++ = + = + = + ++ = + = + = + ++ = + = + = + ++ = + = + = + +
= = = == = = == = = == = = =
+ + ++ + ++ + ++ + + = += += += ++ + + ++ + + ++ + + ++ + + +
3 21 2
4 2 2 2( 2) 1
tan ln 223 2 1 2
and
x x x dx x dxdx x x c
x x x x−−−−+ + ++ + ++ + ++ + + = + = + + += + = + + += + = + + += + = + + +
+ + + ++ + + ++ + + ++ + + +∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫
(6) Find 2
2 2(2 3)
( 1)
x dx
x
++++++++
∫∫∫∫
Solution: we write 2
2 2 2 2 2
2 2 3 2
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 x x
x Ax B x Cx D Ax Bx A C x B D
A B A C B D
A B C D
x
x x x
+ + ++ + ++ + ++ + += += += += ++ + ++ + ++ + ++ + +
+ = + + + + = + + + + ++ = + + + + = + + + + ++ = + + + + = + + + + ++ = + + + + = + + + + +∴ = = + = + =∴ = = + = + =∴ = = + = + =∴ = = + = + =
= = = == = = == = = == = = =
++++ = += += += ++ + ++ + ++ + ++ + +
Mathematics For Engineering
64
2
2 2 2 2 2(2 3) 2 1
( 1) ( 1) ( 1)
xdx dx dx
x x x
++++ = += += += ++ + ++ + ++ + ++ + +
∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫
22 2
22
2 2 4
1To solve tan sec ,
( 1)
1 sec 1 1cos sin2
2 4( 1) sec
dx put x u dx u dux
u dudx udu u u
x u
==== ⇒⇒⇒⇒ ====++++
= = = += = = += = = += = = +++++
∫∫∫∫
∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫
1 12 2
12
21 1
2 2 2 2 2 2
12
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 xx
xx
x xdx dx dx x x
x x x x
xx c
x
− −− −− −− −
−−−−
− −− −− −− −
−−−−
= + = += + = += + = += + = ++ ++ ++ ++ +
= += += += +++++
++++∴ = + = + +∴ = + = + +∴ = + = + +∴ = + = + ++ + + ++ + + ++ + + ++ + + +
= + += + += + += + +++++
∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫
Indefinite Integration
65
Exercise (7)
Find
1 3: ln
6 3
1 1: ln
5 6
1 2: ln ( 1)( 4)
5
1 4 12 6: 3 ln(1 )
22 (1 ) 2(1 )
2
2
2
4
3
4 3 2
3 2
(1)9
(2)7 6
(3)3 4
(4)(1 )
2 3 3(5)
2 3
xAns c
x
xAns c
x
Ans x x c
Ans x x xx x c
Ans
dx
x
dx
x xdx
x x
x dx
x
x x x xdx
x x x
−−−−++++
++++
++++++++
++++
+ − ++ − ++ − ++ − +
−−−−− − − − +− − − − +− − − − +− − − − +
−−−− − +− +− +− +
−−−− + ++ ++ ++ + − +− +− +− + −−−−
− + − +− + − +− + − +− + − +
− +− +− +− +
∫∫∫∫
∫∫∫∫
∫∫∫∫
∫∫∫∫
∫∫∫∫1 2
: ln3 22 2 3
12: ln( 1)
21
1 42 1: ln( 4) tan
22 2 4
1 12 1: ln ( 1) tan
22 2 1
3
2 2
3 2
2 2
3
2 2
4 3
2(6)
( 1)
2 4(7)
( 4)
1(8)
( 1)
8(9)
xx c
x x x
Ans x cx
xAns x c
x
xAns x x c
x
x dx
x
x x xdx
x
x xdx
x
x x
+ ++ ++ ++ +− +− +− +− +
+ + ++ + ++ + ++ + +++++
−−−−+ + + ++ + + ++ + + ++ + + +++++
−−−−+ − − ++ − − ++ − − ++ − − +++++
++++
+ ++ ++ ++ +
++++
+ −+ −+ −+ −
++++
+ −+ −+ −+ −
∫∫∫∫
∫∫∫∫
∫∫∫∫
3 23 2 2 11
: ln tan2 1 3 3( 1)
1 cosh 2: ln
cos
2
2 2
2
2 1
( )( 1)
sin(10)
cos (1 cos )
x x x xAns c
xx
xAns c
x
x xdx
x x x
x dx
x x
− + −− + −− + −− + −−−−−− + +− + +− + +− + +++++++++
++++++++
+ ++ ++ ++ + + ++ ++ ++ +
++++
∫∫∫∫
∫∫∫∫
Mathematics For Engineering
66
Miscellaneous substitution:
ve integer put( )
(i)Form
nx
dx m ax b uax b
+ + =+ + =+ + =+ + =++++∫∫∫∫
Example(1): Find 2
3 / 2(5 2 )
xdx
x−−−−∫∫∫∫
Solution: Put 5 2x u− =− =− =− = 2dx du∴ − =∴ − =∴ − =∴ − =
(((( ))))
2 2 2
3 / 2 (3 / 2) (3 / 2)
2
(3 / 2) (3 / 2) (3 / 2)
( 3 / 2) ( 1 / 2) ( 1 / 2)
( 1 / 2) (1 / 2) (3 / 2)
3
1 (5 ) 1 25 108 8(5 2 )
1 25 108
125 10
81 2
25 208 3
1 50 220 5 2 5 2
8 35 2
x u u udx du du
x u u
u udu
u u u
u u u du
u u u
x xx
− − −− − −− − −− − −
−−−−
− − − − +− − − − +− − − − +− − − − += == == == =−−−−
−−−−= − += − += − += − +
−−−− = − += − += − += − +
−−−− = − += − += − += − +
= + − − −= + − − −= + − − −= + − − − −−−−
∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫
∫∫∫∫
∫∫∫∫
c++++
ve integer put( )
(ii)Fom n
dx ax bdx m u
xx ax br +++++ =+ =+ =+ =
++++∫∫∫∫
Example(2): Find 3 / 2 5 / 2(3 5)
dxdx
x x ++++∫∫∫∫
Solution:
Put 2
23 5 5 5 5
, 3 ,5 3
x xy y dx dy dx dy x
x x yx
+ −+ −+ −+ −= + == + == + == + = ⇒⇒⇒⇒ ==== ⇒⇒⇒⇒ = ∴ == ∴ == ∴ == ∴ =− −− −− −− −
Indefinite Integration
67
2 2
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 2
2 5 / 2 2 5 / 2 5 / 2
( 1/ 2) ( 3 / 2) ( 5 / 2) (1 / 2) ( 1 / 2) ( 3 / 2)
3
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
x x x
− − − − −− − − − −− − − − −− − − − −
− − − − − +− − − − − +− − − − − +− − − − − += = == = == = == = =
− −− −− −− − = − + = + += − + = + += − + = + += − + = + +
++++−−−− = + − += + − += + − += + − + + ++ ++ ++ +
∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫
∫∫∫∫
ve integer( )
1put ln ln
(iii)Forn
n
dxdx m
x ax b
dx dux n x u n
u x u
++++++++
= → = − → = −= → = − → = −= → = − → = −= → = − → = −
∫∫∫∫
Example(3): Find 3( 2)
dxdx
x x ++++∫∫∫∫
Solution: 3
3 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
u
= ∴ = − = −= ∴ = − = −= ∴ = − = −= ∴ = − = −
−−−−= − = − + = − + += − = − + = − + += − = − + = − + += − = − + = − + +++++++++ ++++
∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫
22
2
1 1,
)
(iv)For
Thequadratic Form( canbechangedto a sum or difference of two squars
dx dxax bx cax bx c
ax bx c
+ ++ ++ ++ ++ ++ ++ ++ +
+ ++ ++ ++ +
∫ ∫∫ ∫∫ ∫∫ ∫
Mathematics For Engineering
68
2
( )(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
3
2 4 3
dx
x x+ ++ ++ ++ +∫∫∫∫
Solution:
2 2 2
12
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
dxx c
x−−−−
= == == == =+ + + + + + − ++ + + + + + − ++ + + + + + − ++ + + + + + − +
= = + += = + += = + += = + ++ ++ ++ ++ +
∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫
∫∫∫∫
Note: To Complete the square of 2x ax++++ we add 2
2a
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+ = + + − = + + − = + −+ = + + − = + + − = + −+ = + + − = + + − = + −+ = + + − = + + − = + −
Example(5): Find 218 6
dx
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
x x x
−−−− ++++= = = += = = += = = += = = +− + + − +− + + − +− + + − +− + + − +
∫ ∫∫ ∫∫ ∫∫ ∫
Indefinite Integration
69
Example(6): Find 24 5
3 3
x
x x
+++++ ++ ++ ++ +
∫∫∫∫
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
+ + = ++ + = ++ + = ++ + = ++ = + ++ = + ++ = + ++ = + + ⇒⇒⇒⇒ = + == + == + == + =
Q
2 2
2 2
22
2 2
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
1
A B x x
x xdx dx
x x x xx
dx dxx x x x
xNow dx x x
x x
And dx dxx x x x
∴ = =∴ = =∴ = =∴ = = ⇒⇒⇒⇒ + = + ++ = + ++ = + ++ = + +
+ + ++ + ++ + ++ + +====+ + + ++ + + ++ + + ++ + + +
++++= += += += ++ + + ++ + + ++ + + ++ + + +
++++ = + += + += + += + ++ ++ ++ ++ +
====+ + + + ++ + + + ++ + + + ++ + + + +
====
∫ ∫∫ ∫∫ ∫∫ ∫
∫ ∫∫ ∫∫ ∫∫ ∫
∫∫∫∫
∫ ∫∫ ∫∫ ∫∫ ∫
2
1
1
2 12
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
dxx
x
x
From in
x xdx x x c
x x
−−−−
−−−−
−−−−
+ ++ ++ ++ +
++++====
++++====
+ ++ ++ ++ +∴ = + + + +∴ = + + + +∴ = + + + +∴ = + + + ++ ++ ++ ++ +
∫∫∫∫
∫∫∫∫
Example(7): Find 2
2 3
3 6 10
x
x x
−−−−
+ ++ ++ ++ +∫∫∫∫
Mathematics For Engineering
70
Solution:
2
2 2
(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
+ + = ++ + = ++ + = ++ + = +− = + +− = + +− = + +− = + + ⇒⇒⇒⇒ = − + == − + == − + == − + =
−−−−∴ = = − − =∴ = = − − =∴ = = − − =∴ = = − − = ⇒⇒⇒⇒ − = − + +− = − + +− = − + +− = − + +
− − + +− − + +− − + +− − + +====+ + + ++ + + ++ + + ++ + + +
∫ ∫∫ ∫∫ ∫∫ ∫
Q
2 2
22
2 2 2
1
2
22
1 (6 6)5 (1)
2 3 6 10 3 6 101 (6 6)
3 6 10 (2)2 3 6 10
5 55
3 10 3 73 6 10 2 2 13 3
5 5 3( 1)sinh (3)
3 7 3 7( 1)
3
(4 5)(2),(3) (1) 3 6
3 3
x dx dx
x x x xx dx
Now x xx xdx dx dx
Andx x x x x x
dx xc
x
xFrom in dx x
x x
−−−−
− +− +− +− += += += += ++ + + ++ + + ++ + + ++ + + +
− +− +− +− + = − + += − + += − + += − + ++ ++ ++ ++ +
= == == == =+ ++ ++ ++ + + + + + ++ + + + ++ + + + ++ + + + +
++++= = += = += = += = ++ ++ ++ ++ +
++++∴ = − +∴ = − +∴ = − +∴ = − ++ ++ ++ ++ +
∫ ∫∫ ∫∫ ∫∫ ∫
∫∫∫∫
∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫
∫∫∫∫
∫∫∫∫15 3( 1)
10 sinh3 7
xx c−−−− +++++ + ++ + ++ + ++ + +
2
1
( ) ( )
y =G(x)2 y =G(x)
1=F(x)
y
(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
Indefinite Integration
71
Example(8): Find (4 1) 3 1
dx
x x+ ++ ++ ++ +∫∫∫∫ case ( )a
Solution: 2 21
3 1 4 2 , ( 1)3
put x y dx ydy x y+ = ∴ = = −+ = ∴ = = −+ = ∴ = = −+ = ∴ = = −
22 2
1 1
(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− −− −− −− −
= = == = == = == = = + ++ ++ ++ + −−−−− +− +− +− + −−−−
− + −− + −− + −− + −= − = − + = += − = − + = += − = − + = += − = − + = ++ ++ ++ ++ +
∫ ∫ ∫ ∫∫ ∫ ∫ ∫∫ ∫ ∫ ∫∫ ∫ ∫ ∫
22 2
1 1
(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
dx
x x− −− −− −− −∫∫∫∫ case ( )b
Solution: 2 2
2 2 4 2
12 5 2 2 , ( 5)
2
(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 1
1 1
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
y yc
x xc
− −− −− −− −
− −− −− −− −
= = −= = −= = −= = −+ + + ++ + + ++ + + ++ + + +
= − += − += − += − +
− −− −− −− −= − += − += − += − +
∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫
Example(10): Find 2( 1) 4 2
dx
x x x+ + ++ + ++ + ++ + +∫∫∫∫ (case ( )c )
Mathematics For Engineering
72
Solution:
(((( )))) (((( ))))
1 2 1
2
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
dx
x x+ −+ −+ −+ −∫∫∫∫ case (d)
Solution: 2
22 2 2
1 4 1,
11 (1 )
x xdx yput y dy x
yx x
− − −− − −− − −− − −= ∴ = == ∴ = == ∴ = == ∴ = =+++++ ++ ++ ++ +
2 2 2
2 2 2 2 2
2
1
2
21
2
(1 ) (1 )
(1 ) 1 ( 4 )(1 ) 1 ( 4 ) 1
11
11 1 14 21 1 1 2 2 2
11 1
1 1sin (2 1)
2 2 1 1 2 2( )
4 2
1 1 3sin
2 2 1
dx x dy x dy
x x x x x x x
ydy
y dy dy
y y y y y yy y
dyy c
y
xc
x
−−−−
−−−−
+ ++ ++ ++ +∴ = = =∴ = = =∴ = = =∴ = = =+ − − + − − −+ − − + − − −+ − − + − − −+ − − + − − −
−−−−++++ ++++− − −− − −− − −− − − = = == = == = == = =− − −− − −− − −− − − −−−−−−−−+ ++ ++ ++ +
− −− −− −− −= = − += = − += = − += = − +− −− −− −− −
− −− −− −− −= += += += + ++++
∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫
∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫
∫∫∫∫
Indefinite Integration
73
(vii)Forn
dx
x ax b++++∫∫∫∫ put
21 2
ln 2ln ,n dx dyx n x y n
x yy
−−−−= ∴ = − == ∴ = − == ∴ = − == ∴ = − =
Example(12): Find 4 3
dx
x x ++++∫∫∫∫
Solution: 4 2
2
14 2 2
12
14ln 2 ln , 4
1 1 1sinh 3
2 2 2 33 3 1 3
1 3sinh
2 3
dx dyput x y x y
x yy
dx dy dyy
x x y y y
cx
−−−−
−−−−−−−−
−−−−
−−−−= = ∴ = − == = ∴ = − == = ∴ = − == = ∴ = − =
− − −− − −− − −− − −∴ = = =∴ = = =∴ = = =∴ = = =+ + ++ + ++ + ++ + +
−−−−= += += += +
∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫
Mathematics For Engineering
74
Exercise(7) Find
3 2 3
2 2 3/ 2
2 3 3/ 2 2 5/ 2 3/ 2
2 3 4
2 2 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
(13)8
x x xdx dx dx
x x x
dx dx dxx x x x x x
dx dx dxx x x x x x
dx dx dxx x x x x x
− + −− + −− + −− + −
− + +− + +− + +− + +
+ − −+ − −+ − −+ − −
+ − − + + −+ − − + + −+ − − + + −+ − − + + −
−−−−
∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫
∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫
∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫
∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫
2 2 2
2 2 2
2 2 2
2 2 2
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
1(25)
8 2
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
dx dx dxx x x x x x
x
− − − − −− − − − −− − − − −− − − − −+ + −+ + −+ + −+ + −
+ − − + + −+ − − + + −+ − − + + −+ − − + + −+ − −+ − −+ − −+ − −
− − − − − −− − − − − −− − − − − −− − − − − −
+ − − + + −+ − − + + −+ − − + + −+ − − + + −
− −− −− −− −
∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫
∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫
∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫
∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫
2 2 2
2 2 2
2 2 2
2
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
dx
− − − −− − − −− − − −− − − −+ + −+ + −+ + −+ + −
+ − − + + −+ − − + + −+ − − + + −+ − − + + −+ − −+ − −+ − −+ − −
− − − − − −− − − − − −− − − − − −− − − − − −
+ − − ++ − − ++ − − ++ − − + + −+ −+ −+ −
∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫
∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫
∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫
∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫
2 2 2
2 2 2 2
(38) (39)1) 5 (2 3) 3 1 (3 1) 3
(40) (41)(3 1) 2 1 ( 2) 3 1
dx dx
x x x x x x
dx dx
x x x x
+ ++ ++ ++ + + − + ++ − + ++ − + ++ − + +
+ + − −+ + − −+ + − −+ + − −
∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫
∫ ∫∫ ∫∫ ∫∫ ∫
Indefinite Integration
75
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
11
( ) ( ) ( ) ( ) ... ( ) ( )
( ) ( )
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
1
( ) ( ) ... ( )
( )
n n nn
k kk
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====
y
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76
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
1
1
inf { ( ) : [ , ]}
sup{ ( ) : [ , ]}k k k
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
11
( ; ) ( )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
11
( ; ) ( )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
y
1M
1m
0a x==== 1x 1nx −−−− x
Indefinite Integration
77
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
11
1 0 1 2 1 2 1
( ; ) ( ) ,
( ) ( ) ( ) ( ) ... ( ) ( )
n
n k k k k kk
n n n
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
( ) ( )( ) ( )
( ; ) ( ; ) ( ; )
n n n
k k k k k k k k kk k k
n
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
11
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
11
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
Mathematics For Engineering
78
1lim ( ; ) lim ( )
n
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 ( )
b n
R k kn ka
f x dx I f xξξξξ→∞→∞→∞→∞ ====
= = ∆= = ∆= = ∆= = ∆∑∑∑∑∫∫∫∫
Corollary: If ( )f x is continuous function on [ , ]a b the
1lim ( ; ) lim ( )
n
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 1n
2n
3n
4n
5n
6n
7n
8n
9n
1
Indefinite Integration
79
( )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
(((( ))))
(((( ))))
(((( ))))
1
11 1
2 2 21 1 1 1
2 2
11 2 12 2 2
2 1 4 2( ) 4 1 4 1 1
2
1 4 2( ; ) ( ) 1
1 1 4 1 14 2 2 1
4 ( 1) 1 2 ( 1. 2 1
2
k kr
k k
k n
n k k kr k
n n n n
k k k k
kx x k kn n n
k kf
n n
kS P f x x f
n n
k kn nn n n
n n n nn
n n
ξξξξ
ξ ξξ ξξ ξξ ξ
ξξξξ
−−−−
−−−−= == == == =
= = = == = = == = = == = = =
−−−−++++ −−−−= = + == = + == = + == = + =
− −− −− −− − = − = − = −= − = − = −= − = − = −= − = − = −
−−−− ∴ = − = −∴ = − = −∴ = − = −∴ = − = −
= − − = − −= − − = − −= − − = − −= − − = − −
+ ++ ++ ++ + = − − == − − == − − == − − =
∑ ∑∑ ∑∑ ∑∑ ∑
∑ ∑ ∑ ∑∑ ∑ ∑ ∑∑ ∑ ∑ ∑∑ ∑ ∑ ∑
2
2
) 21
2 ( 1) 2lim ( ; ) lim 1 2 1 1n
n n
nn
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−−−−= + = + ≤ ≤= + = + ≤ ≤= + = + ≤ ≤= + = + ≤ ≤
16
( ) , 1,2, ...,k kx x k nn−−−−− = =− = =− = =− = =
Mathematics For Engineering
80
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 −−−− −−−−++++ −−−−= = + + + = += = + + + = += = + + + = += = + + + = +
(((( ))))
(((( ))))
11 1
2 21
2 21 1 1
2 2
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
2
k k
k n
n k k kr k
n
k
n n n
k k k
k kf
n n n
kS P f x x f
n n n
kn n n
kn n n
nn 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
2
2
(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−−−−− −− −− −− −∆ = − = = =∆ = − = = =∆ = − = = =∆ = − = = =
Indefinite Integration
81
and
(((( ))))0 1 2
1
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−−−−
= = + = += = + = += = + = += = + = +
−−−−= + = + = + == + = + = + == + = + = + == + = + = + =
let
1
2
2
2 2 2
22 2 2
8 42
2
8 4( ) 4 2 3
64 32 16 16 644 4 3
256 64 128 64 25619
k kr
k
x x kn n
kf
n n
k k kn nn n n
k k kn nn n n
ξξξξ
ξξξξ
−−−− ++++= = + −= = + −= = + −= = + −
= + − += + − += + − += + − +
= + + + − − += + + + − − += + + + − − += + + + − − +
= + + + − −= + + + − −= + + + − −= + + + − −
11 1
23 3 2 2 3
1
8( ; ) ( ) ( ) ( )
152 2048 512 1024 512 2048
n n
n k k k kk k
n
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
1 26
1024 1 512 2048 11 1
2 2
nn n n n
n n n n
= + + + += + + + += + + + += + + + +
+ + − − ++ + − − ++ + − − ++ + − − +
1x 2x 3x 4x 5x 6x 7x 8x 9x 10x 0x
2 10
Mathematics For Engineering
82
where
(((( )))) (((( )))) (((( ))))
(((( ))))
2
1 12
3
1
1 1(1) 1 , (2) 1 2 1
2 6
1(3) 1
2
n n
n
r n n r n n n
r n n
= + = + += + = + += + = + += + = + +
= += += += +
∑ ∑∑ ∑∑ ∑∑ ∑
∑∑∑∑
since
0 1 2{ , , , ..., }np x x x x====
(((( )))) (((( )))) (((( ))))0
8, 0
1024 4040lim ; lim ; 152 2 512
3 3k
k k
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 ( ; )
n n
k k k k k kn nk k
n n
m x x M x x
L P f U P f
− −− −− −− −→∞ →∞→∞ →∞→∞ →∞→∞ →∞= == == == =
→∞ →∞→∞ →∞→∞ →∞→∞ →∞
− −− −− −− −
= == == == =
∑ ∑∑ ∑∑ ∑∑ ∑
and
( )b
l ua
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
Indefinite Integration
83
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
(((( ))))
2 2
2
22
2 2
1 21 1
22 2 2
1
23 3 3
1
2
8( 1) 8( 1) 64( 1)1 1
8 641
64 2 18( ; ) ( )
8 64 128 641
512 1016 512 8
512
k
k
n n
k k kk k
n
k
n
k
k k km f
n n nk
M f kn n
k k nL P f x x m
n n
k kn n n n
k knn n n
n
−−−−= == == == =
====
====
− − −− − −− − −− − − = = − = −= = − = −= = − = −= = − = −
= = −= = −= = −= = −
− + −− + −− + −− + − = − == − == − == − =
= − + −= − + −= − + −= − + −
= − + −= − + −= − + −= − + −
====
∑ ∑∑ ∑∑ ∑∑ ∑
∑∑∑∑
∑∑∑∑
(((( )))) (((( )))) (((( ))))3 3
11016 512 81 2 1
6 2
512 512 24 488lim ( ; ) 8 (1)
3 3 3n
n nnn n n n
nn n
L P f→∞→∞→∞→∞
++++⋅ + + − ⋅ + ⋅ − ⋅⋅ + + − ⋅ + ⋅ − ⋅⋅ + + − ⋅ + ⋅ − ⋅⋅ + + − ⋅ + ⋅ − ⋅
−−−−= − = == − = == − = == − = =
Similarly 2
21
23 3
1
8 64( ; ) 1
512 8 512( 1)(2 1) 8
6
512 488lim ( ; ) 8 (2)
3 3
n
k
n
k
n
U P f kn n
nk n n
nn n
U P f
====
−−−−
→∞→∞→∞→∞
= −= −= −= −
= − = ⋅ + + −= − = ⋅ + + −= − = ⋅ + + −= − = ⋅ + + −
∴ = − =∴ = − =∴ = − =∴ = − =
∑∑∑∑
∑∑∑∑
From (1), (2) the function is integrable on v and 8
0
488( )
3f x dx ====∫∫∫∫
Mathematics For Engineering
84
:)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 ( )
b an
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
x xn
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
11
11
( ; ) ( ; ) ( )( )
( )
( )
n
k k k kk
n
k k kk
U f p L f p x x M m
b ax x w n
b a n
b ab a
εεεε εεεε
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 .
Indefinite Integration
85
:)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
(((( )))) (((( )))) (((( ))))
[[[[ ]]]]
11
1 11 1
1 0 2 1 1
( ; ) ( ; ) ( )( )
( ) ( ) ( ) ( )
( ) ( ) ( ) ( ) ... ( ) ( )
( ) ( )
n
k k k kk
n n
k k k kk k
n n
U f p L f p x x M m
b a b af x f x f x f x
n n
b af x f x f x f x f x f x
nb a
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
a b
1( )f a m==== ( )nM f b==== 1 2M m====
Mathematics For Engineering
86
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
(((( )))) (((( )))) (((( ))))
(((( ))))
2
1 12
3
1
1 11 , 1 2 1
2 6
11
2
(1) (2)
(3)
n n
n
r n n r n n n
r n n
= + = + += + = + += + = + += + = + +
= += += += +
∑ ∑∑ ∑∑ ∑∑ ∑
∑∑∑∑
find the following sums
2
1 1102
1 120 152 2
1 1
( 5) ( 3)
( 3 5) (2 5)
(2 13) ( 2 4)
(4) (5)
(6) (7)
(8) (9)
n n
k kn
k k
k k
k k
k k k
k k k
= == == == =
= == == == =
= == == == =
+ ++ ++ ++ +∑ ∑∑ ∑∑ ∑∑ ∑
+ − −+ − −+ − −+ − −∑ ∑∑ ∑∑ ∑∑ ∑
− − +− − +− − +− − +∑ ∑∑ ∑∑ ∑∑ ∑
By using Darboux’s sums find
1 12 2
0 12 1
2 2
1 01
2 2
0
( ) ( 1)
( 2)
(15) ( 1)
(10) (11)
(12) (13)
(14)b
a
x x dx x dx
x dx x dx
x dx x dx
−−−−− +− +− +− +
−−−−
++++
∫ ∫∫ ∫∫ ∫∫ ∫
∫ ∫∫ ∫∫ ∫∫ ∫
∫ ∫∫ ∫∫ ∫∫ ∫
.
By using Riemann’s sums find the integrals (10)-(15)
Indefinite Integration
87
:Properties of definite integral
Let ( ), ( )f x g x are integrable on the interval [ , ]a b and let [ , ]c a b∈∈∈∈
Then
is an odd function
is an even function0
[ ( ) ( )] ( ) ( )
( ) ( ) ,
( ) ( )
( ) ( ) ( )
0 ( )
( )2 ( ) ( )
(
(1)
(2)
(3)
(4)
(5)
(6)
b b b
a a ab b
a ab a
a bb c b
a a c
aa
a
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 f x dx f x dx
if f x
f x dxf x dx if f x
f
λ λ λλ λ λλ λ λλ λ λ
−−−−
± = ±± = ±± = ±± = ±
= ∈= ∈= ∈= ∈
= −= −= −= −
= += += += +
====
∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫
∫ ∫∫ ∫∫ ∫∫ ∫
∫ ∫∫ ∫∫ ∫∫ ∫
∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫
∫∫∫∫∫∫∫∫
�
0 0
) ( )
( ) ( )(8)
b b
a a
a a
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= −= −= −= −∫∫∫∫ .
Mathematics For Engineering
88
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
1
( )
( ) ( ) ( )
( ) ( ) ( )
b b b
a a ab
ab
a
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( ) ( )
b
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
aa
F f x x a b= ∀ ∈= ∀ ∈= ∀ ∈= ∀ ∈∫∫∫∫ Then aF
differentiable at any point [ , ]c a b∈∈∈∈ at which f continuous ,and ( ) ( )aF c f c′′′′ ====
Indefinite Integration
89
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
c h
c
dxh
++++====∫∫∫∫
(((( ))))
( ) ( ) 1 1( ) ( ) ( )
1 1( ) ( )
c h c ha a
c c
c h
c
F c h F cf c f x dx f c dx
h h h
f x f c dx hh h
ε εε εε εε ε
+ ++ ++ ++ +
++++
+ −+ −+ −+ −− = −− = −− = −− = −
= − ≤ == − ≤ == − ≤ == − ≤ =
∫ ∫∫ ∫∫ ∫∫ ∫
∫∫∫∫
Since 0εεεε >>>> is arbitrary it follows that
0
( ) ( )( ) lim ( )a a
ah
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
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
Mathematics For Engineering
90
( ) ( )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
2 2
sin sin
sin sin
( 3 4) 3 4
( )
( )
( )
y
aa
y
t
a
dx dx y
dy
dx dx y
dy
dx x dx t t
dt
a
b
c
====
= −= −= −= −
+ + = + ++ + = + ++ + = + ++ + = + +
∫∫∫∫
∫∫∫∫
∫∫∫∫
:Example By using fundamental theorem of calculus find .
2
cos
( 3 5)
(i)
(ii)
b
ab
a
x dx
x x dx+ ++ ++ ++ +
∫∫∫∫
∫∫∫∫
Solution:
32 2
3 2 3 2
cos [ sin ]
cos sin sin
3( 3 5) ( 5 )
3 2
( 3 5 ) ( 3 5 )3 2 3 2
(i)
(ii)
b b
a ab
a
b b
a a
dx dx x dx
dx
x dx b a
d xx x dx x x dx
dx
b b a ab a
====
= −= −= −= −
+ + = + ++ + = + ++ + = + ++ + = + +
= + + − + += + + − + += + + − + += + + − + +
∫ ∫∫ ∫∫ ∫∫ ∫
∫∫∫∫
∫ ∫∫ ∫∫ ∫∫ ∫
Indefinite Integration
91
Exercises
Using fundamental theorem of calculus find 10 2
2
2 123
2
2 02 2
2 2 2
2
(4 3)
(4 2) sin
(cos ) (sin )
(1) (2)
(3) (4)
(5) (6)
x
x
o
x dx e dx
x dx x dx
x e dx x x dx
ππππ
π ππ ππ ππ π
ππππ−−−−
++++
++++
+ ++ ++ ++ +
∫ ∫∫ ∫∫ ∫∫ ∫
∫ ∫∫ ∫∫ ∫∫ ∫
∫ ∫∫ ∫∫ ∫∫ ∫
if
Find
2
3 4 2
2 4
1 2
2 2( ) , ( )
10 5
[ ( )] [ ( )]
(7) x x xf x g x
x x x
d df x dx g x dx
dx 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:
( )b
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
Mathematics For Engineering
92
at x b==== we define
0
( ) 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
0
( ) 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
0 0
( ) 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
20 9
dx
x−−−−∫∫∫∫
Indefinite Integration
93
Solution: The integrand is discontinuous at 2x ==== we consider
33 31 1
2 20 0 00 0 0
1
3lim lim sin lim sin
3 39 9
sin 12
dx dx x
x x
εεεεεεεε
ε ε εε ε εε ε εε ε ε
εεεε
ππππ
−−−−−−−−− −− −− −− −
→ → →→ → →→ → →→ → →
−−−−
−−−− = = == = == = == = = − −− −− −− −
= == == == =
∫ ∫∫ ∫∫ ∫∫ ∫
Example(2):
Find 2
0 2dx
x−−−−∫∫∫∫
Solution: The integrand is discontinuous at 3x ==== we consider
22 2 2
00 0 00 0 0
0
1lim lim ln 2 lim ln
2 2 2
1 1lim ln ln
2
dx dxx
x x x
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
( 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
1 4
0 0 0 00 1
1 1 1lim lim
( 1) ( 1) ( 1)
1 1 1 1 1lim lim lim 1 lim
1 1 3
x x x
x x
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
0 4
dx
x
∞∞∞∞
++++∫∫∫∫
Mathematics For Engineering
94
Solution: The upper limit of integration is infinite. We consider
1 1 12 2
0 0 0
1 1lim lim tan lim tan tan 0
2 2 2 2 44 4
bb
b b b
dx dx x b
x x
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
2 12
xe dx−∞−∞−∞−∞
====∫∫∫∫
Example(6):
Find x x
dx
e e
∞∞∞∞
−−−−−∞−∞−∞−∞ ++++∫∫∫∫
Solution: Both limit of integration is infinite. We consider
12 2
1 1
lim lim tan1 1
lim tan tan 02 2
x xa axx x x x aa aa
a a
a
dx e dx e dxe
e e e e
e eπ ππ ππ ππ π
∞ ∞∞ ∞∞ ∞∞ ∞−−−−
−−−− −−−−→∞ →∞→∞ →∞→∞ →∞→∞ →∞−∞ −∞ −−∞ −∞ −−∞ −∞ −−∞ −∞ −
− − −− − −− − −− − −→∞→∞→∞→∞
= = == = == = == = = + + ++ + ++ + ++ + +
= − = − == − = − == − = − == − = − =
∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫
Hence 2x x
dx
e e
ππππ∞∞∞∞
−−−−−∞−∞−∞−∞
====++++
∫∫∫∫ .
Indefinite Integration
95
Exercise(9)
Evaluate the following integrals:
(((( ))))
2 1
1 0
2
2 1
2 22 0
2
22
3 21
1ln
ln
1 1
4 1
11 4
4
(1) (2)
(3) (4)
(5) (6)
(7) (8)
(9) (10)
o
dx x dxx x
dx dx
x x x
dx xdx
x x
x dxdx
xx
x dxdx
x x
∞ ∞∞ ∞∞ ∞∞ ∞
−∞−∞−∞−∞
−−−−∞∞∞∞
−∞ −−∞ −−∞ −−∞ −∞ ∞∞ ∞∞ ∞∞ ∞
−∞−∞−∞−∞
++++ ++++
− −− −− −− −
−−−−++++
−−−−
∫ ∫∫ ∫∫ ∫∫ ∫
∫ ∫∫ ∫∫ ∫∫ ∫
∫ ∫∫ ∫∫ ∫∫ ∫
∫ ∫∫ ∫∫ ∫∫ ∫
∫ ∫∫ ∫∫ ∫∫ ∫
Evaluation of some definite integrals: Example(1):
Find 2
0sinm x dx
ππππ
∫∫∫∫
Solution:
2 2 1
0 0sin sin sinm mx dx x x dx
π ππ ππ ππ π
−−−−====∫ ∫∫ ∫∫ ∫∫ ∫ use integration by parts
let 1
2
21 2 220 0
sin sin
( 1)sin cos cos
cos sin ( 1)sin cos
m
m
m mm
u x dv x dx
du m x x dx v x
I x x m x x dx
ππππππππ
−−−−
−−−−
− −− −− −− −
= == == == =
= − = −= − = −= − = −= − = −
∴ = − + −∴ = − + −∴ = − + −∴ = − + − ∫∫∫∫
Mathematics For Engineering
96
since
if is even then
2 2 2
0
2 22
0 0
2 2
0
2
0
20
0
2
0
0 ( 1) sin (1 sin )
( 1) sin ( 1) sin
( 1) ( 1) sin
1
( 1).( 3)( 5)... 3.1.( 1).( 2)... 4.2
2
(sin
m
m m
mm m
m m
m
mm
m x x dx
m xdx m x dx
I m I m xdx
mI I
mm m m
m I Im m m
I dx
I x dx
ππππ
π ππ ππ ππ π
ππππ
ππππ
ππππ
ππππ
−−−−
−−−−
−−−−
−−−−
= + − −= + − −= + − −= + − −
= − − −= − − −= − − −= − − −
+ − = −+ − = −+ − = −+ − = −
−−−−∴ =∴ =∴ =∴ =
− − −− − −− − −− − −====− −− −− −− −
= == == == =
= == == == =∴∴∴∴
∫∫∫∫
∫ ∫∫ ∫∫ ∫∫ ∫
∫∫∫∫
∫∫∫∫
∫∫∫∫
[[[[ ]]]]since
is even positive integer
if is odd then 1
22
1 00
1).( 3)( 5)... 3.1,
.( 1).( 2)... 4.2 2
( 1).( 3)( 5)... 4.2.( 1).( 2)... 5.3
sin cos 1
m
m m mm m m
m
m m mm I I
m m m
I dx x
ππππ ππππ
ππππ− − −− − −− − −− − −− −− −− −− −
− − −− − −− − −− − −====− −− −− −− −
= = − == = − == = − == = − =∫∫∫∫
is odd positive integer
2
0
( 1).( 3)( 5)... 4.2sin
.( 1).( 2)... 5.3m
mm m m
I x dxm m m
m
ππππ− − −− − −− − −− − −= == == == =∴∴∴∴ − −− −− −− −∫∫∫∫
Indefinite Integration
97
2 26 76 7
0 0
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:
2 2 1
0 0cos cos cosm mx dx x x dx
π ππ ππ ππ π
−−−−====∫ ∫∫ ∫∫ ∫∫ ∫ use integration by parts
let 1
2
21 2 220 0
2 2 2
0
2 22
0 0
cos cos
( 1)cos sin sin
cos sin ( 1)cos sin
0 ( 1) cos (1 cos )
( 1) cos ( 1) cos
m
m
m mm
m
m m
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
π ππ ππ ππ π
−−−−
−−−−
− −− −− −− −
−−−−
−−−−
= == == == =
= − − == − − == − − == − − =
∴ = + −∴ = + −∴ = + −∴ = + −
= + − −= + − −= + − −= + − −
= − − −= − − −= − − −= − − −
∫∫∫∫
∫∫∫∫
∫ ∫∫ ∫∫ ∫∫ ∫
if is even then
2 2
0
2
0
( 1) ( 1) cos
1
( 1).( 3)( 5)... 3.1.( 1).( 2)... 4.2
mm m
m m
m
I m I m xdx
mI I
mm m m
m I Im m m
ππππ
−−−−
−−−−
+ − = −+ − = −+ − = −+ − = −
−−−−∴ =∴ =∴ =∴ =
− − −− − −− − −− − −====− −− −− −− −
∫∫∫∫
Mathematics For Engineering
98
since2
00 2
I dx
ππππππππ= == == == =∫∫∫∫
is even positive integer
2
0
( 1).( 3)( 5)... 3.1cos ,
.( 1).( 2)... 4.2 2m
mm m m
I x dxm m m
m
ππππππππ− − −− − −− − −− − −= == == == =∴∴∴∴ − −− −− −− −∫∫∫∫
[[[[ ]]]]since
if is odd then 1
22
1 00
( 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
2
0
( 1).( 3)( 5)... 4.2cos
.( 1).( 2)... 5.3m
mm m m
I x dxm m m
m
ππππ− − −− − −− − −− − −= == == == =∴∴∴∴ − −− −− −− −∫∫∫∫
2 26 76 7
0 0
5.3.1 6.4.2cos . , cos
6.4.2 2 7.5.3I x dx I x dx
π ππ ππ ππ πππππ= = = == = = == = = == = = =∫ ∫∫ ∫∫ ∫∫ ∫
2
0
221 1 1 2
00
sin .cos
1 1sin .cos sin .cos
m n
m n m n
x x dx
nx x x x dx
m n m n
ππππ
ππππππππ+ − + −+ − + −+ − + −+ − + −−−−− = += += += + + ++ ++ ++ +
∫∫∫∫
∫∫∫∫
Indefinite Integration
99
2 1 2
0
2
0
221 1 2
00
2 2
0
10 sin .cos (1)
sin .cos
1 1sin .cos sin .cos
10 sin .cos (2)
m n
m n
m n m n
m n
nx x dx
m n
and x x dx
mx x x x dx
m n m n
mx x dx
m n
ππππ
ππππ
ππππππππ
ππππ
+ −+ −+ −+ −
− + −− + −− + −− + −
−−−−
−−−−= += += += +++++
− −− −− −− − = += += += + + ++ ++ ++ +
−−−−= += += += +++++
∫∫∫∫
∫∫∫∫
∫∫∫∫
∫∫∫∫
applying the formula (2) in the R.H.S. of (1) the formula (1) will reduce to
2 2 2 2
0 0
( 1)( 1)sin .cos sin .cos (1)
( )( 2)m n m nn m
x x dx x x dxm n m n
π ππ ππ ππ π
− −− −− −− −− −− −− −− −====+ + −+ + −+ + −+ + −∫ ∫∫ ∫∫ ∫∫ ∫
and by repeating of the formula we get
2
0
2 4 4
0
sin .cos
( 1)( 1) ( 3)( 3). sin .cos
( )( 2) ( 4)( 6)
m n
m n
x x dx
n m n mx x dx
m n m n m n m n
ππππ
ππππ
− −− −− −− −− − − −− − − −− − − −− − − −====+ + − + − + −+ + − + − + −+ + − + − + −+ + − + − + −
∫∫∫∫
∫∫∫∫
if m and n are both even
2 2
0 0
( 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
π ππ ππ ππ π
ππππ
− − − −− − − −− − − −− − − −====+ + − + −+ + − + −+ + − + −+ + − + −
− − − −− − − −− − − −− − − −====+ + − + −+ + − + −+ + − + −+ + − + −
∫ ∫∫ ∫∫ ∫∫ ∫
Mathematics For Engineering
100
Solved examples:
2 6 4
0
5.3.1.3.1sin .cos .
10.8.6.4.2 2 2x x dx
ππππππππ====∫∫∫∫
2 8
0
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.
2 2
0 0
( 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:
2 5 3
0
4.2.2 1 1sin .cos .
8.6.4 2 24x x dx
ππππ
= == == == =∫∫∫∫
2 7
0
6.4.2.1 16cos ( 8, 0)
7.5.3.1 35x dx m n
ππππ
= = = == = = == = = == = = =∫∫∫∫
if m is odd and n is even.
2 2
0 0
( 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:
2 5 4
0
4.2.3.1 8sin .cos .1
9.7.5.3 315x x dx
ππππ
= == == == =∫∫∫∫
Indefinite Integration
101
if m is even and n is odd.
2 2
0 0
( 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:
2 4 7
0
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
1
0( ) (1)
(1) ( 1) ( ) !
1(2) ( )
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
2
0
1 1( ) ( )
2 2sin .cos2
2 ( )2
m n
m n
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
0 2
9 5 31 7 12 8 2 2 2 2 2 2
0
( ) ( ) . . . . 3sin .cos
1.5.4.3.2.1 5122 ( )
( ) ( ) . . . . 35sin
2 (5) 2.4.3.2.1 256
(1)
(2)
m nx x dx
x dx
ππππ
ππππ
π ππ ππ ππ π ππππ
π ππ ππ ππ π ππππ
+ ++ ++ ++ +
Γ ΓΓ ΓΓ ΓΓ Γ= = == = == = == = =
ΓΓΓΓ
Γ ΓΓ ΓΓ ΓΓ Γ= = == = == = == = =
ΓΓΓΓ
∫∫∫∫
∫∫∫∫
Mathematics For Engineering
102
2 5 3
0
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
2 4 7
0
(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
(3)
(4)
(5)
(6)
x x dx
x dx
x x dx
x x d
ππππ
ππππ
ππππ
ππππ
ππππππππ
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)
Indefinite Integration
103
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
( )n
k kk
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
x b==== x a====
( )y f x====
y
1kx x −−−−==== kx x====
1( )kf x −−−−
Mathematics For Engineering
104
(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
1( )
2
b
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
1
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:
3 32
1 1
26( )
3A f x dx x dx= = == = == = == = =∫ ∫∫ ∫∫ ∫∫ ∫
Example(2): Find the area above the x-axis and under the parabola 24y x x= −= −= −= −
Indefinite Integration
105
Solution: 3 3
2
1 1
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
2 2
0
646 2
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
0 0 0
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= = + = + + == = + = + + == = + = + + == = + = + + =∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫
Mathematics For Engineering
106
Exercise(11)
(1) Find the area bounded as follows: 2
2
2
2
2
, 0, 2, 6 [ . : 39]
224 , 0, 1, 3 [ . : ]
3
1 , 10 [ . : 36]
1259 , 3 [ . : ]
6cos , sin [ . : ]
2cos cos 2 1, 2sin sin 2 [ . : 6 ]
9 (3 )
(1)
(2)
(3)
(4)
(5)(6)
(7)
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
2
32 2 2 2
2
3 3 3
8 3[ . : ]
5
2( ) [ . : ]
38
4 2 2 [ . : ]3
3( sin ), (1 cos ) [ . : ]
2
(8)
(9)
(10)
(11)
aAns
ay x a x Ans
y x from x to x Ans
ax a t t y a t Ans
x y a
= −= −= −= −
= − = − == − = − == − = − == − = − =
= − = −= − = −= − = −= − = −
+ =+ =+ =+ =
(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
Indefinite Integration
107
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= == == == =
Then
[[[[ ]]]]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
where
2 21 1 1
11
1 1
( ) ( ) ( )( ) ( )
( )( ) ( ) 2 ( )
2
lim lim 2 ( ) 2 ( )
k k k k k k k k k k
k kk k k k k k k k
bn n
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==== ∫∫∫∫
Mathematics For Engineering
108
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
4 4
256(2 ) 2 (2 )
8 15y
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
2 3
02
2
02
3 2
0
(1 cos ) (1 cos ) (1 cos )
(1 3cos 3cos cos )
5 33cos cos 2 (1 sin )cos
2 2
5 3 13sin sin 2 (sin sin ) 5
2 4 3
t t t
t t tt
tt
t
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
ππππ
ππππ
π π ππ π ππ π ππ π π
ππππ
ππππ
π ππ ππ ππ π
= = == = == = == = =
= = == = == = == = =====
========
====
= = − − = −= = − − = −= = − − = −= = − − = −
= − + − == − + − == − + − == − + − =
= − + − −= − + − −= − + − −= − + − −
= − + − − == − + − − == − + − − == − + − − =
∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫
∫∫∫∫
∫∫∫∫
Indefinite Integration
109
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
0 02
2
02
2 2 3 3
0
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
t t
t tt
t
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
π ππ ππ ππ π
ππππ
ππππ
π ππ ππ ππ π
ππππ
π ππ ππ ππ π
= == == == =
= == == == =====
====
= = − − −= = − − −= = − − −= = − − −
= − + − + −= − + − + −= − + − + −= − + − + −
= − + + + + + + == − + + + + + + == − + + + + + + == − + + + + + + =
∫ ∫∫ ∫∫ ∫∫ ∫
∫∫∫∫
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= = = = −= = = = −= = = = −= = = = −
2 2
2 4 2
2 2
25616, 0, 8; [ . ]
34
(1 ); [ . ]35
4 9 36; [ . 16 ]
(2)
(3)
(4)
x y y x x axis Ans
y x x x axis Ans
x y x axis Ans
ππππ
ππππ
ππππ
− = = = −− = = = −− = = = −− = = = −
= − −= − −= − −= − −
+ = −+ = −+ = −+ = −
Find the volume generated by revolving the the given plane area about the given line, using the shell method.
2
3
2
2 , 0, 0, 5; [ .625 ]
320, 0, 2; 8 [ . ]
75
5 6, 0; [ . ]6
(5)
(6)
(7)
y x y x x y axis Ans
y x y x y Ans
y x x y y axis Ans
ππππππππ
ππππ
= = = = −= = = = −= = = = −= = = = −
= = = == = = == = = == = = =
= − + = −= − + = −= − + = −= − + = −
Mathematics For Engineering
110
Find the volume generated by revolving the the given plane area about the given line, using any appropriate method.
2
2
1sin 2 ; [ . ]
4
sin , 1 cos ; [ .5 ]
642cos cos 2 1, 2sin sin 2 ; [ . ]
3
(8)
(9)
(10)
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
k
yP P x x y y x y x
x− − −− − −− − −− − − ∆∆∆∆= − + − = ∆ + ∆ = + ∆= − + − = ∆ + ∆ = + ∆= − + − = ∆ + ∆ = + ∆= − + − = ∆ + ∆ = + ∆ ∆∆∆∆
2 2
1lim 1 1
bnk
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
2b
a
dx dyAB L dt
dt dt ∴ = = +∴ = = +∴ = = +∴ = = +
∫∫∫∫
if the equation of the curve given in the polar form ( )r f θθθθ==== then
2
1
22 dr
AB L r dd
θθθθ
θθθθθθθθ
θθθθ ∴ = = +∴ = = +∴ = = +∴ = = +
∫∫∫∫
Example(1):
Find the length of the arc of the curve 32y x==== from 0x ==== to 5x ==== .
Indefinite Integration
111
Solution:
(((( ))))
12
2
5352
0 0
32
9 11 1 4 9
4 2
1 1 2 1 3354 9 . . 4 9
2 2 3 9 27
dyx
dx
y x x
L x dx x
====
′′′′+ = + = ++ = + = ++ = + = ++ = + = +
= + = + == + = + == + = + == + = + =
∫∫∫∫
Example(2):
Find the length of the arc of the curve 3
123x y
−−−−==== from 0y ==== to 4y ==== .
Solution:
(((( ))))
12
2
4342
0 0
92
81 11 1 4 81
4 2
1 1 2 1 84 81 . . 4 81 (82 82 1)
2 2 3 81 234
dxy
dy
dxy y
dy
L y dy x
====
+ = + = ++ = + = ++ = + = ++ = + = +
= + = + = −= + = + = −= + = + = −= + = + = −
∫∫∫∫
Example(3): Find the length of the arc of the curve 424 48xy x= += += += + from 2x ==== to
4x ==== . Solution:
4
2
2 4 2 4 4 2
4 4 4
4 4 2 4 8 42 2
16
8
( 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
−−−−====
− −− −− −− − ∴ + = + = +∴ + = + = +∴ + = + = +∴ + = + = +
= + − = + − += + − = + − += + − = + − += + − = + − +
Mathematics For Engineering
112
8 4 42 2
44 44 2 2 3
22 2 2
1 132 256 ( 16)
8 8
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
2 22 4 2
4 4 42 2 2 3/ 2
00 0
2 , 3
4 9 4 9
1 2 14 9 18 4 9 . (4 9 )
18 3 18
8(37 37 1)
27
dx dyt t
dt dt
dx dyt t t t
dt dt
L t t dt t t dt t
units
= == == == =
∴ + = + = +∴ + = + = +∴ + = + = +∴ + = + = +
= + = + = += + = + = += + = + = += + = + = +
= −= −= −= −
∫ ∫∫ ∫∫ ∫∫ ∫
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
2
22
0 0
1 cos , sin
(1 cos ) sin 1 2cos cos sin
2(1 cos 4sin 2sin2 2
2sin 2 2cos 82 2
dx dyd d
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
θ θθ θθ θθ θθθθθ
= − == − == − == − =
∴ + = − + = − + +∴ + = − + = − + +∴ + = − + = − + +∴ + = − + = − + +
= − = == − = == − = == − = =
= = − == = − == = − == = − =
∫∫∫∫
Indefinite Integration
113
Example(6): Find the circumference of cardoid (1 cos )r a θθθθ= −= −= −= − Solution:
22 2 2 2 2 2 2
2 2
2 22 2 22 2 2
0 0 0 0
(1 cos )
sin
1(1 cos ) sin (2 2cos ) 2 (1 cos )
2
4 sin2
4 sin 2 sin 4 cos 82 2 2
r a
dra
d
drr a a a a
d
a
drL r a d a d a a
d
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]
units]
units]
to
to
to
to
to
from
from
from
from
from
3 2
3
2 3
2
[ .(104 13 125) / 27
[ .17 / 12
1[ .3 2 ln (2 2)
2
[ .14
8 1 8
6 3 1 2
ln 1 2 2
27 4( 2) (2,0) (11,6 3)
ln(1 ) 1/ 4
(1)
(2)
(3)
(4)
(5)
Ans
Ans
Ans
Ans
y x x x
xy x x x
y x x x
y x
y x x
−−−−
− + +− + +− + +− + +
= = == = == = == = =
= + = == + = == + = == + = =
= = == = == = == = =
= −= −= −= −
= − == − == − == − = units]
units]
units]
tofrom
1[ .ln(21 / 5)
24[ . 2( 1)
[ .16
3 / 4
cos , sin 0 4
2cos cos 2 1, 2sin sin 2
(6)(7)
t t
Ans
Ans e
Ans
x
x e t y e t t t
x yθ θ θ θθ θ θ θθ θ θ θθ θ θ θ
−−−−
−−−−
====
= = = == = = == = = == = = == + + = += + + = += + + = += + + = +
Mathematics For Engineering
114
(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:
2
2
2 2 1
2 1
b b
xa a
d
c
dyS ydL y dx
dx
dxy dy
dy
π ππ ππ ππ π
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:
2
2
2 2 1
2 1
b b
ya a
d
c
dyS xdL x dx
dx
dxx dy
dy
π ππ ππ ππ π
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:
2
1
2 2
2 2t
xAB t
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:
2
1
2 2
2 2t
yAB t
dx dyS xdL x dt
dt dtπ ππ ππ ππ π = = += = += = += = +
∫ ∫∫ ∫∫ ∫∫ ∫
Indefinite Integration
115
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
xa
S y y dxππππ ′′′′= += += += +∫∫∫∫ Now
[[[[ ]]]]
2
2
2
6 612 2 12
1236 3
12
3 31 12 1 12 12 36
y x yy yy x
yx x
xy y x x x
x x x
′ ′′ ′′ ′′ ′==== ⇒⇒⇒⇒ ==== ⇒⇒⇒⇒ = == == == =
′′′′∴ = =∴ = =∴ = =∴ = =
′′′′+ = + = + = ++ = + = + = ++ = + = + = ++ = + = + = +
22 1 2 12 36 24(2 2 1)b b
xa a
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
2 1d
yc
dxS x dy
dyππππ
= += += += +
∫∫∫∫ Now
3 2
23 4
3
1 1 9
dxx y y
dy
dxx y y
dy
==== ⇒⇒⇒⇒ ====
+ = ++ = ++ = ++ = +
13 4 3 4
02 1 9 2 1 9 (10 10 1)
27
d
yc
S y y dy y y dyπππππ ππ ππ ππ π∴ = + = + = −∴ = + = + = −∴ = + = + = −∴ = + = + = −∫ ∫∫ ∫∫ ∫∫ ∫
Mathematics For Engineering
116
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
2 1d
xc
dxS y dy
dyππππ
= += += += +
∫∫∫∫ Now
2
2
2
2 2 22 2 2
2
2 2 4 4 2
4 2ln
12ln
4
1 2 1 12
4 2
1 (1 ) 11 1 4 (1 )
4 2
1 14 1 2 2 1
2 2
y x y
x y y
dx yy
dy y y
dx yx y y
dy yy
y y y y yy y
+ =+ =+ =+ =
= −= −= −= −
−−−−= − == − == − == − =
−−−−+ = + = + −+ = + = + −+ = + = + −+ = + = + −
= + − + = + += + − + = + += + − + = + += + − + = + +
22 2
2 3
1
332 3
1 1
1 ( 1) 1 1( 1)
2 2 2
2 12 1
2
1 321
3 3
d
c
yy y
y y y
dxS y dy y y dy
dy y
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
2 1b
xa
dyS y dx
dxππππ = += += += +
∫∫∫∫ Now
Indefinite Integration
117
2 2 4 2 2 22 2 2 2 4 2
2 2
2 2 3
2 3
2
2 3 2 2 3 2 2 3 22
4 2 2 2 2 42 2 44
2
2 3 2 2 2 22
2 2 2 4
( )8
8 8
16 2 4
2 4
16
(2 4 ) (2 4 ) (2 4 )
256 32 ( )256
8
(2 4 ) (3 2 )1 1
32 ( ) 8
a x x x a xa y a x x y
a a
a yy a x x
a x xy
a y
a x x a x x a x xy
a y a a x xa x xa
a
a x x a xy
a a x x
− −− −− −− −= −= −= −= − ⇒⇒⇒⇒ = == == == =
′′′′∴ = −∴ = −∴ = −∴ = −
−−−−′′′′ ====
− − −− − −− − −− − −′′′′ = = == = == = == = = −−−−−−−−
− −− −− −− −′′′′+ = + =+ = + =+ = + =+ = + =−−−−
Q
2 2 2
2 2 2 2 2 22
2 2 2 20 0
( )
( ) (3 2 )2 1 2 .
8 8 ( )
a a
x
a a x
x a x a xS y y dx dx
a a a xπ ππ ππ ππ π
−−−−
− −− −− −− −′′′′∴ = + =∴ = + =∴ = + =∴ = + =−−−−
∫ ∫∫ ∫∫ ∫∫ ∫
2 2 2 22 2
0 0
2 2 2 24 4
2 20
1 12 (3 2 ) . (3 2 )( 4 )
48 4
1 (3 2 ). 9
4 2 44 32
a a
a
x a x dx a x x dxa a
a x aa a
a a
ππππππππ
π π ππ π ππ π ππ π π
−−−−= − = − −= − = − −= − = − −= − = − −
− − −− − −− − −− − − = = − == = − == = − == = − =
∫ ∫∫ ∫∫ ∫∫ ∫
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
2 1b
xa
dyS y dx
dxππππ = += += += +
∫∫∫∫ Now
2 22
2 2
11, 16
16 4 21 24 16 2 16
x yy x
dy x xdx x x
+ = = −+ = = −+ = = −+ = = −
− −− −− −− −∴ = =∴ = =∴ = =∴ = =− −− −− −− −
Mathematics For Engineering
118
squareunits
2 2 2 2
2 2
2 2 22 2
2
24 42
4 44
2 1
4
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
x
dy x x xdx x x
dy x xy x x
dx x
dyS y dx x dx
dx
x xx
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
2b
xa
dx dyS y d
d dπ θπ θπ θπ θ
θ θθ θθ θθ θ = += += += +
∫∫∫∫ Now
3 3
2 2
2 22 4 2 2 4 2 2 2 2
2 23 2 2 2 2 4
2
cos , sin
3 cos sin , 3 sin cos
9 cos sin 9 sin cos 9 sin cos
sin 9 sin cos 3 sin cos
2b
xa
x a y a
dx dya a
d d
dx dya a a
d d
dx dyy a a a
d d
dx dyS y
d d
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
02 2
/ 20
2(2 ) 3 sin cos
12 12sin5
5 5
d a d
a a
ππππ
ππππ
θ π θ θ θθ π θ θ θθ π θ θ θθ π θ θ θ
π ππ ππ ππ πθθθθ
====
= == == == =
∫∫∫∫
Indefinite Integration
119
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
2 2 2
0
22 3 2
0
( sin ), (1 cos )
(1 cos ), sin
(1 cos ) sin 2 sin2
2 4 (1 cos )sin2
648 sin
2 3
b
xa
x a t t y a t
dx dya t a t
dt dt
dx dy tdL dt a t a t a dt
dt dt
dx dy tS y dt a a t dt
dt dt
ta dt a
ππππ
ππππ
π ππ ππ ππ π
π ππ ππ ππ π
= − = −= − = −= − = −= − = −
= − == − == − == − =
= + = − + == + = − + == + = − + == + = − + =
∴ = + = −∴ = + = −∴ = + = −∴ = + = −
= == == == =
∫ ∫∫ ∫∫ ∫∫ ∫
∫∫∫∫
Exercise(15)
Find the area of the surface of revaluation generated by revolving the given arc about the given axis.
from to
from to
from to
3
2 2 2
20 2; [ .4 1 . .]
10 3; [ . 982 82 1) / 9 . .]
31
0 2; [ . 9 82 ln( 82 9) . .]2
18 (1 ), ; [ . . .]
4
(1)
(2)
(3)
(4)
(5)
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
ππππ
ππππ
ππππ
= = = − += = = − += = = − += = = − +
= = = − −= = = − −= = = − −= = = − −
= = = − + += = = − + += = = − + += = = − + +
= − −= − −= − −= − −
====2 2 2
( sin ), (1 cos );
[ .64 ( 4) . .]
1282cos cos 2 , 2sin sin 2 [ . . .]
5(6)
a y a x axis
Ans a e e s u
x y Ans s u
θ θ θθ θ θθ θ θθ θ θ
ππππππππθ θ θ θθ θ θ θθ θ θ θθ θ θ θ
−−−−
− = − −− = − −− = − −− = − −
− +− +− +− +
= − = −= − = −= − = −= − = −