1. The document discusses indefinite integration and provides formulas for integrating common functions. 2. Formulas are given for integrating trigonometric, inverse trigonometric, exponential, logarithmic, and other functions. 3. Examples of integrating various functions using the formulas are also provided.

1. Calculus
of
Integration
Prof. Dr. M. Abul-Ez
Mathematics Department
Faculty of Science
Sohag University

2. 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) 2
d d d
x x x x
dx 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) 1
1
(5) ln
(6)
ln
(7)
m
m
x
x
x x
d
f x f x c
dx
u v dx udx vdx
udx udx
x
x dx c m
m
dx
x c
x
a
a 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
xdx x c
xdx x c
xdx x c
x dx x c
= − += − += − += − +
= += += += +
= += += += +
= += += += +
∫∫∫∫
∫∫∫∫
∫∫∫∫
∫∫∫∫

3. Indefinite Integration
3
(12) sec ln sec tan
(13) cosec ln csc cot
xdx x x c
dx x x c
= + += + += + += + +
= − += − += − += − +
∫∫∫∫
∫∫∫∫
2
2
(14) sec tan
(15) cosec cot
(16) sec tan sec
(17) cosec cot cosec
xdx x c
dx x c
x xdx x c
x xdx x c
= += += += +
= − += − += − += − +
= += += += +
= − += − += − += − +
∫∫∫∫
∫∫∫∫
∫∫∫∫
∫∫∫∫
Integration tends to inverse of Trigonometric functions:
1
2 2 2 1
1
2 2 2
1
1
2 2 2 1
1
2 2 2
1
sin
(18)
1
cos
1
tan
(19)
1
cot
1
sec
(20)
1
cosec
1
coth
(21)
1
ln
2
bx
c
dx b a
bxa b x c
b a
bx
c
dx ab a
bxa b x c
ab a
bx
c
dx a a
dx
bxx b x a c
a a
bx
c
ab adx
dx
bx ab x a c
ab bx a
−−−−
−−−−
−−−−
−−−−
−−−−
−−−−
−−−−

++++
==== 
−−−−−−−− ++++


++++
==== 
−−−−++++  ++++


++++
==== 
−−−−−−−− ++++


++++
==== 
−−−−−−−− ++++
++++
∫∫∫∫
∫∫∫∫
∫∫∫∫
∫∫∫∫
1
2 2 2
2 2
12 2
1
tanh
(22)
1
ln
2
ln( )
(23)
sinh
bx
c
ab adx
a bxa b x c
ab a bx
x x a cdx
x
cx a
a
−−−−
−−−−



++++
==== 
++++−−−−  ++++
 −−−−
 + + ++ + ++ + ++ + +
==== 
++++++++

∫∫∫∫
∫∫∫∫

4. Mathematics For Engineering
4
2 2
12 2
2
2 2 2 2 1
2
2 2 2 2 1
2
2 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 cdx
x
cx a
a
a x
a x x a x c
a
a x
x a x x a c
a
a x
x 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 cos2 )
2
1
(5)cos (1 cos2 )
2
1
(6)cos cos cos( ) cos( )
2
1
(7)sin sin cos( ) cos( )
2
1
(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 cos2 ) cos2
2 2 2
1 1 1
(2) cos (1 cos2 ) cos2
2 2 2
(3) sec tan ,
x dx x dx x x c
x dx x dx x x c
x dx x c
    
= − = − += − = − += − = − += − = − +    
    
    
= + = + += + = + += + = + += + = + +    
    
= += += += +
∫ ∫∫ ∫∫ ∫∫ ∫
∫ ∫∫ ∫∫ ∫∫ ∫
∫∫∫∫

5. 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( )
2
1 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 x
c
a b a b
ax bxdx a b x a b x dx
a b x a b x
c
a b a b
ax bx dx a b x a b x dx
= + + −= + + −= + + −= + + −
    + −+ −+ −+ −
= + += + += + += + +    
+ −+ −+ −+ −    
= − − += − − += − − += − − +
    − +− +− +− +
= + += + += + += + +    
− +− +− +− +    
= + + −= + + −= + + −= + + −
====
∫ ∫∫ ∫∫ ∫∫ ∫
∫ ∫∫ ∫∫ ∫∫ ∫
∫ ∫∫ ∫∫ ∫∫ ∫
1 cos( ) cos( )
2 ( ) ( )
a b x a b x
c
a b a b
    − + −− + −− + −− + −
+ ++ ++ ++ +    
+ −+ −+ −+ −    
Solved Examples:
5 6
1 4
3 3 3
1
1
1
:
6
3
4
1 ( )
( )
( 1)
1 1
ln ( )
( )
1 1 ( )
( ) , 1
( 1)( )
Exampl(1)
Exampl(2):
Exampl(3):
Exampl(4):
Exampl(5):
n
n
n
n
n
x dx x c
x dx x dx x c
ax b
ax b dx c
a n
dx ax b c
ax b a
ax b
dx ax b dx c n
a nax b
++++
− +− +− +− +
−−−−
= += += += +
= = += = += = += = +
++++
+ = ++ = ++ = ++ = +
++++
= + += + += + += + +
++++
++++
= + = + ≠= + = + ≠= + = + ≠= + = + ≠
− +− +− +− +++++
∫∫∫∫
∫ ∫∫ ∫∫ ∫∫ ∫
∫∫∫∫
∫∫∫∫
∫ ∫∫ ∫∫ ∫∫ ∫

6. Mathematics For Engineering
6
1 3 1 3 3 5
2 2 2 2 2 2
2
3 5 2 3 2
4 3
1 2
2 2
(1 ) ( )
3 5
Find ( ) ( 2) , ( ) ( ) 2
2
Exampl(6):
Exampl(7):
Exampl(8):
dx ax b c
aax b
x x dx x x dx x dx x dx x x c
x dx
i x x dx ii dx iii x x dx
x
= + += + += + += + +
++++
− = − = − = − +− = − = − = − +− = − = − = − +− = − = − = − +
+ ++ ++ ++ +
++++
∫∫∫∫
∫ ∫ ∫ ∫∫ ∫ ∫ ∫∫ ∫ ∫ ∫∫ ∫ ∫ ∫
∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫
3 2
3 5 2 5 6 3 6
1 3 32
34 4 4
44 3
3 3
3 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 du
ii 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 3
2 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 2
3
3 3
1 2 1
ln 2 3
2 3 2 2 3 2
1 6 1
ln(1 2 )
6 61 2 1 2
5 ( 1) 4 ( 1) 4
( 1) ( 1) ( 1) ( 1)
4
1 4ln 1
( 1)
Exampl(10):
Exampl(11):
Exampl(12):
dx dx
x c
x x
x dx x dx
x c
x x
x x x
dx dx dx dx
x x x x
dx
x x c
x
= = − += = − += = − += = − +
− −− −− −− −
− −− −− −− −
= − = − += − = − += − = − += − = − +
− −− −− −− −
+ + + ++ + + ++ + + ++ + + +
= = += = += = += = +
+ + + ++ + + ++ + + ++ + + +
    
= + = + + += + = + + += + = + + += + = + + +    
++++    
∫ ∫∫ ∫∫ ∫∫ ∫
∫ ∫∫ ∫∫ ∫∫ ∫
∫ ∫ ∫ ∫∫ ∫ ∫ ∫∫ ∫ ∫ ∫∫ ∫ ∫ ∫
∫∫∫∫

7. Indefinite Integration
7
2 2 2 2 2
2 2 3
1 1
tan 2ln sec
2 2
1 1
(sec ) sec ( ) ln sec tan
2 2
1
sin cos sin (sin ) sin
3
Example(13):
Example(14):
Example(15):
xdx x c
x x dx x d x x x c
x xdx xd x x c
= += += += +
= = + += = + += = + += = + +
= = += = += = += = +
∫∫∫∫
∫ ∫∫ ∫∫ ∫∫ ∫
∫ ∫∫ ∫∫ ∫∫ ∫
2 3 2 3 2 4 6
2 4 6
3 7 11
2 2 2
5 9 13
2 2 2
5 5 6
:
(1 ) (1 ) (1 3 3 )
1 3 3
( )
3 3
6 6 2
2
5 9 13
1
( 1) ( 1) ( 1) ( 1)
6
Example(16)
Example(17): x x x x x
x x x x x
dx dx dx
x x x
x x x
dx
x x x x
dx
x dx x dx x dx
x
x x x x c
e e dx e d e e c
+ + + + ++ + + + ++ + + + ++ + + + +
= == == == =
= + + += + + += + + += + + +
= + + += + + += + + += + + +
= + + + += + + + += + + + += + + + +
+ = + + = + ++ = + + = + ++ = + + = + ++ = + + = + +
∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫
∫∫∫∫
∫ ∫ ∫ ∫∫ ∫ ∫ ∫∫ ∫ ∫ ∫∫ ∫ ∫ ∫
∫ ∫∫ ∫∫ ∫∫ ∫
3 3 3
2 2
2
2 2
2 2
2 5 2
2 5 2 5
2 4
1 1
(3 )
3 3ln
1 ( 1) 1
ln( 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 x
x
x x
x x
x x
x x
x
a dx a dx a c
a
e d e
dx dx e c
e e
e dx d e
e d e
e e
e
c
−−−−
−−−−
= = += = += = += = +
−−−−
= = − += = − += = − += = − +
− −− −− −− −
−−−−
= = − −= = − −= = − −= = − −
− −− −− −− −
−−−−
= += += += +
−−−−
∫ ∫∫ ∫∫ ∫∫ ∫
∫ ∫∫ ∫∫ ∫∫ ∫
∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫
4 4
1 1
sin4 cos4 , (19) cos3 sin3
4 3
cos sin cos ( sin )
Example(21):
Example(22):
x dx x c x dx x c
x x dx x x dx
−−−−
= + = += + = += + = += + = +
= − −= − −= − −= − −
∫ ∫∫ ∫∫ ∫∫ ∫
∫ ∫∫ ∫∫ ∫∫ ∫

8. Mathematics For Engineering
8
3 2 3 4
5 2 5 2
5 6
5 2 4 2
2 2 2
1
tan sec tan (tan ) tan
4
cot cosec cot ( cosec )
1
cot (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 1
sin 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 1
cos 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
= −= −= −= −
==== ⇒⇒⇒⇒ = −= −= −= −
∴ = − = − +∴ = − = − +∴ = − = − +∴ = − = − +
= − += − += − += − +
∫∫∫∫
∫∫∫∫
∫∫∫∫
∫∫∫∫

9. Indefinite Integration
9
4 3
3 4
2
2
sec tan sec (sec tan )
1
sec (sec ) sec
4
sin sin 1
tan sec sec
cos coscos
cos cos 1
cot cosec cot
sin sinsin
Example(27):
Example(28):
Example(29):
Example(30)
x x dx x x x dx
x d x c
x x
dx dx x x dx x c
x xx
x x
dx 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 tan
sec sec
sec tan sec tan
ln sec tan
:
Example(31):
Exampl
dx x dx x dx
x x x
dx x dx
x x
xdx x x dx
x x c
x x x x x
x dx x dx dx
x x x x
x x c
− −− −− −− −
= == == == =
++++ −−−−
= −= −= −= −
= −= −= −= −
= − + += − + += − + += − + +
+ ++ ++ ++ +
= == == == =
+ ++ ++ ++ +
= + += + += + += + +
∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫
∫ ∫∫ ∫∫ ∫∫ ∫
∫ ∫∫ ∫∫ ∫∫ ∫
∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫
2
4
cosec cot
cosec cosec
cosec cot
cosec cosec cot
ln cosec cot
cosec cot
int
( ) sin (2 3cos ) ,
sin sin
( ) , ( )
(2 3cos ) (2 3cos )
e(32):
Example(33):
x x
x dx x dx
x x
x x x
dx x x c
x x
FindThefolowing egrals
i I x x dx
x dx x dx
ii J iii K
x x
−−−−
====
−−−−
−−−−
= = − += = − += = − += = − +
−−−−
= += += += +
= == == == =
++++ ++++
∫ ∫∫ ∫∫ ∫∫ ∫
∫∫∫∫
∫∫∫∫
∫∫∫∫
3
2
3
2
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
= += += += + ⇒⇒⇒⇒ = −= −= −= −
− −− −− −− −            
∴ = + = =∴ = + = =∴ = + = =∴ = + = =             
            
−−−−    
= + += + += + += + +    
    
∫∫∫∫
∫ ∫∫ ∫∫ ∫∫ ∫

10. Mathematics For Engineering
10
(((( ))))
4 3
4 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 du
ii J u c x c
x u
xdx du
iii 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
(tan3 sec3 ) (tan 3 2tan3 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) 2tan3 sec3 sec 3
2sec 3 2tan 3 sec3 1
2 2
tan3 sec3
3 3
cos (sin )
1
sec (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 c
a
e x dx
= − + += − + += − + += − + +
= + −= + −= + −= + −
= + − += + − += + − += + − +
= = += = += = += = +
= = += = += = += = +
====
∫∫∫∫
∫∫∫∫
∫ ∫∫ ∫∫ ∫∫ ∫
∫ ∫∫ ∫∫ ∫∫ ∫
∫∫∫∫
[[[[ ]]]]
cos cos
22
(cos )
lnsin )cot lnsin cot
1 1
lnsin
2 2
Example(40): (
x x
e d x e c
x x dx let y x dy xdx
I ydy y c x c
− = − +− = − +− = − +− = − +
==== ⇒⇒⇒⇒ ====
∴ = = + = +∴ = = + = +∴ = = + = +∴ = = + = +
∫∫∫∫
∫∫∫∫
∫∫∫∫

11. 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 )
1
3ln sec tan tan tan 1 tan
2
1
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 2
1 cos2sec2 cot 2 1 cos2
sin2 sin 2
1 cos2 2sin2
sin2 1 1 1
ln ln 1 cos2
1 cos 2 2 2 2
dx xdx
dx
xx x x
x x
let u x du xdx
x dx du
u 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
1
2
1
2
2tan 3sec tan sec
1
2 2ln sec 3tan tan
2
sin
39
1
tan
5 55
Example(44):
Example(45):
x x dx x dx x x dx
x x x x c
dx x
dx c
x
dx x
c
x
−−−−
−−−−
+ + ++ + ++ + ++ + +
= − + + + += − + + + += − + + + += − + + + +
= += += += +
−−−−
= += += += +
++++
∫ ∫ ∫ ∫∫ ∫ ∫ ∫∫ ∫ ∫ ∫∫ ∫ ∫ ∫
∫∫∫∫
∫∫∫∫

12. Mathematics For Engineering
12
3 2 3
2
2
2
3
(1 cos ) (1 3cos 3cos cos )
3
1 3cos (1 cos2 ) cos cos
2
3 3
1 3cos cos2 cos (1 sin )
2 2
5 3
4cos cos2 sin cos
2 2
5 3 1
4sin sin2 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 )
3
1 3sin (1 cos2 ) sin sin
2
3 3
1 3sin cos2 sin (1 cos )
2 2
5 3
4sin cos2 sin cos
2 2
5 3 1
4cos 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 1
2 22 3
2
4 9
1 1 2 2 1 2
. tan tan
4 4 3 3 6 34 9
Example(48):
dx
find
x
dx dx x x
c c
x x
− −− −− −− −
++++
= = + = += = + = += = + = += = + = +
++++ ++++
∫∫∫∫
∫ ∫∫ ∫∫ ∫∫ ∫
(((( ))))
1 1
2 2 24 2
9 3
1
2 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 x
c
x x x
dx dx x
c
x x
− −− −− −− −
−−−−
= = = = += = = = += = = = += = = = +
+ ++ ++ ++ + ++++
= = += = += = += = +
+ ++ ++ ++ +
∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫
∫ ∫∫ ∫∫ ∫∫ ∫

13. Indefinite Integration
13
(((( )))) (((( ))))
1
2 2 2
1
2 2
2 2 3
1 3
6 3 2 3 2
2
where f(x)is linear
2 1 2
sec
3 34 9 2 (2 ) (3)
( ) 1 ( )
Accordingto sec
( ) ( )
1 3 1 ( ) 1
sin
3 3 31 1 ( ) 1 ( )
sin
cos
Example(49):
Example(50):
Example(51):
dx dx x
c
x x x x
f x f x
c
a a
f x f x a
x dx x dx d x
x c
x x x
xdx
−−−−
−−−−
−−−−
= = += = += = += = +
− −− −− −− −
′′′′
= += += += +
−−−−
= = = += = = += = = += = = +
− − −− − −− − −− − −
++++
∫ ∫∫ ∫∫ ∫∫ ∫
∫∫∫∫
∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫
1
2
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 1
ta
2 2cos 1 1
Example(52):
d x
x c
x xdx x xdx d x
x x
x c
let u x du x xdx
x xdx du
u
−−−−
−−−−
= − = − += − = − += − = − += − = − +
++++
−−−−
= == == == =
+ + ++ + ++ + ++ + +
= += += += +
==== ⇒⇒⇒⇒ ====
−−−−
∴ = =∴ = =∴ = =∴ = =
+ ++ ++ ++ +
∫ ∫∫ ∫∫ ∫∫ ∫
∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫
∫ ∫∫ ∫∫ ∫∫ ∫
1 1 2
1
2 2 2
1
2 2 2
3 2
2 2
2
1
1
n tan (cos )
2
1 4 1 4
sin
4 4 525 16 5 (4 )
( 2)
sin
24 ( 2) 2 ( 2)
3 4 3 4
(3 4)
1 1
3
4 4tan
2
Example(53):
Example(54):
Example(55):
Example
u c x c
dx dx x
c
x x
dx dx x
c
x x
x x x
dx x dx
x x
x
x 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
x
c−−−−
= == == == =
+ + + + + + ++ + + + + + ++ + + + + + ++ + + + + + +
++++
= += += += +
∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫

14. Mathematics For Engineering
14
2 2 2
1
2 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) 2
2 25 4 5 4 5 4
1 ( 2 2) 1 2
2 25 4 5 4
1 ( 2 2
2
Example(57):
Example(58):
dx dx dx
x x x x x x
dx x
c
x
x x x
dx dx dx
x x x x x x
x
dx dx
x 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 dx
x x x
x
x x c
d
x x x
dx
d
x A x x B A x B
dx
A B x x
−−−−
+ =+ =+ =+ =
− − − +− − − +− − − +− − − +
− +− +− +− +
= − − + += − − + += − − + += − − + +
− − = − −− − = − −− − = − −− − = − −
+ = − − + = − − ++ = − − + = − − ++ = − − + = − − ++ = − − + = − − +
− −− −− −− −
∴ = = + = − − +∴ = = + = − − +∴ = = + = − − +∴ = = + = − − +
∫ ∫∫ ∫∫ ∫∫ ∫
1
2
2 2
1
2
2 2 2
2 1
2
( 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)
xx
dx dx
x x x x
x dx dx x dx
x x x x x x
dx x
x x c
x
−−−−
−−−−
−−−−
− − +− − +− − +− − +++++
====
− − − −− − − −− − − −− − − −
− −− −− −− − − − −− − −− − −− − −
= + == + == + == + =
− − − − − −− − − − − −− − − − − −− − − − − −
− +− +− +− +
+ = − − + ++ = − − + ++ = − − + ++ = − − + +
− +− +− +− +
∫ ∫∫ ∫∫ ∫∫ ∫
∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫
∫∫∫∫
2
2 3
9 12 8
1 39
2 3 (18 12) , 3 12
9 9
Example(59):
x
dx
x x
put x A x B A B A
++++
− +− +− +− +
+ = − ++ = − ++ = − ++ = − + ⇒⇒⇒⇒ = = + == = + == = + == = + =
∫∫∫∫

15. Indefinite Integration
15
2 2
2 2
2 2
2 2
2 1
2 3 1 (18 12) 39
99 12 8 9 12 8
1 (18 12) 1 39
9 99 12 8 9 12 8
1 (18 12) 1 39
9 99 12 8 (9 12 4) 4
1 (18 12) 1 39
9 99 12 8 (3 2) 4
1 39 1 1 (3
ln 9 12 8 . . tan
9 9 3 2
x x
dx dx
x x x x
x
dx dx
x x x x
x
dx dx
x x x x
x
dx dx
x x x
x
x x −−−−
+ − ++ − ++ − ++ − +
∴ =∴ =∴ =∴ =
− + − +− + − +− + − +− + − +
−−−−
= += += += +
− + − +− + − +− + − +− + − +
−−−−
= += += += +
− + − + +− + − + +− + − + +− + − + +
−−−−
= += += += +
− + − +− + − +− + − +− + − +
−−−−
= − + += − + += − + += − + +
∫ ∫∫ ∫∫ ∫∫ ∫
∫ ∫∫ ∫∫ ∫∫ ∫
∫ ∫∫ ∫∫ ∫∫ ∫
∫ ∫∫ ∫∫ ∫∫ ∫
2)
2
c++++
[[[[ ]]]]
[[[[ ]]]]
2
2 2 2 2
2 2
2 2
2
2 (4 2 )
4
1 1
, 4, 2 (4 2 ) 8
2 2
(4 2 ) 82 1 1 (4 2 ) 1 8
2 2 24 4 4 4
1 (4 2 ) 1 8
2 24 4 (4 4 )
1 (4 2 ) 1 8
4
2 24 4 ( 2)
Example(60):
x
dx put x A x B
x x
A B x x
xx x
dx dx dx dx
x x x x x x x x
x dx dx
x x x x
x dx dx
x
x x x
++++
+ = − ++ = − ++ = − ++ = − +
−−−−
− −− −− −− −
∴ = = + = − +∴ = = + = − +∴ = = + = − +∴ = = + = − +
− +− +− +− ++ − − −+ − − −+ − − −+ − − −
= = −= = −= = −= = −
− − − −− − − −− − − −− − − −
− −− −− −− −
= −= −= −= −
− − − +− − − +− − − +− − − +
− −− −− −− −
= − = − −= − = − −= − = − −= − = − −
− − −− − −− − −− − −
∫∫∫∫
∫ ∫ ∫ ∫∫ ∫ ∫ ∫∫ ∫ ∫ ∫∫ ∫ ∫ ∫
∫ ∫∫ ∫∫ ∫∫ ∫
∫ ∫∫ ∫∫ ∫∫ ∫
2 11 ( 2)
sin
2 2
x
x c−−−− −−−−
− +− +− +− +
Exercise(1)
Integrate the following functions with respect to x :
3
4 3
2 2 3 7
2
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 x
x
x x
x
x
x x x x
x x
− + − + + +− + − + + +− + − + + +− + − + + +
+ −+ −+ −+ −
−−−−
++++
+ −+ −+ −+ −
+ ++ ++ ++ +

16. Mathematics For Engineering
16
2 2
2
2
3 4
2
3 sin tan 2
2 2
5
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 x
xx x
x x x
x x
x x x
a x x
x e
x x
x e
x x x e e
xx
−−−−
+ −+ −+ −+ −
++++++++
+ + ++ + ++ + ++ + +
++++
++++ + ++ ++ ++ +
−−−−
++++
−−−− ++++
+ + ++ + ++ + ++ + +
++++
2 3 4
5 2 4 2
2
2
2
(25)(cos sin ) (26)sin cos (27)cos sin
sin8
(28)tan sec (29)cot cosec (30)
9 sin 4
sin 1 cos 2 cot ((1 cosec )
(31) (32) (33)
2 sin 2 cosec(1 cos )
sec 3 1 1
(34) (35) (36)
1 sin2 1 cos2(1 tan3 )
(1
(37)
x
e
x x x x x x
x
x x x
x
x x x x
x x xx
x
x xx
−−−−
++++
+ ++ ++ ++ +
++++++++
− −− −− −− −++++
3 5
2
2 2
22 4
cot ) (1 tan) 1
(38) (39)
1 sin2 1 cos 2 5
sec sin cos
(40) (41) (42)
1 cos 24 tan 1
x
x
x
x x x
x e x x
xx e
+ ++ ++ ++ +
− −− −− −− − ++++
++++− −− −− −− −
22 2
2 22
2 2
tan cot 1
(43) (44) (45)
2 8cos 4 sin 4
1 2 3 1
(46) (47) (48)
6 13 3 4 32 8
2 1
(49) (50)
27 6 12 4
x x
x xx x
x x
x x x xx x
x x
x x x x
+ −+ −+ −+ −− −− −− −− −
− −− −− −− −
+ + − ++ + − ++ + − ++ + − ++ −+ −+ −+ −
−−−−
+ − + −+ − + −+ − + −+ − + −

17. 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 x
x x
x x
x x
x x x x
x x x x
x e e x
x e e
x e e x
x e e
x e e e e
x 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) sinh2
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 cosh2 1
2
1
(11) cosh cosh2 1
2
(12) cosh sinh
(13) cosh sinh
2tanh
(14)tanh2
1 tanh
x
x
x x x
x x x
x x
x x
x x e
x x e
x
x
x
−
=
= +
= −
= +
+ =
− =
=
+
we can proof this lows by using the definition
in the following we stat integration formula for hyperbolic functions

18. 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
xdx x c
xdx x c
xdx x c
xdx x c
xdx x c
xdx x c
x x dx x c
x x dx x c
dx
bb x a
−−−−
= += += += +
= += += += +
= += += += +
= += += += +
= += += += +
= − += − += − += − +
=− +=− +=− +=− +
= − += − += − += − +
====
++++
∫∫∫∫
∫∫∫∫
∫∫∫∫
∫∫∫∫
∫∫∫∫
∫∫∫∫
∫∫∫∫
∫∫∫∫
∫∫∫∫
1
1
2 2 2
1
2 2 2
1
2 2 2
2
1
(10) cosh
1
tanh
(11)
1
ln
2
1
coth
(12)
1
ln
2
Solved Examples:
1 cosh cosh
(1) sech
cosh cosh 1 si
bx
c
a
dx bx
c
b ab x a
bx
c
dx ab a
a bxa b x
c
ab a bx
bx
c
dx ab a
bx ab x a c
ab bx a
x x
xdx 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 2
sinh2
2 2 3
dx x c
x
xdx x dx x x c
xdx x x dx x dx x x dx
x
x c
−−−−
= += += += +
= − = − += − = − += − = − += − = − +
= + = += + = += + = += + = +
= + += + += + += + +
∫∫∫∫
∫ ∫∫ ∫∫ ∫∫ ∫
∫ ∫ ∫ ∫∫ ∫ ∫ ∫∫ ∫ ∫ ∫∫ ∫ ∫ ∫

19. Indefinite Integration
19
2 2
3 3
5 5
3 3
8 2 8 2
(4) cosh2 .
2
1 1 1
( ) ( )
2 2 3
(5) sinh5
2
1 1 1 1
( ) ( )
2 2 8 2
x x
x x
x x x x
x x
x x
x x x x
e e
e x dx e dx
e e dx e e c
e e
e x dx e dx
e e dx e e c
−−−−
− −− −− −− −
−−−−
− −− −− −− −
++++
====
= + = − += + = − += + = − += + = − +
−−−−
====
= − = + += − = + += − = + += − = + +
∫ ∫∫ ∫∫ ∫∫ ∫
∫∫∫∫
∫ ∫∫ ∫∫ ∫∫ ∫
∫∫∫∫
Solved Examples
[[[[ ]]]]
2 2
1
2
2
2
1
: sinh3 cosh3
3
1 cosh cosh
: sech
cosh cosh 1 sinh
(sinh )
tan (sinh )
1 sinh
1 1 1
: sinh cosh2 1 sinh2
2 2 2
: cosh 3
Example(1)
Example(2)
Example(3)
Example(4)
xdx x c
x x
x dx dx dx dx
x x x
d x
x c
x
x dx x dx x x c
x d
−−−−
= += += += +
= = == = == = == = =
++++
= = += = += = += = +
++++
    
= − = − += − = − += − = − += − = − +    
    
∫∫∫∫
∫ ∫ ∫ ∫∫ ∫ ∫ ∫∫ ∫ ∫ ∫∫ ∫ ∫ ∫
∫∫∫∫
∫ ∫∫ ∫∫ ∫∫ ∫
∫∫∫∫ [[[[ ]]]]
2 2
4 2 2
2 2 2 3
1 1 1
cosh6 1 sinh6
2 2 6
1
: tanh 5 1 sech 5 tanh5
5
: sech 1 tanh sech
1
sech 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
xdx x xdx x x c
    
= + = + += + = + += + = + += + = + +    
    
        = − = − += − = − += − = − += − = − +             
    = −= −= −= −
    
= − = − += − = − += − = − += − = − +
∫∫∫∫
∫ ∫∫ ∫∫ ∫∫ ∫
∫ ∫∫ ∫∫ ∫∫ ∫
∫ ∫∫ ∫∫ ∫∫ ∫
3 2
2
2
3
: sinh 4 sinh 4 (sinh4 )
(cosh 4 1)(sinh4 )
(cosh 4 (sinh4 ) (sinh4 )
1 1
cosh 4 cosh4
12 4
Example(7) xdx x x dx
x x dx
x x dx x dx
x x c
====
= −= −= −= −
= −= −= −= −
= − += − += − += − +
∫ ∫∫ ∫∫ ∫∫ ∫
∫∫∫∫
∫ ∫∫ ∫∫ ∫∫ ∫

20. Mathematics For Engineering
20
3 3
3 3
4 2
4 2
3 3
2 2 2 3 3
5
5
( ) 1
: cosh3 ( )
2 2
1 1
( )
2 2 4 2
( ) 1
: sinh3 ( )
2 2
1 1
( )
2 2 5
Example(8)
Example(9)
x x
x x x x x
x x
x x
x x
x x x x x
x
x x x
e e
e xdx e dx e e e dx
e e
e e dx c
e e
e xdx e dx e e e dx
e
e e dx e c
−−−−
−−−−
−−−−
−−−−
−−−−
−−−−
− −− −− −− −
++++
= = += = += = += = +
    
= + = + += + = + += + = + += + = + +        −−−−    
−−−−
= = −= = −= = −= = −
    
= + = + += + = + += + = + += + = + +        
    
∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫
∫∫∫∫
∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫
∫∫∫∫
(((( ))))
1
2
1 1
: sinh
2 2
1 1
2 2
1 1
2 2
1 1
2 2
cosh sinh
Try to solve sinh by parts
: sinh
16
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
− −− −− −− −
−−−−
− −− −− −− −
− −− −− −− −
−−−−
    = − = −= − = −= − = −= − = −
    
= −= −= −= −
            = − − − − += − − − − += − − − − += − − − − +
            
            = − − − += − − − += − − − += − − − +
            
= − += − += − += − +
====
++++
∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫
∫ ∫∫ ∫∫ ∫∫ ∫
∫∫∫∫
∫∫∫∫
1
2
3 2
2
2
3
4
: cosh
525
: sinh 4 sinh 4 (sinh4 )
(cosh 4 1)(sinh4 )
(cosh 4 (sinh4 ) (sinh4 )
1 1
cosh 4 cosh4
12 4
Example(12)
Example(13)
c
dx x
c
x
xdx x x dx
x x dx
x x dx x dx
x x c
−−−−
++++
= += += += +
−−−−
====
= −= −= −= −
= −= −= −= −
= − += − += − += − +
∫∫∫∫
∫ ∫∫ ∫∫ ∫∫ ∫
∫∫∫∫
∫ ∫∫ ∫∫ ∫∫ ∫

21. Indefinite Integration
21
Exercise(2)
Integrate the following functions:
2 2
2 2 3
2
5 2
3 4 5 2
4 2
2
(1)sinh3 (2)cosesh (3) sinh
(4)sinh cosh (5) sinh3 (6) cosh3
sech
(7) sinh (3 2cosh ) (8) (9)(cosh sinh )
3 5tanh
(10)sinh cosh (11)cosh sinh (12)tanh sech
sinh8
(13)coth cosech (14)
9 sinh
x
x x x x
x x e x e x
x
x x x x
x
x x x x x x
x
x
+ ++ ++ ++ +
−−−−
++++ 2
2
2
3 5 2
2
sinh cosh
(15)
4 1 cosh 2
sinh 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 x
x x x x
x x xx
x
x xx
x x
x x x
++++
+ ++ ++ ++ +
++++++++
− +− +− +− +++++
+ ++ ++ ++ +
− −− −− −− − −−−−

22. Mathematics For Engineering
22
Methods of Integration:
(1) Integration by parts
When u and v are differentiable functions then
( )
( )
d uv udv vdu
udv d uv vdu
= += += += +
= −= −= −= −
and by integrate ( ) (1)udv d uv vdu= −= −= −= −∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫
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 x
xe dx∫∫∫∫
Solution:
If we chose x
u e==== to be differentiated and dv xdx==== to be
integrated
2 21 1
2 2
x x x
xe 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
xe dx xe e dx
= == == == =
= == == == =
= −= −= −= −∫ ∫∫ ∫∫ ∫∫ ∫
Note that : The integral in the right side x
e dx∫∫∫∫ is simple than the
integral x
xe dx∫∫∫∫ Finally x x
I xe e c= − += − += − += − + .

23. Indefinite Integration
23
Example (2): Find: 2
lnx xdx∫∫∫∫
Solution:
Consider the partition 2
lnu x dv x dx= == == == =
Then
3
1
3
x
du dx v
x
= == == == =
Substitute in the rule we have:
3 3 3 3 3
21 1
ln ln ln
3 3 3 3 3 9
x x x x x
I x dx x x dx x c
x
∴ = − = − = − +∴ = − = − = − +∴ = − = − = − +∴ = − = − = − +∫ ∫∫ ∫∫ ∫∫ ∫
Try to solve ln ,n
nI x xdx n= ∈= ∈= ∈= ∈∫∫∫∫
Example (3): Find 1x x dx++++∫∫∫∫
Solution:
Let 1u x dv xdx= = += = += = += = +
3
2
2
(1 )
3
du dx v x∴ = = +∴ = = +∴ = = +∴ = = +
by using partition rule
( )udv d uv vdu= −= −= −= −∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫
3 3 3 5
2 2 2 2
2 2 2 2 2
(1 ) (1 ) (1 ) ( )( )(1 )
3 3 3 3 5
x x
I x x dx x x c∴ = + − + = + − + +∴ = + − + = + − + +∴ = + − + = + − + +∴ = + − + = + − + +∫∫∫∫
(try with thenew partation what hapen)
trytosolve
1
( ) ,n
u x dv xdx
x ax b dx n
= + == + == + == + =
+ ∈+ ∈+ ∈+ ∈∫∫∫∫
Example (4): Find sinx xdx∫∫∫∫
Solution:
Let sinu x dv xdx= == == == =
cosdu dx v x∴ = = −∴ = = −∴ = = −∴ = = −
by using partition rule
( )udv d uv vdu= −= −= −= −∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫
sin cos cos cos sinI x xdx x x xdx x x x c∴ = = − + = − + +∴ = = − + = − + +∴ = = − + = − + +∴ = = − + = − + +∫ ∫∫ ∫∫ ∫∫ ∫

24. Mathematics For Engineering
24
(try with thepartation what hapen)
try to solve is positive integer
sin
sinn
n
u x dv xdx
I x xdx n
= == == == =
==== ∫∫∫∫
Example (5): Find cosx x dx∫∫∫∫
Solution:
let cos
sin
u x dv xdx
du dx v x
= == == == =
∴ = =∴ = =∴ = =∴ = =
sin sin sin cosI x x xdx x x x c∴ = − = − +∴ = − = − +∴ = − = − +∴ = − = − +∫∫∫∫
trytosolve is positive integercosn
nI x xdx n==== ∫∫∫∫
Example (6): Find 2
sinx xdx∫∫∫∫
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 xdx
we can solve x xdx by parts as in example
x x dx let u x dv xdx
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 xdx x x x x x c
x x x x x c
∴ = − + = − + − +∴ = − + = − + − +∴ = − + = − + − +∴ = − + = − + − +
= − + − += − + − += − + − += − + − +
∫∫∫∫
Example (7): Find 2
cosx xdx∫∫∫∫
Solution:
2
cos
2 sin
u x dv xdx
du xdx v x
= == == == =
∴ = =∴ = =∴ = =∴ = =

25. 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 x
x 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 xe dx x e xe e c
= == == == =
∴ = =∴ = =∴ = =∴ = =
∴ = − = + − +∴ = − = + − +∴ = − = + − +∴ = − = + − +∫∫∫∫
where is a positive integerm x
mtryto solve I x e dx m==== ∫∫∫∫
Example (9): Find 1
sin xdx−−−−
∫∫∫∫
Solution:
1
2
sin
1
u x dv dx
dx
du let v x
x
−−−−
= == == == =
∴ = =∴ = =∴ = =∴ = =
−−−−
1 1
2 2
1 2
sin sin
21 1
xdx xdx
I udv uv vdu x x x x
x x
− −− −− −− −
= = − − = −= = − − = −= = − − = −= = − − = −
− −− −− −− −
∫ ∫ ∫ ∫∫ ∫ ∫ ∫∫ ∫ ∫ ∫∫ ∫ ∫ ∫
1 1 2
2
1 2 1
sin sin .2 1
2 21
xdx
x x x x x c
x
− −− −− −− −−−−−
= + = + − += + = + − += + = + − += + = + − +
−−−−
∫∫∫∫
try tosolve where is a positive integer1
sinx xdx m−−−−
∫∫∫∫
Example (10): Find 1
tan x dx−−−−
∫∫∫∫
Solution
1
2
tan
1
u x dv dx
dx
du let v x
x
−−−−
= == == == =
∴ = =∴ = =∴ = =∴ = =
++++
1 1 1
2 2
1 2
tan tan tan
21 1
xdx xdx
udv xdx x x x x
x x
− − −− − −− − −− − −
∴ = = − = −∴ = = − = −∴ = = − = −∴ = = − = −
+ ++ ++ ++ +
∫ ∫ ∫ ∫∫ ∫ ∫ ∫∫ ∫ ∫ ∫∫ ∫ ∫ ∫

26. Mathematics For Engineering
26
1 21
tan ln(1 )
2
x x x c−−−−
= − + += − + += − + += − + +
Example (11): Find sin , cosax ax
I e bxdx J e bxdx= == == == =∫ ∫∫ ∫∫ ∫∫ ∫
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
1
cos
ax
ax
let u e dv bxdx
du ae dx v bx
b
= == == == =
−−−−
∴ = =∴ = =∴ = =∴ = =
(((( ))))
where
1 1
sin cos cos
cos cos
cos (1)
cos
ax ax ax
ax
ax
ax
ax
I e bxdx e bx bx ae dx
b b
e a
bx e bxdx
b b
e a
I bx J
b b
J e bxdx
− −− −− −− −    
= = −= = −= = −= = −     
    
−−−−
= += += += +
−−−−
∴ = +∴ = +∴ = +∴ = +
====
∫ ∫∫ ∫∫ ∫∫ ∫
∫∫∫∫
∫∫∫∫
Similarly taking the second integral J
let cos
1
sin
1 1 1
sin sin . sin sin
1
sin (2)
ax
ax
ax ax ax ax
ax
u e dv bxdx
du ae dx v bx
b
a
J e bx bx ae dx e bx e bx dx
b b b b
a
J e bx I
b b
= == == == =
∴ = =∴ = =∴ = =∴ = =
∴ = − = −∴ = − = −∴ = − = −∴ = − = −
= −= −= −= −
∫ ∫∫ ∫∫ ∫∫ ∫
from (1),(2) we get
2
2 2
2
2 2
1
cos sin cos sin
cos sin
ax ax
ax ax
ax
ax
e a a e a a
I bx e bx I bx e bx I
b b b b b b b
a e a
I I bx e bx
bb b
− −− −− −− −    
∴ = + − = + −∴ = + − = + −∴ = + − = + −∴ = + − = + −    
    
−−−−
+ = ++ = ++ = ++ = +

27. Indefinite Integration
27
2 2 2
2 2 2
2
2 2 2
2 2 2 2
(1 ) ( ) cos sin
cos sin
cos sin
ax
ax
ax
ax
ax ax
a a b e a
I I bx e bx
bb b b
b e a
I bx e bx
bb a b
b a
e 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 1
sin sin cos
1
sin cos
1
1 sin cos
co
ax
ax
ax
ax ax
ax
ax
ax
ax
ax
e
I e bxdx b bx a bx c
b a
a a e a
J e bx I e bx bx J
b b b b b b
ae a
e bx bx J
b b b
a a a b ae
J J J J e bx bx
bb b b b
J e
= = − + += = − + += = − + += = − + +
++++
    −−−−
= − = − += − = − += − = − += − = − +    
        
= + −= + −= + −= + −
            ++++
+ = + = = ++ = + = = ++ = + = = ++ = + = = +            
                        
∴ =∴ =∴ =∴ =
∫∫∫∫
[[[[ ]]]]2 2
s sin cos
ax
e
bxdx b bx a bx c
b a
= + += + += + += + +
++++
∫∫∫∫
in the integrals sinh , coshax ax
I e bxdx J e bxdx= == == == =∫ ∫∫ ∫∫ ∫∫ ∫ we use the
diffination of the hyperbolic functions sinh , coshbx bx as a functions
of x
e then
( ) ( )
( ) ( )
sinh ,
2 2
cosh ,
2 2
bx bx a b x a b x
ax ax
bx bx a b x a b x
ax ax
e e e e
Ih e bxdx e dx dx
e e e e
Jh e bxdx e dx dx
− + −− + −− + −− + −
− + −− + −− + −− + −
            − −− −− −− −
= = == = == = == = =            
            
            
            + ++ ++ ++ +
= = == = == = == = =            
            
            
∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫
∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫
Example (12): Find 5 22
(1 )
dx
I
x
====
++++
∫∫∫∫

28. Mathematics For Engineering
28
Solution:
consider
5 2
5 2
3 2
3 2
2
2
2
2
(1 )
(1 )
(1 )
(1 )
dx
I x dx
x
dx
J 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 x
I J
dx dx
J 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 2
1 2
3 2
1 2
3 2 2
2
2
2
2
(1 )
(1 )
(1 )
(1 ) 2
3 3 (1 )
dx x x K J
x
J x x c
x
x x x
I c
x
−−−−
−−−−
−−−−
−−−−
= + + −= + + −= + + −= + + −
= + = += + = += + = += + = +
++++
++++
∴ = + +∴ = + +∴ = + +∴ = + +
++++

29. Indefinite Integration
29
Example (13): Find 3
sec 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 xdx
u x dv xdx
du x xdx v x
x dx x x x xdx x x x x dx
x x x x dx x x xdx xdx
x dx x x xdx x x
====
= == == == =
= == == == =
∴ = − = − −∴ = − = − −∴ = − = − −∴ = − = − −
= − − = − += − − = − += − − = − += − − = − +
= + == + == + == + =
∫ ∫∫ ∫∫ ∫∫ ∫
∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫
∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫
∫ ∫∫ ∫∫ ∫∫ ∫
3
ln sec tan
1 1
sec sec tan ln sec tan
2 2
x x c
x dx x x x x c
+ + ++ + ++ + ++ + +
∴ = + + +∴ = + + +∴ = + + +∴ = + + +∫∫∫∫
Example (14): Find 2 2
I x a dx= += += += +∫∫∫∫
Solution:
2 2 2 2
2 2
2 2 2 2
2 2 2 2 2 2
2 2 2 2
2 2 2
2 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
xdx
du v x
x a
x dx x a a dx
I x a dx x x a x x a
x a x a
x a dx a dx
x x a
x a x a
x
x x a x a dx a
a
x
x a dx x x a a
a
x
x a dx x
−−−−
−−−−
= + = + == + = + == + = + == + = + =
= == == == =
++++
+ −+ −+ −+ −
= + = + − = + −= + = + − = + −= + = + − = + −= + = + − = + −
+ ++ ++ ++ +
++++
= + − += + − += + − += + − +
+ ++ ++ ++ +
= + − + += + − + += + − + += + − + +
∴ + = + +∴ + = + +∴ + = + +∴ + = + +
∴ + =∴ + =∴ + =∴ + =
∫∫∫∫
∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫
∫ ∫∫ ∫∫ ∫∫ ∫
∫∫∫∫
∫∫∫∫
∫∫∫∫
2
2 1
sinh
2
a x
a c
a
−−−−
+ + ++ + ++ + ++ + +

30. Mathematics For Engineering
30
Example (15): Find 2 2
I a x dx= −= −= −= −∫∫∫∫
Solution:
2 2
2 2
2 2 2 2
2 2 2 2 2 2
2 2 2 2
2 2 2
2 2
2 2 2 2
2 2 2 2 2 1
2 2 2 2 2 1
2
2 2 2 2 1
( )
( )
sin
2 sin
sin
2 2
let u a x dv dx
xdx
du v x
a x
x dx a x a dx
I x a dx x a x x a x
a x a x
a x dx a dx
x a x
a x a x
x
x a x a x dx a
a
x
a x dx x a x a
a
x a x
a 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
xe x e xe e x
x x x x
x x x x x
−−−−
− − − −− − − −− − − −− − − −
2
2 1 2
3
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
x
x x x x x
x
x x x x
x x x x x
−−−−
− − −− − −− − −− − −
++++

31. 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:
1
m m m-2
1 1
If I sin I sin cos I(1) then show thatm m m
x dx x x
m m
−−−−− −− −− −− −
= = += = += = += = +∫∫∫∫
proof:
1 1
m
1
2
1 2 2
m
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
1
m 2 m
1
m 2
1
m 2
( 1) sin
I sin cos ( 1) ( 1)I
1 ( 1) I sin cos ( 1)
1 ( 1)
I sin cos
m
m
m
m
m
m
m
xdx m xdx
x x m I m
m x x m I
m
x x I
m m
−−−−
−−−−
−−−−
−−−−
−−−−
−−−−
−−−−
− −− −− −− −
= − + − − −= − + − − −= − + − − −= − + − − −
∴ + − = − + −∴ + − = − + −∴ + − = − + −∴ + − = − + −
− −− −− −− −
= += += += +
∫∫∫∫
similarly we can prove that
1
m m m-2
1 1
If I cos I cos sin I(2) then show thatm m m
x dx x x
m m
−−−− −−−−
= = += = += = += = +∫∫∫∫
[[[[ ]]]] [[[[ ]]]]
m m m-1
m m m-1
m m m-1
1
If I I I
1
If 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
m
x e dx x e
a a
m
x a dx x a
a a
x dx x x m
= = −= = −= = −= = −
= = −= = −= = −= = −
= = −= = −= = −= = −
∫∫∫∫
∫∫∫∫
∫∫∫∫

32. Mathematics For Engineering
32
m
1
m m-22
1
m m m-2
2
m m m-2
If I sin
( 1)
I cos sin I
tan
If I tan I I
1
sec tan 2
If I sec I I
1 1
(6)
then show that
(7) then show that
(8) then show that
m
m
m
m
m
m
m
x ax dx
x m m m
ax x ax
a aa
x
dx
m
x x m
dx
m m
−−−−
−−−−
−−−−
====
− −− −− −− −
= + −= + −= + −= + −
= = −= = −= = −= = −
−−−−
−−−−
= = += = += = += = +
− −− −− −− −
∫∫∫∫
∫∫∫∫
∫∫∫∫
m
2
m m-2
n
n
If I cosec
cosec cot 2
I I
1 1
If I cos , sin
I sin . cos
(9) then show that
(10)
show that
m
m
n n
n
n n
n n n
x dx
x x m
m m
x bx dx J x bx dx
x bx nJ J x bx n I
−−−−
====
− −− −− −− −
= += += += +
− −− −− −− −
= == == == =
= − = −= − = −= − = −= − = −
∫∫∫∫
∫ ∫∫ ∫∫ ∫∫ ∫
Exercise(4)
If
if and prove that
find
Find 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 n
n n
n n
n n n n
m n m m
m n
m n
n n
n 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
− −− −− −− −
− +− +− +− +
= == == == =
= − = −= − = −= − = −= − = −
====
= == == == =
∫ ∫∫ ∫∫ ∫∫ ∫
∫∫∫∫
∫ ∫∫ ∫∫ ∫∫ ∫
provethat
Find thereducionformulafor
are positive integer.
1
2
1
2
2 2
sin cos ( 1) .
cos sin ( 1) .
1-1sin sin( 1) sin sin
1 ,
,
(4)
(5)
n
n n
n
n n
n
m
I x x n I
nJ x x n J
n nx n x dx x nx
n
x x dx
m n
−−−−
−−−−
−−−−
−−−−
= + −= + −= + −= + −
= − + −= − + −= − + −= − + −
+ =+ =+ =+ =∫∫∫∫
    ++++
    ∫∫∫∫

33. Indefinite Integration
33
find
find
find
find
Findthereducionformulafor
Findthereducionformulafor
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 2
1 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
xdx xdx
dx
a b x
I
∫∫∫∫
− −− −− −− −− −− −− −− −= + −= + −= + −= + −
−−−−+ − − − −+ − − − −+ − − − −+ − − − −
∫ ∫∫ ∫∫ ∫∫ ∫
==== ∫∫∫∫
++++

34. 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 cos2
2
1
(5) cos 1 cos2
2
(6) sin2 2sin cos
(7) 1 cos2 2sin
(8)1 cos2 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
+ =+ =+ =+ =
+ =+ =+ =+ =
+ =+ =+ =+ =
= −= −= −= −
= += += += +
====
− =− =− =− =
+ =+ =+ =+ =
= − + += − + += − + += − + +
= − − += − − += − − += − − +
[[[[ ]]]]
1
11) cos cos cos( ) cos( )
2
x y x y x y= − + += − + += − + += − + +
(1)Integrals in the form sinm
x dx∫∫∫∫
if m is odd positive integer
1
sin sin (sin )m m
x dx x xdx−−−−
====∫ ∫∫ ∫∫ ∫∫ ∫ 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 1
sin 1 cos2 ,cos 1 cos2
2 2
x x x x= − = += − = += − = += − = +

35. Indefinite Integration
35
Also we can use the following reduction formula
1
m m-2
1 1
I sin sin cos Im m m
xdx x x
m m
−−−−− −− −− −− −
= = += = += = += = +∫∫∫∫
Example (1): Find 3
sin xdx∫∫∫∫
Solution:
3 2 2
3 2 3 3
sin sin (sin ) (1 cos )(sin )
cos sin
1 1
sin (1 ) cos cos
3 3
xdx x xdx x xdx
let y x dy xdx
xdx y dy y y x x c
= = −= = −= = −= = −
= = −= = −= = −= = −
            
∴ = − − = − − = − − +∴ = − − = − − = − − +∴ = − − = − − = − − +∴ = − − = − − = − − +            
            
∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫
∫ ∫∫ ∫∫ ∫∫ ∫
Example (2): Find 5
sin xdx∫∫∫∫
Solution:
5 4 2 2
5 2 2 2 2 3
2 3
sin sin (sin ) (1 cos ) (sin )
cos sin
1
sin (1 ) (1 2 )
3
1
cos cos cos
3
xdx x xdx x xdx
let y x dy xdx
xdx y dy y y dy y y y
x x x c
= = −= = −= = −= = −
= = −= = −= = −= = −
    
∴ = − − = − − + = − − +∴ = − − = − − + = − − +∴ = − − = − − + = − − +∴ = − − = − − + = − − +    
    
    
= − − + += − − + += − − + += − − + +    
    
∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫
∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫
Example (3): Find 4
sin xdx∫∫∫∫
Solution:
4 2 2 2
2
1
(3) sin sin sin (1 cos2 )
4
1 1 1
1 2cos2 cos 2 1 2cos2 (1 cos4 )
4 4 2
1 3 1 1 3 1
2cos2 cos4 sin2 sin4
4 2 2 4 2 8
xdx x xdx x dx
x x dx x x dx
x
x x dx x x c
= = −= = −= = −= = −
        = − + = − + −= − + = − + −= − + = − + −= − + = − + −             
            
= − − = − − += − − = − − += − − = − − += − − = − − +            
            
∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫
∫ ∫∫ ∫∫ ∫∫ ∫
∫∫∫∫

36. Mathematics For Engineering
36
(2)Integrals in the form cosm
x dx∫∫∫∫
(i) if m is odd positive integer 1
cos cos (cos )m m
x dx x xdx−−−−
====∫ ∫∫ ∫∫ ∫∫ ∫ 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 1
sin 1 cos2 ,cos 1 cos2
2 2
x x x x= − = += − = += − = += − = +
(iii) we can use the successive formula.
1
m m-2
1 1
I cos cos sin Im m m
xdx x x
m m
−−−− −−−−
= = += = += = += = +∫∫∫∫
Example (4): Find 3
cos xdx∫∫∫∫
Solution:
3 2 2
3 2 3 3
cos cos (cos ) (1 sin )(cos )
sin cos
1 1
cos (1 ) sin sin
3 3
xdx x xdx x xdx
let y x dy xdx
xdx y d y y y c x c
= = −= = −= = −= = −
= == == == =
            
∴ = − = − + = − +∴ = − = − + = − +∴ = − = − + = − +∴ = − = − + = − +            
            
∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫
∫ ∫∫ ∫∫ ∫∫ ∫
Example (5): Find 5
cos xdx∫∫∫∫
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 1
sin sin sin
3 3
xdx x xdx x xdx
let y x dy xdx
xdx y dy y y dy
y y y x x x c
= = −= = −= = −= = −
= == == == =
∴ = − = − +∴ = − = − +∴ = − = − +∴ = − = − +
            
= − + = − + += − + = − + += − + = − + += − + = − + +            
            
∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫
∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫
Example (6): Find 4
cos xdx∫∫∫∫
Solution:
4 2 2 21
cos cos cos (1 cos2 )
4
x dx x x dx x dx= = += = += = += = +∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫

37. Indefinite Integration
37
21 1 1
1 2cos2 cos 2 1 2cos2 (1 cos4 )
4 4 2
1 3 1 1 3 1
2cos2 cos4 sin2 sin4
4 2 2 4 2 8
x x dx x x dx
x
x 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) 2
tan tan (tan ) (sec 1) (tan )
sec sec tan
m m m
x dx x xdx x x dx
put y x dy x xdx
− −− −− −− −
= = −= = −= = −= = −
= → == → == → == → =
∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫
(ii)if m is even odd positive integer use
2 2 2 2 2
tan tan (sec 1) tan sec tanm m m m
x dx x x dx x xdx xdx− − −− − −− − −− − −
= − = −= − = −= − = −= − = −∫ ∫ ∫ ∫∫ ∫ ∫ ∫∫ ∫ ∫ ∫∫ ∫ ∫ ∫
(reduction formula)
Example (7): Find 5
tan xdx∫∫∫∫
Solution:
5 3 2 3 2
3 2 3
3 2 2
3 2 2
4 2
2
tan , sec
tan tan (tan ) tan (sec 1)
tan sec tan
tan sec tan (sec 1)
tan sec (tan sec tan )
1 1
tan tan ln sec
4 2
let y x dy xdx
xdx x x dx x x dx
x xdx xdx
x xdx x x dx
x xdx x x x dx
x x x c
= == == == =
= = −= = −= = −= = −
= −= −= −= −
= − −= − −= − −= − −
= − −= − −= − −= − −
= − + += − + += − + += − + +
∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫
∫ ∫∫ ∫∫ ∫∫ ∫
∫ ∫∫ ∫∫ ∫∫ ∫
∫ ∫∫ ∫∫ ∫∫ ∫
Example (8):Find 6
tan xdx∫∫∫∫
Solution:
6 4 2 4 2
tan tan (tan ) tan (sec 1)x dx x x dx x x dx= = −= = −= = −= = −∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫

38. 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 1
tan 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-2
tan
If I tan Then I I
( 1)
m
m
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 2
cot cot (cot ) (cosec 1) (cot )
cosec cosec cot
m m m
x dx x xdx 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 xdx xdx
−−−−
− −− −− −− −
= −= −= −= −
= −= −= −= −
∫ ∫∫ ∫∫ ∫∫ ∫
∫ ∫∫ ∫∫ ∫∫ ∫
In general we can use the reduction formula
1
m m m-2
cot
If I cot Then I I
( 1)
m
m
axdx
a m
−−−−
= = − −= = − −= = − −= = − −
−−−−
∫∫∫∫
Example (9): Find 4
cot 3xdx∫∫∫∫
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
= = −= = −= = −= = −
= −= −= −= −
∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫
∫ ∫∫ ∫∫ ∫∫ ∫

39. Indefinite Integration
39
2 2 2
2 2 2
3
cot 3 cosec 3 (cosec 3 1)
cot 3 cosec 3 cosec 3
1 1
cot 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) 2
sec sec (sec ) (tan 1) (sec )
sec sec tan
m m m
x dx x xdx 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 xdx
put y x and x x
−−−−
−−−−
====
= = += = += = += = +
∫ ∫∫ ∫∫ ∫∫ ∫
to get the reduction formula
2
m m-2
sec tan 2
I sec I
1 1
m
m x m
dx
m m
−−−−
−−−−
= = += = += = += = +
− −− −− −− −
∫∫∫∫
Example (10): Find 4
sec 2xdx∫∫∫∫
Solution:
4 2 2 2 2
2 2 2 3
(10) sec 2 sec 2 (sec 2 ) sec 2 (1 tan 2 )
1 1
sec 2 sec 2 tan 2 tan2 tan 2
2 6
xdx x x dx x x dx
x dx x xdx x x c
= = += = += = += = +
= + = + += + = + += + = + += + = + +
∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫
∫ ∫∫ ∫∫ ∫∫ ∫
(6)Integrals in the form: cosecm
x dx∫∫∫∫
(i) if m is odd positive integer then m-1 is even

40. Mathematics For Engineering
40
1 2 ( 1) 2
cosec cosec (cosec ) (cot 1) (cosec )
cosec cosec cot
m m m
x dx x xdx 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 xdx
put y x and x x
−−−−
−−−−
====
= = += = += = += = +
∫ ∫∫ ∫∫ ∫∫ ∫
to get the reduction formula
Example (11): Find 6
cosec axdx∫∫∫∫
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 1
cot cot cot
3 5
axdx ax ax dx ax ax dx
ax ax ax dx
ax ax ax ax ax dx
ax ax ax c
a a a
= = += = += = += = +
= + += + += + += + +
= + += + += + += + +
−−−−
= − − += − − += − − += − − +
∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫
∫∫∫∫
∫∫∫∫
(7)Integrals in the form cos sinm n
x x dx∫∫∫∫ , , are positive integersm n
If m is an odd
1
cos sin cos sin (cos ) and put sinm n m n
x x dx x x xdx y x−−−−
= == == == =∫ ∫∫ ∫∫ ∫∫ ∫
If n is an odd
1
cos sin cos sin (sin ) andput cosm n m n
x x dx x x xdx y x−−−−
= == == == =∫ ∫∫ ∫∫ ∫∫ ∫
if m and n are both an even we use
[[[[ ]]]] [[[[ ]]]]2 21 1
sin 1 cos2 ,cos 1 cos2
2 2
x x x x= − = += − = += − = += − = +
Example (12): Find 2 2
sin cosx xdx∫∫∫∫
Solution:
(((( ))))
2
22 2 1
(12) sin cos sin cos sin2
2
x x dx x x dx x dx
    
= == == == =     
    
∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫

41. Indefinite Integration
41
21 1 1 1 1
sin 2 (1 cos4 ) ( sin4 )
4 4 2 8 4
x dx x dx x x c= = − = − += = − = − += = − = − += = − = − +∫ ∫∫ ∫∫ ∫∫ ∫
Example (13): Find 3 4
sin cosax axdx∫∫∫∫
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 1
cos cos
5 7
ax axdx ax ax axdx
ax ax axdx
ax ax axdx
ax axdx ax axdx
ax ax c
a a
====
= −= −= −= −
= −= −= −= −
= −= −= −= −
−−−−
= + += + += + += + +
∫ ∫∫ ∫∫ ∫∫ ∫
∫∫∫∫
∫∫∫∫
∫∫∫∫
Example (14): Find 4 2
sin cosax ax dx∫∫∫∫
Solution:
[[[[ ]]]]
2
4 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
1
sin 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 c
a a
x ax ax c
a a
    
−−−−    
    
    
= − − += − − += − − += − − +    
    
= − − += − − += − − += − − +
∫ ∫∫ ∫∫ ∫∫ ∫

42. Mathematics For Engineering
42
The reduction formula for cos sinm n
x 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
d
x 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.to x 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 n
m n m n
x x
n x x dx m n x x dx
n I m n I
n
I x x dx x x I
m n m n
m n
+ −+ −+ −+ −
−−−−
−−−−
+ −+ −+ −+ −
−−−−
    
    
= − − + += − − + += − − + += − − + +
= − − + += − − + += − − + += − − + +
−−−−
= = −= = −= = −= = −
+ ++ ++ ++ +∴∴∴∴
∫ ∫∫ ∫∫ ∫∫ ∫
∫∫∫∫
Example(15):
4 6 5 5
4,6 4,4
5 5 5 3
4,2
5 5 5 3 5
4,0
4
4
1 1
sin cos sin cos
10 2
1 1 1 3
sin cos sin cos
10 2 8 8
1 1 1 3 1 1
sin cos sin cos sin cos
10 2 8 8 6 6
3 1 1
sin sin2 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
= = −= = −= = −= = −
    
= − −= − −= − −= − −    
    
        
= − − −= − − −= − − −= − − −        
        
= = − += = − += = − += = − +
∫∫∫∫
∫∫∫∫

43. Indefinite Integration
43
5 5
4,6
5 3 5
5 5 5 3 5
1
sin cos
10
1 1 3 1 1 3 1 1
sin cos sin cos sin2 sin4
2 8 8 6 6 8 4 23
1 1 3 3
sin cos sin cos sin cos
10 16 32 256
1 3
sin2 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
d
x 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 n
m n m n
m n m n
m n m n
x x m n x x dx m x x dx
x x m n I m I
m n
I x x dx x x I
m m
m n
+ − ++ − ++ − ++ − +
+ −+ −+ −+ −
++++
+ ++ ++ ++ +
++++
     = − + + + += − + + + += − + + + += − + + + +
    
= − + + + += − + + + += − + + + += − + + + +
+ ++ ++ ++ +
= = += = += = += = +
+ ++ ++ ++ +∴∴∴∴
∫ ∫∫ ∫∫ ∫∫ ∫
∫∫∫∫
Example(16):

44. Mathematics For Engineering
44
3 2
4, 3 ( 2, 3)4 3
( 2, 3)3 2
1 5
sin cos
3 3sin cos
1 5
33sin cos
dx
I x x I
x x
I
x x
− −− −− −− −
− − − −− − − −− − − −− − − −
− −− −− −− −
−−−−
= = += = += = += = +
−−−−
= += += += +
∫∫∫∫
1 2
2, 3 0, 3 0, 32
3 2 2
0, 3 3
2 3
0, 3
2, 3
1
(sin ) cos 3 3
sin cos
sec sec sec sec tan sec tan
cos
sec tan sec (1 sec ) sec tan sec sec
1 1
sec tan ln sec tan
2 2
1
sin
I x x I I
x x
dx
I xdx x xdx 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 3
sec tan ln sec tan
2 2cos
1 5 5 5
sec tan ln sec tan
3 33sin cos 3sin cos
x x x x
x x
I x x x x c
x 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
d
x 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 n
m n m n
x x
n x x dx m x x dx
n I m I
m
I x x dx x x I
n n
m n
− +− +− +− +
− +− +− +− +
− +− +− +− +
− +− +− +− +
− =− =− =− =
    
    
= − + + −= − + + −= − + + −= − + + −
= − + + −= − + + −= − + + −= − + + −
− −− −− −− −
= = += = += = += = +
+ ++ ++ ++ +∴∴∴∴
∫ ∫∫ ∫∫ ∫∫ ∫
∫∫∫∫

45. Indefinite Integration
45
Example(17):
6 5
5 7
(6, 8) (4, 6) (4, 6)8 7
3
3 5
(4, 6) (2, 4) (2, 4)5
3
(2, 4) (0, 2) (0, 2)3
(0, 2) 2
sin 1 5 1 sin 5
sin cos
7 7 7 7cos cos
1 3 sin 3
sin (cos )
5 5 55cos
1 1 sin 1
sin (cos )
3 3 33cos
cos
x dx x
I x x I I
x x
x
I x x I I
x
x
I x x I I
x
dx
I
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 1
tan , tan
3 53cos 5cos 5cos
sin 1 sin sin sin 1
tan
7 7cos cos 7cos 7cos
x dx x
x x x
I x I x
x x x
x dx x x x
I x
x x x x
− −− −− −− −
−−−−
= == == == =
∴ = − = − +∴ = − = − +∴ = − = − +∴ = − = − +
−−−−
= = − + −= = − + −= = − + −= = − + −
∫ ∫∫ ∫∫ ∫∫ ∫
∫∫∫∫
we can solve this example without reduction formulas
6
6 2 7
8
sin 1
tan sec tan
7cos
x dx
x 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
d
x 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 n
m n m n
x x n x x dx m x x dx
n I m I
n
I x x dx x x I
m m
m n
+ − + −+ − + −+ − + −+ − + −
+ −+ −+ −+ −
+ −+ −+ −+ −
+ −+ −+ −+ −
     = − − + += − − + += − − + += − − + +
    
= − − + += − − + += − − + += − − + +
−−−−
= = += = += = += = +
+ ++ ++ ++ +∴∴∴∴
∫ ∫∫ ∫∫ ∫∫ ∫
∫∫∫∫

46. 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 3
cos sin
5 5 5sin 5sin
1 1 cos 1
cos sin
3 3 33sin
csec cot
sin
cos 1
cot
33sin
cos
sin
x dx x
I x x I I
x x
x
I x x I I
x
dx
I x dx x
x
x
I
x
x dx
I
x
−−−−
− − −− − −− − −− − −
−−−−
− − −− − −− − −− − −
−−−−
−−−−
−−−−
−−−−
= = − − = −= = − − = −= = − − = −= = − − = −
−−−−
= + = −= + = −= + = −= + = −
− −− −− −− −
= = = −= = = −= = = −= = = −
−−−−
∴ = +∴ = +∴ = +∴ = +
====
∫∫∫∫
∫ ∫∫ ∫∫ ∫∫ ∫
3
) 5 3
cos cos 1
cot
55sin 5sin
x x
c
x x
−−−−
= + − += + − += + − += + − +∫∫∫∫
we can solve this example without reduction formulas
4
4 2 5
6
cos 1
cot cosec cot
5sin
x dx
x x dx x c
x
−−−−
= = += = += = += = +∫ ∫∫ ∫∫ ∫∫ ∫
(8)Integrals in the form sec tanm n
x x dx∫∫∫∫ , , are positive integersm n
If n is an odd and m either an even or odd
1 1
sec tan sec tan (sec tan )
andput sec
m n m n
x x dx x x x xdx
y x
− −− −− −− −
====
====
∫ ∫∫ ∫∫ ∫∫ ∫
If m is an even ,n either an even or odd
2 2
sec tan sec tan (sec ) andput tanm n m n
x x dx x x xdx y x−−−−
= == == == =∫ ∫∫ ∫∫ ∫∫ ∫
Example (19): Find 4 3
sec 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
====
= −= −= −= −
= −= −= −= −
∫ ∫∫ ∫∫ ∫∫ ∫
∫∫∫∫
∫∫∫∫

47. Indefinite Integration
47
5 3 6 4 6 4
tan3 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 2
sec tanx x dx∫∫∫∫
Solution:
3 2 3 2 5 3
5 3
5 3 2 3
5
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
3
5 3
3 2
3
2
2 3
sec sec )
1
sec 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 )
1
sec 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 3
3 3 3
3
3
1 1
(1),(2) (3) sec tan sec tan 3
4 4
1 3
sec tan sec tan ln sec tan
4 2
1 3
sec 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
            = + − = −= + − = −= + − = −= + − = −
            
    
= − + += − + += − + += − + +    
    
    
= − + + += − + + += − + + += − + + +    
    

48. Mathematics For Engineering
48
Example (21): Find 3 3
sec 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 c
a a a
====
= −= −= −= −
= −= −= −= −
= ∴ == ∴ == ∴ == ∴ =
            
∴ = − = − = − +∴ = − = − = − +∴ = − = − = − +∴ = − = − = − +            
            
∫ ∫∫ ∫∫ ∫∫ ∫
∫∫∫∫
∫∫∫∫
∫∫∫∫
(9)Integrals in the form cosec cotm n
x x dx∫∫∫∫ , , are positive integersm n
If n is an odd and m either an even or odd
1 1
cosec cot cosec cot (cosec cot )
andput cosec
m n m n
x x dx x x x xdx
y x
− −− −− −− −
====
====
∫ ∫∫ ∫∫ ∫∫ ∫
If m is an even ,n either an even or odd
2 2
cosec cot cosec cot (cosec )
andput cot
m n m n
x x dx x x xdx
y x
−−−−
====
====
∫ ∫∫ ∫∫ ∫∫ ∫
Example (22): Find 3 3
cot 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 1
cot 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 axdx
ax ax dx y y dy y y
a a
====
= −= −= −= −
= −= −= −= −
= = −= = −= = −= = −
∴ = − − = − +∴ = − − = − +∴ = − − = − +∴ = − − = − +
= −= −= −= −
∫ ∫∫ ∫∫ ∫∫ ∫
∫∫∫∫
∫∫∫∫
∫ ∫∫ ∫∫ ∫∫ ∫
3 51
cosec cosec
3 5
ax ax c
a a
+ ++ ++ ++ +

49. Indefinite Integration
49
Example (23): Find 3 4
cot 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 1
cot cosec ( y ) y
4 6
1 1
cot 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 y
a a
ax
a 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
xdx xdx xdx
x xdx xdx
x xdx
x xdx
x xdx
∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫
∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫
∫∫∫∫
∫∫∫∫
∫∫∫∫

50. 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 xdx
x
∫ ∫∫ ∫∫ ∫∫ ∫
∫ ∫∫ ∫∫ ∫∫ ∫
∫ ∫∫ ∫∫ ∫∫ ∫
∫ ∫∫ ∫∫ ∫∫ ∫
∫ ∫∫ ∫∫ ∫∫ ∫
∫ ∫∫ ∫∫ ∫∫ ∫
3 5
3 2
3
3
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
x
dx 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
∫∫∫∫
∫∫∫∫

51. 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:
2
2 2 2 2 2
2
2
2 2 2 2 2 2 2
2
For sin , sin cos
( sin ) 1 sin cos
(i)
a a a
a b x put x x dx d
b bb
a
a 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
− = = =− = = =− = = =− = = =
∴ − = − = − =∴ − = − = − =∴ − = − = − =∴ − = − = − =
2
2 2 2 2 2 2
2
2
2 2 2 2 2 2 2 2
2
For tan , tan sec
( tan ) 1 tan sec
(ii)
a a a
a b x put x x dx d
b bb
a
a 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
+ = = =+ = = =+ = = =+ = = =
∴ + = + = + =∴ + = + = + =∴ + = + = + =∴ + = + = + =
2
2 2 2 2 2
2
2
2 2 2 2 2 2 2
2
For sec , sec sec tan
( sec ) sec 1 tan
(iii)
a a a
b x a put x x dx d
b bb
a
b 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
2
2 2
2
2 2 2
(iv)For , tan
sin cos 2
2
sin 2sin cos ,
2 2 1
1
cos cos sin ,
2 2 1
1 1 1
sec (1 tan ) (1 )
2 2 2 2 2
dx dx x
put u
a b x a b x
x x u
x
u
x x u
x
u
x x
du dx dx u dx
====
± ±± ±± ±± ±
∴ = =∴ = =∴ = =∴ = =
++++
−−−−
= − == − == − == − =
++++
= = + = += = + = += = + = += = + = +
∫ ∫∫ ∫∫ ∫∫ ∫

52. Mathematics For Engineering
52
2
2 2 2
2 2 1
, sin , cos
1 1 1
du u u
dx x x
u u u
−−−−
∴ = = =∴ = = =∴ = = =∴ = = =
+ + ++ + ++ + ++ + +
2
2
2 2
2
2 2 2
2
2 2 2
(v)For , tanh
sinh cosh 2
2
sinh 2sinh cosh ,
2 2 1
1
cosh cosh sinh ,
2 2 1
1 1 1
sech (1 tanh ) (1 )
2 2 2 2 2
2 2 1
, sinh , cosh
1 1 1
dx dx x
put u
a b x a b x
x x u
x
u
x x u
x
u
x x
du dx dx u dx
du u u
dx x x
u u u
====
± ±± ±± ±± ±
∴ = =∴ = =∴ = =∴ = =
−−−−
++++
= + == + == + == + =
−−−−
= = − = −= = − = −= = − = −= = − = −
++++
∴ = = =∴ = = =∴ = = =∴ = = =
− − −− − −− − −− − −
∫ ∫∫ ∫∫ ∫∫ ∫
Example(1): Find
2
2
4
x dx
x −−−−
∫∫∫∫
Solution:
(((( ))))
2 2 2 2
2 2
3
2
2 2 2 2
2sec 2sec tan ,
1
4 4sec 4 2tan , tan sec 1 4
2
4sec (2sec tan )
4sec
2tan4
2sec tan 2ln sec tan
1 1 1
2. . 4 2ln 4 4 2ln 4
2 2 2 4 2 2 4
put x dx d
x x
x dx d
d
x
x x x x
x 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 θ
θθθθ
θ θ θ θθ θ θ θθ θ θ θθ θ θ θ
= ∴ == ∴ == ∴ == ∴ =
− = − = = − = −− = − = = − = −− = − = = − = −− = − = = − = −
∴ = =∴ = =∴ = =∴ = =
−−−−
= + += + += + += + +
            
= − + + − = − + + − += − + + − = − + + − += − + + − = − + + − += − + + − = − + + − +            
            
∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫

53. Indefinite Integration
53
Example(2): Find
2
9
xdx
x−−−−
∫∫∫∫
Solution:
[[[[ ]]]]
(((( ))))
2 2 2
2 2
2
2
2
1 1 2
3sin 3cos , 9 9 (3sin ) 3 1 sin 3cos
9sin 3cos 1
9 sin 9 1 cos2
3cos 29
9 1 9 1
sin2 2sin cos
2 2 2 2
9 9 9
sin sin 9
2 3 3 3 2 3 2
let x dx d x
x dx d
d d
x
x x x x x
x 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 θ
θθθθ
θ θ θ θ θθ θ θ θ θθ θ θ θ θθ θ θ θ θ
Example(3): Find
2
9 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 1
cosec ln cosec cot (1)
3 3
2 3 9 4
2 3tan tan ,cot , cosec
3 2 2
1
ln cosec
3
x dx d
x
dx d d d
x x
d
x x
x
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 θ
θ 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
θθθθ
2
1 9 4 3
cot ln
3 2 2
x
c
x x
++++
− = − +− = − +− = − +− = − +θθθθ
Example(3): Find
5 4cos
dx
x++++∫∫∫∫
Solution:

54. Mathematics For Engineering
54
(((( ))))
(((( ))))
2
2 2
2
2
2 2 2
2
2
1 1
2
tan
2
1
cos cos sin ,
2 2 1
2
1
2 2
5 4cos 1 5(1 ) 4(1 )
(1 ) 5 4
1
tan
2 2 2 2tan tan
3 3 3 39
x
put u
x x u
x
u
du
dx
u
dx du du
x u u u
u
u
x
du u
c
u
− −− −− −− −
====
−−−−
∴ = − =∴ = − =∴ = − =∴ = − =
++++
====
++++
= == == == =
++++     −−−− + + −+ + −+ + −+ + −
+ ++ ++ ++ +        ++++    
    
    
= = = += = = += = = += = = +    
++++     
    
∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫
∫∫∫∫
Example(3): Find
12 13sin
dx
x++++∫∫∫∫
Solution:
2 2
2
2
2
2 2
2 2
tan sin 2sin cos ,
2 2 2 1 1
2 2
12 13sin 2 12(1 ) 13(2 )1 12 13
1
2
(2 3)(3 2)12 12 26 6 13 6
2 1 1 1
ln 2 3
15 2 3 5 3 2 15 1
x x x u du
put u x dx
u u
dx du du
x u u uu
u
du du du
u uu u u u
du du
u
u u
= ∴ = = == ∴ = = == ∴ = = == ∴ = = =
+ ++ ++ ++ +
= == == == =
++++             + ++ ++ ++ +    + ++ ++ ++ +              ++++    
= = == = == = == = =
+ ++ ++ ++ +            + + + ++ + + ++ + + ++ + + +
            
− −− −− −− −
= + = + += + = + += + = + += + = + +
+ ++ ++ ++ +
∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫
∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫
∫ ∫∫ ∫∫ ∫∫ ∫ ln 3 2
5
1 1
ln 2tan 3 ln 3tan 2
15 2 15 2
u
x x
c
++++
−−−−
= + + + += + + + += + + + += + + + +

55. Indefinite Integration
55
Exercise(6)
Prove that
2 3/ 2 2
2 2
2
2 2
22 2 2
2 2 2
2
1
2 2 3/ 2 2 2
2 2 2
2 2
2 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)
x
dx c
x x
x x
dx x c
x c
a x
dx c
a xx a x
x
x dx x x x c
x x x
dx c
aa x a x
x
x dx x x x c
x a
dx 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
ln
2
(8)
(4 ) 12(4 )
(9)
( ) ( )
9
(10)
99
a x a a
c
x a a
x dx x
c
x x
dx x
c
a x a a x
dx x
c
xx x
+ −+ −+ −+ −
+ ++ ++ ++ +
+ ++ ++ ++ +
= += += += +
− −− −− −− −
= += += += +
++++ ++++
−−−−
= − += − += − += − +
−−−−
∫∫∫∫
∫∫∫∫
∫∫∫∫
2
2 2
2
2
2 2 2 2 2 5/ 2 2 2 3/ 2
1
(11) 16 8ln 16
216
1
(12) ( ) ( )
5 3
x dx
x x x x c
x
a
x a x dx a x a x c
    = − + + − += − + + − += − + + − += − + + − +
        −−−−
− = − − − +− = − − − +− = − − − +− = − − − +
∫∫∫∫
∫∫∫∫

56. Mathematics For Engineering
56
2
2
2 3/ 2 2
(13) ln( 2 4 13)
4 13
2
(14)
(4 ) 4 4
dx
dx x x x c
x x
dx x
c
x x x x
= − + − + += − + − + += − + − + += − + − + +
− +− +− +− +
−−−−
= += += += +
−−−− −−−−
∫∫∫∫
∫∫∫∫
1
2 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 4
1 1
(17) cos (2 1)cos 1
4 4
dx x x
x
x x
x x dx x x x x c
x x dx x x x x c
−−−−
− −− −− −− −
− −− −− −− −
= + += + += + += + +
+ ++ ++ ++ +
= − + − += − + − += − + − += − + − +
= − − − += − − − += − − − += − − − +
∫∫∫∫
∫∫∫∫
∫∫∫∫
sech 1
2 2 2
1
2 2 2
2
2
2
2
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 bx
c
a ax a b x
dx bx
c
ab aa b x
dx x
c
xx x
xdx x x x
x x c
x x
x
x x dx x x x c
dx
x
−−−−
−−−−
−−−−
−−−−
= += += += +
−−−−
= += += += +
++++
    + −+ −+ −+ −
    = += += += +
    ++++     
    − + − −− + − −− + − −− + − −
    = − − + += − − + += − − + += − − + +
    − −− −− −− −     
−−−−
− = + − − +− = + − − +− = + − − +− = + − − +
====
++++
∫∫∫∫
∫∫∫∫
∫∫∫∫
∫∫∫∫
∫∫∫∫
∫∫∫∫
12 1
tan (1 2tan )
23 3
x−−−−     
++++    
    
find
2
2 3/ 2 2
2
2 3/ 2
(24) (25)
(2 4) 4
(26) (27) 4 13
( 4 5)
x dx dx
x x x
dx
x x dx
x x
++++ −−−−
+ ++ ++ ++ +
− −− −− −− −
∫ ∫∫ ∫∫ ∫∫ ∫
∫ ∫∫ ∫∫ ∫∫ ∫

57. 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 x
dx x dx
dx
x x
++++++++
−−−−
− +− +− +− +
∫ ∫∫ ∫∫ ∫∫ ∫
∫ ∫∫ ∫∫ ∫∫ ∫
sec
(32) (33)
cosh 2sinh 3 2tan
sec (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) (43
4 5cos
dx x dx
x x x
xdx x dx
x x x
dx dx
x x
dx dx
x x
dx dx
x x x x
dx
x
+ −+ −+ −+ −
++++
+ − ++ − ++ − ++ − +
+ −+ −+ −+ −
+ −+ −+ −+ −
+ −+ −+ −+ −
++++
∫ ∫∫ ∫∫ ∫∫ ∫
∫ ∫∫ ∫∫ ∫∫ ∫
∫ ∫∫ ∫∫ ∫∫ ∫
∫ ∫∫ ∫∫ ∫∫ ∫
∫ ∫∫ ∫∫ ∫∫ ∫
∫∫∫∫ )
5 3cos
(44) (45)
4 5sin 5 3sin
(46) (47)
4 5cosh 5 3cosh
(48) (49)
4 5sinh 5 3sinh
dx
x
dx dx
x x
dx dx
x x
dx dx
x x
−−−−
+ −+ −+ −+ −
+ −+ −+ −+ −
+ −+ −+ −+ −
∫∫∫∫
∫ ∫∫ ∫∫ ∫∫ ∫
∫ ∫∫ ∫∫ ∫∫ ∫
∫ ∫∫ ∫∫ ∫∫ ∫

58. Mathematics For Engineering
58
Integration by Partial Fraction:
A function
( )
( )
( )
f x
F x
g 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 2
1 1
x x
x
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 n
ax 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
A
ax 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 2
2
...
( ) ( )
n
n
AA 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 2
ax bx c+ ++ ++ ++ + occurring once in the
denominator of a proper rational fraction, there corresponds a single

59. Indefinite Integration
59
partial fraction of the form 2
Ax B
ax bx c
++++
+ ++ ++ ++ +
, where ,A B is a constants to
be determined.
CASE 4.Repeated linear factors
To each linear factor 2
ax 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 2
2 2 2 2
...
( ) ( )
n n
n
A 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 2
1
(1)
2 24
A B
x xx
= += += += +
− +− +− +− +−−−−
and clear the fraction to obtain
2 2
1 ( 2) ( 2)
2 24 4
A B A x B x
x 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
= += += += +
= = −= = −= = −= = −

60. Mathematics For Engineering
60
method 2:Short method : Substitute in (2) the value of 2x ==== to find
A,where
1
1 4
4
A A= → == → == → == → =
Substitute in (2) the value of 2x = −= −= −= − to find B ,where
1
1 4
4
B B
−−−−
= − → == − → == − → == − → =
substitute in (1) the value of ,A B we have:
2
2
1 (1/ 4) ( 1/ 4) 1 1 1
2 2 4 2 24
1 1 1 1 1
and
4 2 2 4 2 4 24
1 1 2
ln ( 2) ln ( 2) ln
4 4 2
x x x xx
dx dx dx
x x x xx
x
x 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 C
x 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 C
x x x x x x
A x x Bx x Cx x
x 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

61. Indefinite Integration
61
2
1 ( ) ( 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 2
6 10 2 15 36
21 3 2
ln ln 2 ln 3 ln
6 10 15 3
x dx dx dx dx
x x xx x x
x
x 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 C
x 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 x
x 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= −= −= −= − ⇒⇒⇒⇒ ==== ⇒⇒⇒⇒ ====

62. 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)
x
x xx x x x
+ −+ −+ −+ −
∴ = + +∴ = + +∴ = + +∴ = + +
+ −+ −+ −+ −− − + −− − + −− − + −− − + −
3 2 2
(3 5) 1/ 2 1/ 2 4
1 11 ( 1)
1 1 4 1 1 4
ln 1 ln 1 ln
2 2 ( 1) 2 1 1
x dx
and dx dx dx
x xx x x x
x
x 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 x
x x
x x x x x x
x A B C
x 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 2
1 2 1 2
( 1)( 1)
x
x xx x x
+ − −+ − −+ − −+ − −
= + += + += + += + +
−−−−−−−−
4 3
3 2 2 2
2
2
2
( 1) 1 2 1 2
( 1)( 1)
2 1 2 1 1
2ln 2ln 1
( 1) 2
1 1
2ln
2 1
x x x x
dx x dx x dx
x xx x x x x
x dx x x x
x x xx
x
x c
x x
             − − − + − −− − − + − −− − − + − −− − − + − −
= − = − + += − = − + += − = − + += − = − + +             
−−−−− −− −− −− −                 
    −−−−
+ + + = + − − −+ + + = + − − −+ + + = + − − −+ + + = + − − −    −−−−    
= − + += − + += − + += − + +
−−−−
∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫
∫∫∫∫

63. 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 2
3 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 2
1 2
4 2 2 2
( 2) 1
tan ln 2
23 2 1 2
and
x x x dx x dx
dx 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
+ + ++ + ++ + ++ + +
= += += += +
+ + ++ + ++ + ++ + +
+ = + + + + = + + + + ++ = + + + + = + + + + ++ = + + + + = + + + + ++ = + + + + = + + + + +
∴ = = + = + =∴ = = + = + =∴ = = + = + =∴ = = + = + =
= = = == = = == = = == = = =
++++
= += += += +
+ + ++ + ++ + ++ + +

64. Mathematics For Engineering
64
2
2 2 2 2 2
(2 3) 2 1
( 1) ( 1) ( 1)
x
dx dx dx
x x x
++++
= += += += +
+ + ++ + ++ + ++ + +
∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫
2
2 2
2
2
2 2 4
1
To solve tan sec ,
( 1)
1 sec 1 1
cos sin2
2 4( 1) sec
dx put x u dx u du
x
u du
dx udu u u
x u
==== ⇒⇒⇒⇒ ====
++++
= = = += = = += = = += = = +
++++
∫∫∫∫
∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫
1 1
2 2
1
2
2
1 1
2 2 2 2 2 2
1
2
1 1 1 1 1
tan 2sin cos tan
2 4 2 2 1 1
1 1
tan
2 2 1
(2 3) 2 1 1 1
2tan tan
2 2( 1) ( 1) ( 1) 1
5 1
tan
2 2 ( 1)
x
x u u x
x x
x
x
x
x x
dx dx dx x x
x x x x
x
x c
x
− −− −− −− −
−−−−
− −− −− −− −
−−−−
= + = += + = += + = += + = +
+ ++ ++ ++ +
= += += += +
++++
++++
∴ = + = + +∴ = + = + +∴ = + = + +∴ = + = + +
+ + + ++ + + ++ + + ++ + + +
= + += + += + += + +
++++
∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫

65. 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
x
Ans c
x
x
Ans c
x
Ans x x c
Ans x x x
x x c
Ans
dx
x
dx
x x
dx
x x
x dx
x
x x x x
dx
x x x
−−−−
++++
++++
++++
++++
++++
+ − ++ − ++ − ++ − +
−−−−
− − − − +− − − − +− − − − +− − − − +
−−−− − +− +− +− +
    
    
    −−−−
    
    
    + ++ ++ ++ +
    
        − +− +− +− +
    
    
    −−−−     
− + − +− + − +− + − +− + − +
− +− +− +− +
∫∫∫∫
∫∫∫∫
∫∫∫∫
∫∫∫∫
∫∫∫∫
1 2
: ln
3 22
2 3
12
:ln( 1)
2
1
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)
x
x c
x x x
Ans x c
x
x
Ans x c
x
x
Ans x x c
x
x dx
x
x x x
dx
x
x x
dx
x
x x
+ ++ ++ ++ +
− +− +− +− +
+ + ++ + ++ + ++ + +
++++
−−−−
+ + + ++ + + ++ + + ++ + + +
++++
−−−−
+ − − ++ − − ++ − − ++ − − +
++++
    
    
        
    
    
++++     
    + ++ ++ ++ +
    
++++     
        + −+ −+ −+ −
        
++++         
+ −+ −+ −+ −
∫∫∫∫
∫∫∫∫
∫∫∫∫
3 2
3 2 2 11
: ln tan
2 1 3 3( 1)
1 cosh 2
:ln
cos
2
2 2
2
2 1
( )( 1)
sin
(10)
cos (1 cos )
x x x x
Ans c
xx
x
Ans c
x
x x
dx
x x x
x dx
x x
− + −− + −− + −− + −−−−−
− + +− + +− + +− + +
++++++++
++++
++++
    + ++ ++ ++ +
    
    + ++ ++ ++ +     
    
    
    ++++     
∫∫∫∫
∫∫∫∫

66. Mathematics For Engineering
66
Miscellaneous substitution:
ve integer put
( )
(i)For
m
n
x
dx m ax b u
ax b
+ + =+ + =+ + =+ + =
++++
∫∫∫∫
Example(1): Find
2
3/ 2
(5 2 )
x
dx
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 10
8 8(5 2 )
1 25 10
8
1
25 10
8
1 2
25 20
8 3
1 50 2
20 5 2 5 2
8 35 2
x u u u
dx du du
x u u
u u
du
u u u
u u u du
u u u
x x
x
− − −− − −− − −− − −
−−−−
− − − − +− − − − +− − − − +− − − − +
= == == == =
−−−−
    −−−−
= − += − += − += − +    
        
−−−−     = − += − += − += − +
    
−−−−     
= − += − += − += − +    
    
    
= + − − −= + − − −= + − − −= + − − −
−−−−
∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫
∫∫∫∫
∫∫∫∫
c++++

ve integer put
( )
(ii)Fo m n
dx ax b
dx m u
xx ax b
r
++++
+ =+ =+ =+ =
++++
∫∫∫∫
Example(2): Find 3/ 2 5 / 2
(3 5)
dx
dx
x x ++++
∫∫∫∫
Solution:
Put
2
2
3 5 5 5 5
, 3 ,
5 3
x x
y y dx dy dx dy x
x x yx
+ −+ −+ −+ −
= + == + == + == + = ⇒⇒⇒⇒ ==== ⇒⇒⇒⇒ = ∴ == ∴ == ∴ == ∴ =
− −− −− −− −

67. Indefinite Integration
67
2 2
3/ 2 5/ 2 4 5/ 2(3/ 2) (5/ 2) 5/ 2
1
3 5 5(3 5) 5 ( )
dx x dy x dy
xx x x yx x
x
−−−−
= == == == =
++++++++ −−−−
∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫
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 6
125 (3 5) (3 5)
dy y dy y y dy
x y y y
y y y dy y y y
x x x
c
x x x
− − − − −− − − − −− − − − −− − − − −
− − − − − +− − − − − +− − − − − +− − − − − +
= = == = == = == = =
− −− −− −− −     = − + = + += − + = + += − + = + += − + = + +
    
    ++++−−−−     = + − += + − += + − += + − +
    + ++ ++ ++ +
    
∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫
∫∫∫∫
ve integer
( )
1
put ln ln
(iii)For n
n
dx
dx m
x ax b
dx du
x n x u n
u x u
++++
++++
= → = − → = −= → = − → = −= → = − → = −= → = − → = −
∫∫∫∫
Example(3): Find 3
( 2)
dx
dx
x x ++++
∫∫∫∫
Solution:
3
3 3
1
3ln ln ,
3
1 1 1 2
ln(1 2 ) ln(1 )
1 3 1 2 6 6( 2) 3 ( 2)
dx du
put x x u
u x u
dx du du
dx u c
ux x xu
u
= ∴ = − = −= ∴ = − = −= ∴ = − = −= ∴ = − = −
−−−−
= − = − + = − + += − = − + = − + += − = − + = − + += − = − + = − + +
++++++++ ++++
∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫
22
2
1 1
,
)
(iv)For
Thequadratic Form( canbechangedto a sum or
difference of two squars
dx dx
ax bx cax bx c
ax bx c
+ ++ ++ ++ ++ ++ ++ ++ +
+ ++ ++ ++ +
∫ ∫∫ ∫∫ ∫∫ ∫

68. 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 n
dx
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 n
dx
ax bx c
++++
+ ++ ++ ++ +
∫∫∫∫
Example(4): Find 2
3
2 4 3
dx
x x+ ++ ++ ++ +
∫∫∫∫
Solution:
2 2 2
1
2
3 3 3
2 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
dx
x c
x
−−−−
= == == == =
+ + + + + + − ++ + + + + + − ++ + + + + + − ++ + + + + + − +
= = + += = + += = + += = + +
+ ++ ++ ++ +
∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫
∫∫∫∫
Note: To Complete the square of 2
x ax++++ we add
2
2
a    
    
    
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 2
a a a a a a
x ax x ax x ax x+ = + + − = + + − = + −+ = + + − = + + − = + −+ = + + − = + + − = + −+ = + + − = + + − = + −
Example(5): Find
2
18 6
dx
x x− −− −− −− −
∫∫∫∫
Solution:
2 2 2
7 6 7 (6 ) 7 (6 9 9)
dx dx dx
x x x x x x
= == == == =
− − − + − + + −− − − + − + + −− − − + − + + −− − − + − + + −
∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫
1
2 2 2
( 3)
sin
416 (6 9) 4 ( 3)
dx dx x
c
x x x
−−−− ++++
= = = += = = += = = += = = +
− + + − +− + + − +− + + − +− + + − +
∫ ∫∫ ∫∫ ∫∫ ∫

69. Indefinite Integration
69
Example(6): Find 2
4 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
2
2
2 2
2 13
, 4 5 (2/ 3)(6 1) (13/ 3)
3 3
4 5 (2/ 3)(6 1) (13/ 3)
3 3 3 3
2 (6 1) 13 1
(1)
3 33 3 3 3
2 (6 1) 2
ln 3 3 (2)
3 33 3
13 1 13 1
1 1 1 353 3
3( 1) 3( )
3 3 36 36
1
A B x x
x x
dx dx
x x x x
x
dx dx
x x x x
x
Now dx x x
x x
And dx dx
x x x x
∴ = =∴ = =∴ = =∴ = = ⇒⇒⇒⇒ + = + ++ = + ++ = + ++ = + +
+ + ++ + ++ + ++ + +
====
+ + + ++ + + ++ + + ++ + + +
++++
= += += += +
+ + + ++ + + ++ + + ++ + + +
++++
= + += + += + += + +
+ ++ ++ ++ +
====
+ + + + ++ + + + ++ + + + ++ + + + +
====
∫ ∫∫ ∫∫ ∫∫ ∫
∫ ∫∫ ∫∫ ∫∫ ∫
∫∫∫∫
∫ ∫∫ ∫∫ ∫∫ ∫
2
1
1
2 1
2
3 1
1 359
( ) ( )
6 36
1
6( )
13 6 6. tan
9 35 35
26 (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
dx
x
x
x
From in
x x
dx x x c
x x
−−−−
−−−−
−−−−
+ ++ ++ ++ +
++++
====
++++
====
+ ++ ++ ++ +
∴ = + + + +∴ = + + + +∴ = + + + +∴ = + + + +
+ ++ ++ ++ +
∫∫∫∫
∫∫∫∫
Example(7): Find
2
2 3
3 6 10
x
x x
−−−−
+ ++ ++ ++ +
∫∫∫∫

70. 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 x
dx dx
x x x x
+ + = ++ + = ++ + = ++ + = +
− = + +− = + +− = + +− = + + ⇒⇒⇒⇒ = − + == − + == − + == − + =
−−−−
∴ = = − − =∴ = = − − =∴ = = − − =∴ = = − − = ⇒⇒⇒⇒ − = − + +− = − + +− = − + +− = − + +
− − + +− − + +− − + +− − + +
====
+ + + ++ + + ++ + + ++ + + +
∫ ∫∫ ∫∫ ∫∫ ∫
Q
2 2
2
2
2 2 2
1
2
2
2
1 (6 6)
5 (1)
2 3 6 10 3 6 10
1 (6 6)
3 6 10 (2)
2 3 6 10
5 5
5
3 10 3 73 6 10 2 2 1
3 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 x
x dx
Now x x
x x
dx dx dx
And
x x x x x x
dx x
c
x
x
From in dx x
x x
−−−−
− +− +− +− +
= += += += +
+ + + ++ + + ++ + + ++ + + +
− +− +− +− +
= − + += − + += − + += − + +
+ ++ ++ ++ +
= == == == =
+ ++ ++ ++ + + + + + ++ + + + ++ + + + ++ + + + +
++++
= = += = += = += = +
+ ++ ++ ++ +
++++
∴ = − +∴ = − +∴ = − +∴ = − +
+ ++ ++ ++ +
∫ ∫∫ ∫∫ ∫∫ ∫
∫∫∫∫
∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫
∫∫∫∫
∫∫∫∫
15 3( 1)
10 sinh
3 7
x
x c−−−− ++++
+ + ++ + ++ + ++ + +
2
1
( ) ( )
y =G(x)
2y =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)islinearand G(x) isquadratic put
(d)whenF(x)and G(x) are both q
dx
F x G X
∫∫∫∫
2 G(x)
y =
F(x)
uadratic put

71. 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) 1
2
12(4 1) 3 1 4 1(4/ 3)( 1) 1 ( )
4
1 2 3 1 1
coth (2 ) coth (2 3 1) ln
2 2 3 1 1
dx ydy dy dy
x x yy y y
x
y x c
x
− −− −− −− −
= = == = == = == = =
    + ++ ++ ++ + −−−−− +− +− +− + −−−−    
− + −− + −− + −− + −
= − = − + = += − = − + = += − = − + = += − = − + = +
+ ++ ++ ++ +
∫ ∫ ∫ ∫∫ ∫ ∫ ∫∫ ∫ ∫ ∫∫ ∫ ∫ ∫
22 2
1 1
(2/ 3) 1
2
12(4 1) 3 1 4 1(4/ 3)( 1) 1 ( )
4
1 2 3 1 1
coth (2 ) coth (2 3 1) ln
2 2 3 1 1
dx ydy dy dy
x x yy y y
x
y x c
x
− −− −− −− −
= = == = == = == = =
    + ++ ++ ++ + −−−−− +− +− +− + −−−−    
− + −− + −− + −− + −
= − = − + = += − = − + = += − = − + = += − = − + = +
+ ++ ++ ++ +
∫ ∫ ∫ ∫∫ ∫ ∫ ∫∫ ∫ ∫ ∫∫ ∫ ∫ ∫
Example(9): Find 2
(4 9) 2 5
dx
x x− −− −− −− −
∫∫∫∫ case ( )b
Solution:
2 2
2 2 4 2
1
2 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 1
6 6( 2)( 8) 2 8
1 1 1 1
. tan . tan
6 62 2 2 2 2 2
1 2 5 1 2 5
tan tan
6 2 2 12 2 2 2
dy dy dy
y y y y
y y
c
x x
c
− −− −− −− −
− −− −− −− −
= = −= = −= = −= = −
+ + + ++ + + ++ + + ++ + + +
= − += − += − += − +
− −− −− −− −
= − += − += − += − +
∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫
Example(10): Find
2
( 1) 4 2
dx
x x x+ + ++ + ++ + ++ + +
∫∫∫∫ (case ( )c )

72. Mathematics For Engineering
72
Solution:
(((( )))) (((( ))))
1 2 1
2
2 2
1 1 1
1
1 , (1 )
( 1) 4 2 (1 ) 4 (1 ) 2
x y dx y dy x y y
y
dx y dy
x x x y y y y y
− − −− − −− − −− − −
−−−−
− − −− − −− − −− − −
+ = = ∴ = − = −+ = = ∴ = − = −+ = = ∴ = − = −+ = = ∴ = − = −
−−−−
∴ =∴ =∴ =∴ =
+ + ++ + ++ + ++ + + − + − +− + − +− + − +− + − +
∫ ∫∫ ∫∫ ∫∫ ∫
1 1
2 2
1
sin sin
2 2( 1)1 2 2 ( 1)
dy dy y x
c c
xy y y
− −− −− −− −     − − − −− − − −− − − −− − − −    
= = = − + = − += = = − + = − += = = − + = − += = = − + = − +        
++++         + − − −+ − − −+ − − −+ − − −
∫ ∫∫ ∫∫ ∫∫ ∫
Example(11): Find
2 2
(1 ) 1
dx
x x+ −+ −+ −+ −
∫∫∫∫ case (d)
Solution:
2
2
2 2 2
1 4 1
,
11 (1 )
x xdx y
put y dy x
yx x
− − −− − −− − −− − −
= ∴ = == ∴ = == ∴ = == ∴ = =
+++++ ++ ++ ++ +
2 2 2
2 2 2 2 2
2
1
2
2
1
2
(1 ) (1 )
(1 ) 1 ( 4 )(1 ) 1 ( 4 ) 1
1
1
11 1 1
4 21 1 1 2 2 2
1
1 1
1 1
sin (2 1)
2 2 1 1 2 2
( )
4 2
1 1 3
sin
2 2 1
dx x dy x dy
x x x x x x x
y
dy
y dy dy
y y y y y y
y y
dy
y c
y
x
c
x
−−−−
−−−−
+ ++ ++ ++ +
∴ = = =∴ = = =∴ = = =∴ = = =
+ − − + − − −+ − − + − − −+ − − + − − −+ − − + − − −
    −−−−
++++    ++++− − −− − −− − −− − −    = = == = == = == = =
− − −− − −− − −− − − −−−−−−−−
+ ++ ++ ++ +
− −− −− −− −
= = − += = − += = − += = − +
− −− −− −− −
    − −− −− −− −
= += += += +    
++++        
∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫
∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫
∫∫∫∫

73. Indefinite Integration
73
(vii)For
n
dx
x ax b++++
∫∫∫∫ put 2
1 2
ln 2ln ,n dx dy
x n x y n
x yy
−−−−
= ∴ = − == ∴ = − == ∴ = − == ∴ = − =
Example(12): Find
4
3
dx
x x ++++
∫∫∫∫
Solution:
4 2
2
1
4 2 2
1
2
1
4ln 2ln , 4
1 1 1
sinh 3
2 2 2 33 3 1 3
1 3
sinh
2 3
dx dy
put x y x y
x yy
dx dy dy
y
x x y y y
c
x
−−−−
−−−−
−−−−
−−−−
−−−−
= = ∴ = − == = ∴ = − == = ∴ = − == = ∴ = − =
− − −− − −− − −− − −
∴ = = =∴ = = =∴ = = =∴ = = =
+ + ++ + ++ + ++ + +
    −−−−
= += += += +    
        
∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫

74. 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 3
1
(13)
8
x x x
dx dx dx
x x x
dx dx dx
x x x x x x
dx dx dx
x x x x x x
dx dx dx
x 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 6
1 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 6
1 1 1
(22) (23) (24)
2 3 2 5 4 3
1
(25)
8 2
dx dx dx
x x x x x x
x x x
dx dx dx
x x x x x x
x x x
dx dx dx
x x x x x x
dx dx dx
x x x x x x
x
− − − − −− − − − −− − − − −− − − − −
+ + −+ + −+ + −+ + −
+ − − + + −+ − − + + −+ − − + + −+ − − + + −
+ − −+ − −+ − −+ − −
− − − − − −− − − − − −− − − − − −− − − − − −
+ − − + + −+ − − + + −+ − − + + −+ − − + + −
− −− −− −− −
∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫
∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫
∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫
∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫
2 2 2
2 2 2
2 2 2
2
1 1
(26) (27)
12 4 7 6
1 2 3 3 1
(28) (29) (30)
2 3 2 5 4 3
3 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 dx
x x x x x
x x x
dx dx dx
x x x x x x
x x x
dx dx dx
x x x x x x
dx 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
+ ++ ++ ++ + + − + ++ − + ++ − + ++ − + +
+ + − −+ + − −+ + − −+ + − −
∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫
∫ ∫∫ ∫∫ ∫∫ ∫

75. 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
1
1
( ) ( ) ( ) ( ) ... ( ) ( )
( ) ( )
n n n n
n
k k k
k
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 n
n
k k
k
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

76. Mathematics For Engineering
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
1
1
( ; ) ( )
n
k k k
k
L P f m x x −−−−
====
= −= −= −= −∑∑∑∑
The upper sum of ( )f x corresponding to the partition P is defined to
be
1
1
( ; ) ( )
n
k k k
k
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

77. 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
1
1
1 0 1 2 1 2 1
( ; ) ( ) ,
( ) ( ) ( ) ( ) ... ( ) ( )
n
n k k k k k
k
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 1
1 1 1
( ) ( )( ) ( )
( ; ) ( ; ) ( ; )
n n n
k k k k k k k k k
k 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
1
1
lim ( ; ) lim ( )
n
l k k k
n n k
I L P f m x x −−−−
→∞ →∞→∞ →∞→∞ →∞→∞ →∞ ====
    
= = −= = −= = −= = −    
    
∑∑∑∑
The Upper integral of ( )f x on I is the number
1
1
lim ( ; ) lim ( )
n
u k k k
n 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

78. Mathematics For Engineering
78
1
lim ( ; ) lim ( )
n
R n k k
n 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 k
n ka
f x dx I f xξξξξ
→∞→∞→∞→∞ ====
= = ∆= = ∆= = ∆= = ∆∑∑∑∑∫∫∫∫
Corollary:
If ( )f x is continuous function on [ , ]a b the
1
lim ( ; ) lim ( )
n
n k k
n 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 a
n n
−−−−
==== and
1x 2x 3x 4x 5x 6x 7x 8x 9x 10x0x
0 1
n
2
n
3
n
4
n
5
n
6
n
7
n
8
n
9
n
1

79. Indefinite Integration
79
( )
0 , 1k
b a k k
x k n
n n
−−−−
= + = ≤ ≤= + = ≤ ≤= + = ≤ ≤= + = ≤ ≤ and 1
1
( ) , 1,2,...,k kx x k n
n−−−−− = =− = =− = =− = =
0 1 2
1 2
0, , ,..., 1n
n
x x x x
n n n
= = = = == = = = == = = = == = = = =
Choose rξξξξ such that
(((( ))))
(((( ))))
(((( ))))
1
1
1 1
2 2 2
1 1 1 1
2 2
11 2 1
2 2 2
2 1 4 2
( ) 4 1 4 1 1
2
1 4 2
( ; ) ( ) 1
1 1 4 1 1
4 2 2 1
4 ( 1) 1 2 ( 1
. 2 1
2
k k
r
k k
k n
n k k k
r k
n n n n
k k k k
kx x k k
n n n
k k
f
n n
k
S P f x x f
n n
k k
n nn n n
n n n n
n
n n
ξξξξ
ξ ξξ ξξ ξξ ξ
ξξξξ
−−−−
−−−−
= == == == =
= = = == = = == = = == = = =
    −−−−++++ −−−−
= = + == = + == = + == = + =    
    
− −− −− −− −    
= − = − = −= − = − = −= − = − = −= − = − = −    
    
−−−−    
∴ = − = −∴ = − = −∴ = − = −∴ = − = −    
    
    
= − − = − −= − − = − −= − − = − −= − − = − −    
    
+ ++ ++ ++ +    
= − − == − − == − − == − − =    
    
∑ ∑∑ ∑∑ ∑∑ ∑
∑ ∑ ∑ ∑∑ ∑ ∑ ∑∑ ∑ ∑ ∑∑ ∑ ∑ ∑
2
2
) 2
1
2 ( 1) 2
lim ( ; ) lim 1 2 1 1n
n n
nn
n n
S 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
( ) 6
1 , 1k
b a k k
x a k n
n n
−−−−
= + = + ≤ ≤= + = + ≤ ≤= + = + ≤ ≤= + = + ≤ ≤
1
6
( ) , 1,2,...,k kx x k n
n−−−−− = =− = =− = =− = =

80. Mathematics For Engineering
80
0 1 2
6 6 6
1, 1 , 1 2,..., 1 7nx x x x n
n n n
= = + = + ⋅ = + ⋅ == = + = + ⋅ = + ⋅ == = + = + ⋅ = + ⋅ == = + = + ⋅ = + ⋅ =
Choose rξξξξ such that
(((( ))))1 6 11 6 6 3
1 1 1
2 2
k k
r
kx x k k
n n n
ξξξξ −−−−     −−−−++++ −−−−
= = + + + = += = + + + = += = + + + = += = + + + = +    
    
(((( ))))
(((( ))))
1
1 1
2 2
1
2 2
1 1 1
2 2
6 3 24 12
( ) 4 1 4 1 1 3
6 24 12
( ; ) ( ) 3
18 144 72
18 144 72
1 1
144 72 1 72
18 1 18 72 1
2
k k
k n
n k k k
r k
n
k
n n n
k k k
k k
f
n n n
k
S P f x x f
n n n
k
n n n
k
n n n
n
n n
n nn n
ξ ξξ ξξ ξξ ξ
ξξξξ−−−−
= == == == =
====
= = == = == = == = =
−−−−    
= − = + − = + −= − = + − = + −= − = + − = + −= − = + − = + −    
    
    
∴ = − = + −∴ = − = + −∴ = − = + −∴ = − = + −    
    
    
= + −= + −= + −= + −    
    
= + −= + −= + −= + −
    
= + ⋅ + − ⋅ = + + −= + ⋅ + − ⋅ = + + −= + ⋅ + − ⋅ = + + −= + ⋅ + − ⋅ = + + −    
    
∑ ∑∑ ∑∑ ∑∑ ∑
∑∑∑∑
∑ ∑ ∑∑ ∑ ∑∑ ∑ ∑∑ ∑ ∑
1 72
lim ( ; ) lim 18 72 1 18 72 90n
n n
S 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
1
10 2 8
( )k k k
b a
x x x
n n n−−−−
− −− −− −− −
∆ = − = = =∆ = − = = =∆ = − = = =∆ = − = = =

81. Indefinite Integration
81
and
(((( ))))
0 1 2
1
8 16
2, 2 , 2 ,...,
8 1 8 8
2 , 2 ,..., 2 10k k n
x x x
n n
k k n
x x x
n n n−−−−
= = + = += = + = += = + = += = + = +
−−−−
= + = + = + == + = + = + == + = + = + == + = + = + =
let
1
2
2
2 2 2
2
2 2 2
8 4
2
2
8 4
( ) 4 2 3
64 32 16 16 64
4 4 3
256 64 128 64 256
19
k k
r
k
x x k
n n
k
f
n n
k k k
n nn n n
k k k
n nn n n
ξξξξ
ξξξξ
−−−− ++++
= = + −= = + −= = + −= = + −
    
= + − += + − += + − += + − +    
    
    
= + + + − − += + + + − − += + + + − − += + + + − − +    
        
= + + + − −= + + + − −= + + + − −= + + + − −
1
1 1
2
3 3 2 2 3
1
8
( ; ) ( ) ( ) ( )
152 2048 512 1024 512 2048
n n
n k k k k
k k
n
k
S p f x x f f
n
k k k
n n n n n n
ξ ξξ ξξ ξξ ξ−−−−
= == == == =
====
= − == − == − == − =
    
= + + + − −= + + + − −= + + + − −= + + + − −    
    
∑ ∑∑ ∑∑ ∑∑ ∑
∑∑∑∑
2
152 2048 1 1 512
1 2
6
1024 1 512 2048 1
1 1
2 2
n
n n n n
n n n n
        
= + + + += + + + += + + + += + + + +        
        
            
+ + − − ++ + − − ++ + − − ++ + − − +            
            
1x 2x 3x 4x 5x 6x 7x 8x 9x 10x0x
2 10

82. Mathematics For Engineering
82
where
(((( )))) (((( ))))(((( ))))
(((( ))))
2
1 1
2
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 4040
lim ; lim ; 152 2 512
3 3k
k k
n n
x n
x x as n
n
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 1
1 1
lim ( ) lim ( )
lim ( ; ) lim ( ; )
n n
k k k k k k
n nk k
n n
m x x M x x
L P f U P f
− −− −− −− −
→∞ →∞→∞ →∞→∞ →∞→∞ →∞= == == == =
→∞ →∞→∞ →∞→∞ →∞→∞ →∞
            
− −− −− −− −            
            
= == == == =
∑ ∑∑ ∑∑ ∑∑ ∑
and
( )
b
l u
a
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

83. Indefinite Integration
83
0 1 2
8 2.8 8
0, , ,..., ,..., 8k n
k
x x x x x
n 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
2
2
2 2
1 2
1 1
2
2 2 2
1
2
3 3 3
1
2
8( 1) 8( 1) 64( 1)
1 1
8 64
1
64 2 18
( ; ) ( )
8 64 128 64
1
512 1016 512 8
512
k
k
n n
k k k
k k
n
k
n
k
k k k
m f
n n n
k
M f k
n n
k k n
L P f x x m
n n
k k
n n n n
k k
nn n n
n
−−−−
= == == == =
====
====
− − −− − −− − −− − −            
= = − = −= = − = −= = − = −= = − = −        
            
    
= = −= = −= = −= = −    
    
    − + −− + −− + −− + −    
= − == − == − == − =     
    
    
    
= − + −= − + −= − + −= − + −    
    
    
= − + −= − + −= − + −= − + −    
    
====
∑ ∑∑ ∑∑ ∑∑ ∑
∑∑∑∑
∑∑∑∑
(((( ))))(((( ))))
(((( ))))
3 3
11016 512 8
1 2 1
6 2
512 512 24 488
lim ( ; ) 8 (1)
3 3 3n
n nn
n n n n
nn n
L P f
→∞→∞→∞→∞
++++
⋅ + + − ⋅ + ⋅ − ⋅⋅ + + − ⋅ + ⋅ − ⋅⋅ + + − ⋅ + ⋅ − ⋅⋅ + + − ⋅ + ⋅ − ⋅
−−−−
= − = == − = == − = == − = =
Similarly
2
2
1
2
3 3
1
8 64
( ; ) 1
512 8 512
( 1)(2 1) 8
6
512 488
lim ( ; ) 8 (2)
3 3
n
k
n
k
n
U P f k
n n
n
k n n
nn n
U P f
====
−−−−
→∞→∞→∞→∞
    
= −= −= −= −    
    
    
= − = ⋅ + + −= − = ⋅ + + −= − = ⋅ + + −= − = ⋅ + + −    
    
∴ = − =∴ = − =∴ = − =∴ = − =
∑∑∑∑
∑∑∑∑
From (1), (2) the function is integrable on v and
8
0
488
( )
3
f x dx ====∫∫∫∫

84. 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 y
b a
δδδδ
− <− <− <− <
−−−−
Now let n N>>>> be such that
( )
b a
n
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 k
b a
x x
n
δ εδ εδ εδ ε−−−−
−−−−
− = <− = <− = <− = <
Therefore we have
sup{ ( ) ( ): , }
1,2,3,...,
k k kw M m f x f y x y
b a
for k n
εεεε
= − = − ∈= − = − ∈= − = − ∈= − = − ∈
−−−−
====
we have
1
1
1
1
( ; ) ( ; ) ( )( )
( )
( )
n
k k k k
k
n
k k k
k
U f p L f p x x M m
b a
x x w n
b a n
b a
b 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 n
n n
U p L p
→∞ →∞→∞ →∞→∞ →∞→∞ →∞
==== Then ( )f x is integrable on [ , ]a b .

85. 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 k
b a
x x
n−−−−
−−−−
− =− =− =− = for 1,2,3,...,k n====
since ( )f x is increasing function
on 1[ , ]k kx x−−−− then
1( ) ( )k k k km f x and M f x−−−−= == == == =
Therefore
(((( )))) (((( )))) (((( ))))
[[[[ ]]]]
1
1
1 1
1 1
1 0 2 1 1
( ; ) ( ; ) ( )( )
( ) ( ) ( ) ( )
( ) ( ) ( ) ( ) ... ( ) ( )
( ) ( )
n
k k k k
k
n n
k k k k
k k
n n
U f p L f p x x M m
b a b a
f x f x f x f x
n n
b a
f x f x f x f x f x f x
n
b a
f b f a
n
−−−−
====
− −− −− −− −
= == == == =
−−−−
− = − −− = − −− = − −− = − −
− −− −− −− −
= ⋅ − = −= ⋅ − = −= ⋅ − = −= ⋅ − = −                        
−−−−
    = − + − + + −= − + − + + −= − + − + + −= − + − + + −    
−−−−
= −= −= −= −
∑∑∑∑
∑ ∑∑ ∑∑ ∑∑ ∑
Now if 0εεεε >>>> is given , we choose n N∈∈∈∈ such that
(((( ))))( ) ( ) ( )b a f b f a
n
εεεε
− −− −− −− −
>>>> For corresponding partition p we have
[[[[ ]]]]( ) ( ) ( ) ( )n n
b a
U p L p f b f a
n
εεεε
−−−−
− = − <− = − <− = − <− = − < and since εεεε is arbitrary we can
a b
1( )f a m====
( )nM f b====1 2M m====

86. Mathematics For Engineering
86
chose εεεε small such that lim ( ) lim ( )n n
n n
U p L p
→∞ →∞→∞ →∞→∞ →∞→∞ →∞
==== Then ( )f x is
integrable on [ , ]a b .
Exercises
By mathematical induction prove that
(((( )))) (((( ))))(((( ))))
(((( ))))
2
1 1
2
3
1
1 1
1 , 1 2 1
2 6
1
1
2
(1) (2)
(3)
n n
n
r n n r n n n
r n n
= + = + += + = + += + = + += + = + +
    
= += += += +    
    
∑ ∑∑ ∑∑ ∑∑ ∑
∑∑∑∑
find the following sums
2
1 1
10
2
1 1
20 15
2 2
1 1
( 5) ( 3)
( 3 5) (2 5)
(2 13) ( 2 4)
(4) (5)
(6) (7)
(8) (9)
n n
k k
n
k k
k k
k k
k k k
k k k
= == == == =
= == == == =
= == == == =
+ ++ ++ ++ +∑ ∑∑ ∑∑ ∑∑ ∑
+ − −+ − −+ − −+ − −∑ ∑∑ ∑∑ ∑∑ ∑
− − +− − +− − +− − +∑ ∑∑ ∑∑ ∑∑ ∑
By using Darboux’s sums find
1 1
2 2
0 1
2 1
2 2
1 0
1
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)

87. 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 function
0
[ ( ) ( )] ( ) ( )
( ) ( ) ,
( ) ( )
( ) ( ) ( )
0 ( )
( )
2 ( ) ( )
(
(1)
(2)
(3)
(4)
(5)
(6)
b b b
a a a
b b
a a
b a
a b
b c b
a a c
a
a
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 dx
f 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
a
f 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
a
f x dx f c b a= −= −= −= −∫∫∫∫ .

88. 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 a
b
a
b
a
m dx f x dx M dx
m b a f x dx M b a
f u f x dx f v
b 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
a
f 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
a
f 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
a
a
F f x x a b= ∀ ∈= ∀ ∈= ∀ ∈= ∀ ∈∫∫∫∫ Then aF
differentiable at any point [ , ]c a b∈∈∈∈ at which f continuous ,and
( ) ( )aF c f c′′′′ ====

89. 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
1
1
c h
c
dx
h
++++
====∫∫∫∫
(((( ))))
( ) ( ) 1 1
( ) ( ) ( )
1 1
( ) ( )
c h c h
a a
c c
c h
c
F c h F c
f c f x dx f c dx
h h h
f x f c dx h
h h
ε εε εε εε ε
+ ++ ++ ++ +
++++
+ −+ −+ −+ −
− = −− = −− = −− = −
= − ≤ == − ≤ == − ≤ == − ≤ =
∫ ∫∫ ∫∫ ∫∫ ∫
∫∫∫∫
Since 0εεεε >>>> is arbitrary it follows that
0
( ) ( )
( ) lim ( )a a
a
h
F c h F c
F 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
a
F x F a f− =− =− =− = ∫∫∫∫ if and only if ( ) ( ) [ , ]F x f x x a b′′′′ = ∀ ∈= ∀ ∈= ∀ ∈= ∀ ∈ .
Proof:
If ( ) ( )
x
a
F 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

90. Mathematics For Engineering
90
( ) ( )
x
a
F 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
a
F b F a f− =− =− =− = ∫∫∫∫
:Examples
2 2
sin sin
sin sin
( 3 4) 3 4
( )
( )
( )
y
a
a
y
t
a
d
x dx y
dy
d
x dx y
dy
d
x x dx t t
dt
a
b
c
====
= −= −= −= −
+ + = + ++ + = + ++ + = + ++ + = + +
∫∫∫∫
∫∫∫∫
∫∫∫∫
:Example
By using fundamental theorem of calculus find
.
2
cos
( 3 5)
(i)
(ii)
b
a
b
a
x dx
x x dx+ ++ ++ ++ +
∫∫∫∫
∫∫∫∫
Solution:
3
2 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 a
b
a
b b
a a
d
x dx x dx
dx
x dx b a
d x
x x dx x x dx
dx
b b a a
b a
====
= −= −= −= −
+ + = + ++ + = + ++ + = + ++ + = + +
= + + − + += + + − + += + + − + += + + − + +
∫ ∫∫ ∫∫ ∫∫ ∫
∫∫∫∫
∫ ∫∫ ∫∫ ∫∫ ∫

91. Indefinite Integration
91
Exercises
Using fundamental theorem of calculus find
10 2
2
2 1
23
2
2 0
2 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 x
f x g x
x x x
d d
f x dx g x dx
dx dx
++++
= == == == =
+ ++ ++ ++ +
++++∫ ∫∫ ∫∫ ∫∫ ∫
(8) using fundamental theorem of calculus find the area under the
curve of the following functions:
3
5 7( )f x x= += += += + bounded by 0 2 8, ,y x x= = == = == = == = = .
Improper Integrals:
( )
b
a
f 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

92. Mathematics For Engineering
92
at x b==== we define
0
( ) lim ( )
b b
a a
f 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 a
f 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 c
f 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 a
f 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 a
f 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 a
f x dx f x dx
∞∞∞∞
→∞→∞→∞→∞−∞ −−∞ −−∞ −−∞ −
====∫ ∫∫ ∫∫ ∫∫ ∫
provided the limit exist.
Examples:
Example(1):
Find
3
20 9
dx
x−−−−
∫∫∫∫

93. Indefinite Integration
93
Solution:
The integrand is discontinuous at 2x ==== we consider
33 3
1 1
2 20 0 00 0 0
1
3
lim lim sin lim sin
3 39 9
sin 1
2
dx dx x
x x
εεεεεεεε
ε ε εε ε εε ε εε ε ε
εεεε
ππππ
−−−−−−−−
− −− −− −− −
→ → →→ → →→ → →→ → →
−−−−
−−−−            
= = == = == = == = =            
            − −− −− −− −
= == == == =
∫ ∫∫ ∫∫ ∫∫ ∫
Example(2):
Find
2
0 2
dx
x−−−−
∫∫∫∫
Solution:
The integrand is discontinuous at 3x ==== we consider
22 2 2
00 0 00 0 0
0
1
lim lim ln 2 lim ln
2 2 2
1 1
lim ln ln
2
dx dx
x
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 1
lim lim
( 1) ( 1) ( 1)
1 1 1 1 1
lim lim lim 1 lim
1 1 3
x x x
x x
εεεε
ε εε εε εε ε εεεε
εεεε
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
∞∞∞∞
++++
∫∫∫∫

94. Mathematics For Engineering
94
Solution:
The upper limit of integration is infinite. We consider
1 1 1
2 2
0 0 0
1 1
lim 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
2x
e dx
−∞−∞−∞−∞
∫∫∫∫
Solution:
The lower limit of integration is infinite. We consider
[[[[ ]]]]
00 0
2 2 2 21 1 1 1
lim lim lim 1 1 0
2 2 2 2
x x x a
a a ba a
e dx e dx e e
→−∞ →−∞ →−∞→−∞ →−∞ →−∞→−∞ →−∞ →−∞→−∞ →−∞ →−∞−∞−∞−∞−∞
         = = = − = − == = = − = − == = = − = − == = = − = − =             
∫ ∫∫ ∫∫ ∫∫ ∫
Hence
0
2 1
2
x
e dx
−∞−∞−∞−∞
====∫∫∫∫
Example(6):
Find x x
dx
e e
∞∞∞∞
−−−−
−∞−∞−∞−∞ ++++
∫∫∫∫
Solution:
Both limit of integration is infinite. We consider
1
2 2
1 1
lim lim tan
1 1
lim tan tan 0
2 2
x xa ax
x x x x aa aa
a a
a
dx e dx e dx
e
e e e e
e e
π ππ ππ ππ π
∞ ∞∞ ∞∞ ∞∞ ∞
−−−−
−−−− −−−−→∞ →∞→∞ →∞→∞ →∞→∞ →∞−∞ −∞ −−∞ −∞ −−∞ −∞ −−∞ −∞ −
− − −− − −− − −− − −
→∞→∞→∞→∞
    = = == = == = == = =
    + + ++ + ++ + ++ + +
    = − = − == − = − == − = − == − = − =
    
∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫
Hence
2x x
dx
e e
ππππ∞∞∞∞
−−−−
−∞−∞−∞−∞
====
++++
∫∫∫∫ .

95. Indefinite Integration
95
Exercise(9)
Evaluate the following integrals:
(((( ))))
2 1
1 0
2
2 1
2 22 0
2
2
2
3 2
1
1
ln
ln
1 1
4 1
11 4
4
(1) (2)
(3) (4)
(5) (6)
(7) (8)
(9) (10)
o
dx x dx
x x
dx dx
x x x
dx x
dx
x x
x dx
dx
xx
x dx
dx
x x
∞ ∞∞ ∞∞ ∞∞ ∞
−∞−∞−∞−∞
−−−−
∞∞∞∞
−∞ −−∞ −−∞ −−∞ −
∞ ∞∞ ∞∞ ∞∞ ∞
−∞−∞−∞−∞
++++ ++++
− −− −− −− −
−−−−++++
−−−−
∫ ∫∫ ∫∫ ∫∫ ∫
∫ ∫∫ ∫∫ ∫∫ ∫
∫ ∫∫ ∫∫ ∫∫ ∫
∫ ∫∫ ∫∫ ∫∫ ∫
∫ ∫∫ ∫∫ ∫∫ ∫
Evaluation of some definite integrals:
Example(1):
Find
2
0
sinm
x dx
ππππ
∫∫∫∫
Solution:
2 2
1
0 0
sin sin sinm m
x dx x x dx
π ππ ππ ππ π
−−−−
====∫ ∫∫ ∫∫ ∫∫ ∫ use integration by parts
let 1
2
2
1 2 22
0 0
sin sin
( 1)sin cos cos
cos sin ( 1)sin cos
m
m
m m
m
u x dv x dx
du m x x dx v x
I x x m x x dx
ππππ
ππππ
−−−−
−−−−
− −− −− −− −
= == == == =
= − = −= − = −= − = −= − = −
    ∴ = − + −∴ = − + −∴ = − + −∴ = − + −
     ∫∫∫∫

96. Mathematics For Engineering
96
since
if is even then
2
2 2
0
2 2
2
0 0
2
2
0
2
0
2
0
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
m
m m
m m
m
m
m
m x x dx
m xdx m x dx
I m I m xdx
m
I I
m
m m m
m I I
m m m
I dx
I x dx
ππππ
π ππ ππ ππ π
ππππ
ππππ
ππππ
ππππ
−−−−
−−−−
−−−−
−−−−
= + − −= + − −= + − −= + − −
= − − −= − − −= − − −= − − −
+ − = −+ − = −+ − = −+ − = −
−−−−
∴ =∴ =∴ =∴ =
− − −− − −− − −− − −
====
− −− −− −− −
= == == == =
= == == == =∴∴∴∴
∫∫∫∫
∫ ∫∫ ∫∫ ∫∫ ∫
∫∫∫∫
∫∫∫∫
∫∫∫∫
[[[[ ]]]]since
is evenpositive integer
if is odd then 1
2
2
1 0
0
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 m
m m m
m
m m m
m I I
m m m
I dx x
ππππ
ππππ
ππππ− − −− − −− − −− − −
− −− −− −− −
− − −− − −− − −− − −
====
− −− −− −− −
= = − == = − == = − == = − =∫∫∫∫
is odd positive integer
2
0
( 1).( 3)( 5)... 4.2
sin
.( 1).( 2)... 5.3
m
m
m m m
I x dx
m m m
m
ππππ
− − −− − −− − −− − −
= == == == =∴∴∴∴ − −− −− −− −
∫∫∫∫

97. Indefinite Integration
97
2 2
6 7
6 7
0 0
5.3.1 6.4.2
sin . , sin
6.4.2 2 7.5.3
I x dx I x dx
π ππ ππ ππ π
ππππ
= = = == = = == = = == = = =∫ ∫∫ ∫∫ ∫∫ ∫
Example(2):
Find
2
0
cosm
x dx
ππππ
∫∫∫∫
Solution:
2 2
1
0 0
cos cos cosm m
x dx x x dx
π ππ ππ ππ π
−−−−
====∫ ∫∫ ∫∫ ∫∫ ∫ use integration by parts
let 1
2
2
1 2 22
0 0
2
2 2
0
2 2
2
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 m
m
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
ππππ
ππππ
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
m
m m
m m
m
I m I m xdx
m
I I
m
m m m
m I I
m m m
ππππ
−−−−
−−−−
+ − = −+ − = −+ − = −+ − = −
−−−−
∴ =∴ =∴ =∴ =
− − −− − −− − −− − −
====
− −− −− −− −
∫∫∫∫

98. Mathematics For Engineering
98
since
2
0
0 2
I dx
ππππ
ππππ
= == == == =∫∫∫∫
is even positive integer
2
0
( 1).( 3)( 5)... 3.1
cos ,
.( 1).( 2)... 4.2 2
m
m
m m m
I x dx
m m m
m
ππππ
ππππ− − −− − −− − −− − −
= == == == =∴∴∴∴ − −− −− −− −
∫∫∫∫
[[[[ ]]]]since
if is odd then 1
2
2
1 0
0
( 1).( 3)( 5)... 4.2
.( 1).( 2)... 5.3
cos sin 1
m
m m m
m I I
m m m
I dx x
ππππ
ππππ
− − −− − −− − −− − −
====
− −− −− −− −
= = == = == = == = =∫∫∫∫
is odd positive integer
2
0
( 1).( 3)( 5)... 4.2
cos
.( 1).( 2)... 5.3
m
m
m m m
I x dx
m m m
m
ππππ
− − −− − −− − −− − −
= == == == =∴∴∴∴ − −− −− −− −
∫∫∫∫
2 2
6 7
6 7
0 0
5.3.1 6.4.2
cos . , cos
6.4.2 2 7.5.3
I x dx I x dx
π ππ ππ ππ π
ππππ
= = = == = = == = = == = = =∫ ∫∫ ∫∫ ∫∫ ∫
2
0
221 1 1 2
00
sin .cos
1 1
sin .cos sin .cos
m n
m n m n
x x dx
n
x x x x dx
m n m n
ππππ
ππππππππ
+ − + −+ − + −+ − + −+ − + −−−−−    
= += += += +    + ++ ++ ++ +    
∫∫∫∫
∫∫∫∫

99. Indefinite Integration
99
2
1 2
0
2
0
221 1 2
00
2
2
0
1
0 sin .cos (1)
sin .cos
1 1
sin .cos sin .cos
1
0 sin .cos (2)
m n
m n
m n m n
m n
n
x x dx
m n
and x x dx
m
x x x x dx
m n m n
m
x 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 dx
m 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 m
x 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)...1
sin .cos 1
( )( 2)( 4) ... 2
( 1)( 3)...1.( 1)( 3)...1
.
( )( 2)( 4) ... 2 2
m n m m n n
x x dx dx
m n m n m n
m m n n
m n m n m n
π ππ ππ ππ π
ππππ
− − − −− − − −− − − −− − − −
====
+ + − + −+ + − + −+ + − + −+ + − + −
− − − −− − − −− − − −− − − −
====
+ + − + −+ + − + −+ + − + −+ + − + −
∫ ∫∫ ∫∫ ∫∫ ∫

100. Mathematics For Engineering
100
Solved examples:
2
6 4
0
5.3.1.3.1
sin .cos .
10.8.6.4.2 2 2
x x dx
ππππ
ππππ
====∫∫∫∫
2
8
0
7.5.3.1 35
sin . ( 8, 0)
8.6.4.2 2 256
x dx m n
ππππ
π ππ ππ ππ π
= = = == = = == = = == = = =∫∫∫∫
if m and n are both odd.
2 2
0 0
( 1)( 3)...2.( 1)( 3)...2
sin .cos . sin cos
( )( 2)( 4) ... 4
( 1)( 3)...2.( 1)( 3)...2 1
.
( )( 2)( 4) ... 4 2
m n m m m n
x x dx x x dx
m n m n m n
m m n n
m n m n m n
π ππ ππ ππ π
− − − −− − − −− − − −− − − −
====
+ + − + −+ + − + −+ + − + −+ + − + −
− − − −− − − −− − − −− − − −
====
+ + − + −+ + − + −+ + − + −+ + − + −
∫ ∫∫ ∫∫ ∫∫ ∫
solved Examples:
2
5 3
0
4.2.2 1 1
sin .cos .
8.6.4 2 24
x x dx
ππππ
= == == == =∫∫∫∫
2
7
0
6.4.2.1 16
cos ( 8, 0)
7.5.3.1 35
x dx m n
ππππ
= = = == = = == = = == = = =∫∫∫∫
if m is odd and n is even.
2 2
0 0
( 1)( 3)...2.( 1)( 3)...1
sin .cos . sin
( )( 2)( 4) ... 3
( 1)( 3)...2.( 1)( 3)...1
.1
( )( 2)( 4) ... 3
m n m m n n
x x dx x dx
m n m n m n
m m n n
m n m n m n
π ππ ππ ππ π
− − − −− − − −− − − −− − − −
====
+ + − + −+ + − + −+ + − + −+ + − + −
− − − −− − − −− − − −− − − −
====
+ + − + −+ + − + −+ + − + −+ + − + −
∫ ∫∫ ∫∫ ∫∫ ∫
Example:
2
5 4
0
4.2.3.1 8
sin .cos .1
9.7.5.3 315
x x dx
ππππ
= == == == =∫∫∫∫

101. Indefinite Integration
101
if m is even and n is odd.
2 2
0 0
( 1)( 3)...1.( 1)( 3)...2
sin .cos . cos
( )( 2)( 4) ... 3
( 1)( 3)...1.( 1)( 3)...2
.1
( )( 2)( 4) ... 3
m n m m n n
x x dx x dx
m n m n m n
m m n n
m n m n m n
π ππ ππ ππ π
− − − −− − − −− − − −− − − −
====
+ + − + −+ + − + −+ + − + −+ + − + −
− − − −− − − −− − − −− − − −
====
+ + − + −+ + − + −+ + − + −+ + − + −
∫ ∫∫ ∫∫ ∫∫ ∫
Example:
2
4 7
0
3.1.6.4.2 16
sin .cos .1
11.9.7.5.3 5511
x 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 x
x 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 .cos
2
2 ( )
2
m n
m n
x x dx
m n
ππππ + ++ ++ ++ +
Γ ΓΓ ΓΓ ΓΓ Γ
====
+ ++ ++ ++ +
ΓΓΓΓ
∫∫∫∫
Solved examples:
5 5 3 37 1 12
6 4 2 2 2 2 2 2 2
2
0 2
9 5 31 7 12
8 2 2 2 2 2 2
0
( ) ( ) . . . . 3
sin .cos
1.5.4.3.2.1 5122 ( )
( ) ( ) . . . . 35
sin
2 (5) 2.4.3.2.1 256
(1)
(2)
m n
x 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
∫∫∫∫
∫∫∫∫

102. Mathematics For Engineering
102
2
5 3
0
12
7 2
9 5 37 1
0 2 2 2 2 2
5 3 12
5 4 2 2 2
11 9 5 37 1
0 2 2 2 2 2 2
2
4 7
0
(3) (2) 2.1.1 1
sin .cos
2 (5) 2.4.3.2.1 24
( ) (4) .3.2.1. 16
cos
352 ( ) 2. . . .
(3) ( ) 2.1. . 8
sin .cos
3152 ( ) 2. . . . .
sin .cos
(3)
(4)
(5)
(6)
x x dx
x dx
x x dx
x x d
ππππ
ππππ
ππππ
ππππ
ππππ
ππππ
ππππ
ππππ
Γ ΓΓ ΓΓ ΓΓ Γ
= = == = == = == = =
ΓΓΓΓ
Γ ΓΓ ΓΓ ΓΓ Γ
= = == = == = == = =
ΓΓΓΓ
Γ ΓΓ ΓΓ ΓΓ Γ
= = == = == = == = =
ΓΓΓΓ
∫∫∫∫
∫∫∫∫
∫∫∫∫
∫∫∫∫
5 3 1
2 2 2
13 9 5 311 7 1
2 2 2 2 2 2 2
( ) (4) . .3.2.1 16
55112 ( ) 2. . . . . .
x
ππππ
ππππ
Γ ΓΓ ΓΓ ΓΓ Γ
= = == = == = == = =
ΓΓΓΓ
Numerical integrals
Exercise(10)

103. 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 k
n 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 k
k
f x x
====
∆∆∆∆∑∑∑∑ is simply the sum of the areas of n
approximating rectangles and the limit of the sum is ( )
b
a
f 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 the x axis−−−− and the two
ordinates ,x a x b= == == == = is
b
a
ydx∫∫∫∫ .
x
x b====
x a====
( )y f x====
y
1kx x −−−−==== kx x====
1( )kf x −−−−

104. Mathematics For Engineering
104
(2) Area bounded by the curve ( )y f x==== and the y axis−−−− and the two
ordinates ,y a y b= == == == = is
b
a
x 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
a
A 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
a
A y y dx= −= −= −= −∫∫∫∫
(5) Area of parametric curve ( ), ( )x f t y g t= == == == = given by
1
( )
2
b
a
A 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
2
A r d
θθθθ
θθθθ
θθθθ==== ∫∫∫∫
Example(1):
Find the area bounded by the curve 2
y x==== ,x-axis and the two lines
1, 3x x= == == == = .
Solution:
3 3
2
1 1
26
( )
3
A f x dx x dx= = == = == = == = =∫ ∫∫ ∫∫ ∫∫ ∫
Example(2):
Find the area above the x-axis and under the parabola 2
4y x x= −= −= −= −

105. Indefinite Integration
105
Solution:
3 3
2
1 1
32
( ) (4 )
3
A f x dx x x dx= = − == = − == = − == = − =∫ ∫∫ ∫∫ ∫∫ ∫
Example(3):
Find the area bounded by the parabola 2
6y x x= −= −= −= − and 2
2y x x= −= −= −= −
Solution:
The parabolas intersect at the points (0,0)) and(4,8)
The area =the area under 2
6y x x= −= −= −= − - the area under 2
2y x x= −= −= −= − from
0x ==== to 4x ====
Area = (((( )))) (((( ))))
4
2 2
0
64
6 2
3
x x x x dx    − − − =− − − =− − − =− − − =
    ∫∫∫∫
Example(4):
Find the area bounded by the cardoid (1 cos )r a θθθθ= += += += +
Solution:
2 22 2 2
2 2 2 2
0 0 0
1 1 3
(1 cos ) (1 2cos cos )
2 2 2 2
a a
A r d a d d
π π ππ π ππ π ππ π π ππππ
θ θ θ θ θ θθ θ θ θ θ θθ θ θ θ θ θθ θ θ θ θ θ= = + = + + == = + = + + == = + = + + == = + = + + =∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫

106. Mathematics For Engineering
106
Exercise(11)
(1) Find the area bounded as follows:
2
2
2
2
2
, 0, 2, 6 [ .: 39]
22
4 , 0, 1, 3 [ .: ]
3
1 , 10 [ .: 36]
125
9 , 3 [ . : ]
6
cos , sin [ .: ]
2cos cos2 1, 2sin sin2 [ .: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
ππππ
ππππ
= = = == = = == = = == = = =
= − = = == − = = == − = = == − = = =
= + == + == + == + =
= − = += − = += − = += − = +
= == == == =
= − − = −= − − = −= − − = −= − − = −
= −= −= −= −
2
2
3
2 2 2 2
2
3 3 3
8 3
[ .: ]
5
2
( ) [ .: ]
3
8
4 2 2 [ .: ]
3
3
( sin ), (1 cos ) [ .: ]
2
(8)
(9)
(10)
(11)
a
Ans
a
y x a x Ans
y x from x to x Ans
a
x 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 from x a to x b= == == == = x=a

107. 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 a
V 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 2
1 1 1
1
1
1 1
( ) ( ) ( )( ) ( )
( )
( ) ( ) 2 ( )
2
lim lim 2 ( ) 2 ( )
k k k k k k k k k k
k k
k 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 x
x 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
a
V xf x dxππππ==== ∫∫∫∫

108. 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 4
2 2
4 4
256
(2 ) 2 (2 )
8 15
y
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 2
2 2 3
0 0 0
2
2 3
0
2
2
0
2
3 2
0
(1 cos ) (1 cos ) (1 cos )
(1 3cos 3cos cos )
5 3
3cos cos2 (1 sin )cos
2 2
5 3 1
3sin sin 2 (sin sin ) 5
2 4 3
t t t
t t t
t
t
t
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
= = == = == = == = =
= = == = == = == = =
====
====
====
====
= = − − = −= = − − = −= = − − = −= = − − = −
= − + − == − + − == − + − == − + − =
    
= − + − −= − + − −= − + − −= − + − −    
    
    
= − + − − == − + − − == − + − − == − + − − =    
    
∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫
∫∫∫∫
∫∫∫∫

109. 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 0
2
2
0
2
2 2 3 3
0
2 2 ( sin )(1 cos )(1 cos )
2 ( 2 cos cos sin 2sin cos cosh2 sin )
3 1 1 1 1
2 2( sin cos ) ( sin 2 cos2 ) cos sin cos 6
4 2 2 4 3
t t
t t
t
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
2
2 , 0, 0, 5; [ . 2500 ](1) y x y x x x axis Ans ππππ= = = = −= = = = −= = = = −= = = = −
2 2
2 4 2
2 2
256
16, 0, 8; [ . ]
3
4
(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 [ . ]
7
5
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
ππππ
ππππ
ππππ
= = = = −= = = = −= = = = −= = = = −
= = = == = = == = = == = = =
= − + = −= − + = −= − + = −= − + = −

110. 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
1
sin2 ; [ . ]
4
sin , 1 cos ; [ .5 ]
64
2cos cos2 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 π
ππππ
θ 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
2
2 2 2 2
1 1 1( ) ( ) ( ) ( ) 1 k
k k k k k k k k k
k
y
P P x x y y x y x
x− − −− − −− − −− − −
    ∆∆∆∆
= − + − = ∆ + ∆ = + ∆= − + − = ∆ + ∆ = + ∆= − + − = ∆ + ∆ = + ∆= − + − = ∆ + ∆ = + ∆    
∆∆∆∆    
2 2
1
lim 1 1
bn
k
k
n k k a
y dy
AB 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 dy
AB L dt
dt dt
            
∴ = = +∴ = = +∴ = = +∴ = = +            
            
∫∫∫∫
if the equation of the curve given in the polar form ( )r f θθθθ==== then
2
1
2
2 dr
AB L r d
d
θθθθ
θθθθ
θθθθ
θθθθ
    
∴ = = +∴ = = +∴ = = +∴ = = +     
    
∫∫∫∫
Example(1):
Find the length of the arc of the curve
3
2y x==== from 0x ==== to 5x ==== .

111. Indefinite Integration
111
Solution:
(((( ))))
1
2
2
5
35
2
0 0
3
2
9 1
1 1 4 9
4 2
1 1 2 1 335
4 9 . . 4 9
2 2 3 9 27
dy
x
dx
y x x
L x dx x
====
′′′′+ = + = ++ = + = ++ = + = ++ = + = +
    
= + = + == + = + == + = + == + = + =    
        
∫∫∫∫
Example(2):
Find the length of the arc of the curve
3
1
23x y
−−−−
==== from 0y ==== to 4y ==== .
Solution:
(((( ))))
1
2
2
4
34
2
0 0
9
2
81 1
1 1 4 81
4 2
1 1 2 1 8
4 81 . . 4 81 (82 82 1)
2 2 3 81 234
dx
y
dy
dx
y y
dy
L y dy x
====
    
+ = + = ++ = + = ++ = + = ++ = + = +    
    
    
= + = + = −= + = + = −= + = + = −= + = + = −    
        
∫∫∫∫
Example(3):
Find the length of the arc of the curve 4
24 48xy x= += += += + from 2x ==== to
4x ==== .
Solution:
4
2
2 4 2 4 4 2
4 4 4
4 4 2 4 8 4
2 2
16
8
( 16) 64 ( 16)
1 1
64 64 64
1 1
64 ( 16) 64 32 256
8 8
dy x
dx x
dy x x x
dx x x x
x x x x x
x x
−−−−
====
− −− −− −− −    
∴ + = + = +∴ + = + = +∴ + = + = +∴ + = + = +    
    
= + − = + − += + − = + − += + − = + − += + − = + − +

112. Mathematics For Engineering
112
8 4 4
2 2
44 4
4 2 2 3
2
2 2 2
1 1
32 256 ( 16)
8 8
1 1 1 1 16 17
( 16) ( 16 )
8 8 3 68
x x x
x x
L x dx x x dx x units
xx
−−−−
= + + = += + + = += + + = += + + = +
    
= + = + = − == + = + = − == + = + = − == + = + = − =    
    
∫ ∫∫ ∫∫ ∫∫ ∫
Example(4):
Find the length of the arc of the curve 2 3
,x t y t= == == == = from 0t ==== to 4t ==== .
Solution:
2
2 2
2 4 2
4 4 42 2 2 3/ 2
00 0
2 , 3
4 9 4 9
1 2 1
4 9 18 4 9 . (4 9 )
18 3 18
8
(37 37 1)
27
dx dy
t t
dt dt
dx dy
t 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 2
2 2 2 2
2
22
0 0
1 cos , sin
(1 cos ) sin 1 2cos cos sin
2(1 cos 4sin 2sin
2 2
2sin 2 2cos 8
2 2
dx dy
d d
dx dy
d d
L d units
ππππππππ
θ θθ θθ θθ θ
θ θθ θθ θθ θ
θ θ θ θθ θ θ θθ θ θ θθ θ θ θ
θ θθ θθ θθ θ
θ θθ θθ θθ θ
θθθθ
θ θθ θθ θθ θ
θθθθ
= − == − == − == − =
            
∴ + = − + = − + +∴ + = − + = − + +∴ + = − + = − + +∴ + = − + = − + +            
            
= − = == − = == − = == − = =
    
= = − == = − == = − == = − =    
    
∫∫∫∫

113. Indefinite Integration
113
Example(6):
Find the circumference of cardoid (1 cos )r a θθθθ= −= −= −= −
Solution:
2
2 2 2 2 2 2 2
2 2
2 22 2 2
2 2 2
0 0 0 0
(1 cos )
sin
1
(1 cos ) sin (2 2cos ) 2 (1 cos )
2
4 sin
2
4 sin 2 sin 4 cos 8
2 2 2
r a
dr
a
d
dr
r a a a a
d
a
dr
L r a d a d a a
d
πππππ π ππ π ππ π ππ π π
θθθθ
θθθθ
θθθθ
θ θ θ θθ θ θ θθ θ θ θθ θ θ θ
θθθθ
θθθθ
θ θ θθ θ θθ θ θθ θ θ
θ θθ θθ θθ θ
θθθθ
= −= −= −= −
∴ =∴ =∴ =∴ =
    
+ = − + = − = −+ = − + = − = −+ = − + = − = −+ = − + = − = −    
    
====
            
= + = = = − == + = = = − == + = = = − == + = = = − =            
            
∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫
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)
2
4
[ . 2( 1)
[ .16
3/ 4
cos , sin 0 4
2cos cos2 1, 2sin sin2
(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
−−−−
−−−−
====
= = = == = = == = = == = = =
= + + = += + + = += + + = += + + = +

114. 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
x
a a
d
c
dy
S ydL y dx
dx
dx
y 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
y
a a
d
c
dy
S xdL x dx
dx
dx
x 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 2
t
x
AB t
dx dy
S ydL y dt
dt dt
π 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 2
t
y
AB t
dx dy
S xdL x dt
dt dt
π ππ ππ ππ π
            
= = += = += = += = +            
            
∫ ∫∫ ∫∫ ∫∫ ∫

115. 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 2
2 1
b
x
a
S y y dxππππ ′′′′= += += += +∫∫∫∫ Now
[[[[ ]]]]
2
2
2
6 6
12 2 12
12
36 3
12
3 3
1 12 1 12 12 36
y x yy y
y x
y
x x
x
y y x x x
x x x
′ ′′ ′′ ′′ ′==== ⇒⇒⇒⇒ ==== ⇒⇒⇒⇒ = == == == =
′′′′∴ = =∴ = =∴ = =∴ = =
′′′′+ = + = + = ++ = + = + = ++ = + = + = ++ = + = + = +
2
2 1 2 12 36 24(2 2 1)
b b
x
a 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 3
x y==== from 0y ==== to 1y ==== .
Solution:
We shall use
2
2 1
d
y
c
dx
S x dy
dy
ππππ
    
= += += += +     
    
∫∫∫∫ Now
3 2
2
3 4
3
1 1 9
dx
x y y
dy
dx
x y y
dy
==== ⇒⇒⇒⇒ ====
    
+ = ++ = ++ = ++ = +    
    
1
3 4 3 4
0
2 1 9 2 1 9 (10 10 1)
27
d
y
c
S y y dy y y dy
ππππ
π ππ ππ ππ π∴ = + = + = −∴ = + = + = −∴ = + = + = −∴ = + = + = −∫ ∫∫ ∫∫ ∫∫ ∫

116. 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 1
d
x
c
dx
S y dy
dy
ππππ
    
= += += += +     
    
∫∫∫∫ Now
2
2
2
2 2 2
2 2 2
2
2 2 4 4 2
4 2ln
1
2ln
4
1 2 1 1
2
4 2
1 (1 ) 1
1 1 4 (1 )
4 2
1 1
4 1 2 2 1
2 2
y x y
x y y
dx y
y
dy y y
dx y
x y y
dy yy
y y y y y
y y
+ =+ =+ =+ =
    = −= −= −= −
    
         −−−−
= − == − == − == − =         
             
     −−−−
+ = + = + −+ = + = + −+ = + = + −+ = + = + −    
    
= + − + = + += + − + = + += + − + = + += + − + = + +
2
2 2
2 3
1
33
2 3
1 1
1 ( 1) 1 1
( 1)
2 2 2
2 1
2 1
2
1 32
1
3 3
d
c
y
y y
y y y
dx
S y dy y y dy
dy y
y dy y y
ππππ
ππππ
ππππ
π 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 4
8a y a x x= −= −= −= − about the x-axis .
Solution:
We shall use
2
2 1
b
x
a
dy
S y dx
dx
ππππ
    
= += += += +     
    
∫∫∫∫ Now

117. Indefinite Integration
117
2 2 4 2 2 2
2 2 2 2 4 2
2 2
2 2 3
2 3
2
2 3 2 2 3 2 2 3 2
2
4 2 2 2 2 42 2 4
4
2
2 3 2 2 2 2
2
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 x
a y a x x y
a a
a yy a x x
a x x
y
a y
a x x a x x a x x
y
a y a a x xa x x
a
a
a x x a x
y
a a x x
− −− −− −− −
= −= −= −= − ⇒⇒⇒⇒ = == == == =
′′′′∴ = −∴ = −∴ = −∴ = −
−−−−
′′′′ ====
− − −− − −− − −− − −
′′′′ = = == = == = == = =
     −−−−−−−−
        
    
− −− −− −− −
′′′′+ = + =+ = + =+ = + =+ = + =
−−−−
Q
2 2 2
2 2 2 2 2 2
2
2 2 2 2
0 0
( )
( ) (3 2 )
2 1 2 .
8 8 ( )
a a
x
a a x
x a x a x
S y y dx dx
a a a x
π ππ ππ ππ π
−−−−
− −− −− −− −
′′′′∴ = + =∴ = + =∴ = + =∴ = + =
−−−−
∫ ∫∫ ∫∫ ∫∫ ∫
2 2 2 2
2 2
0 0
2 2 2 2
4 4
2 2
0
1 1
2 (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 dx
a a
a x a
a a
a a
ππππ
ππππ
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
1
16 4
x y
+ =+ =+ =+ = about the x-axis .
Solution:
We shall use
2
2 1
b
x
a
dy
S y dx
dx
ππππ
    
= += += += +     
    
∫∫∫∫ Now
2 2
2
2 2
1
1, 16
16 4 2
1 2
4 16 2 16
x y
y x
dy x x
dx x x
+ = = −+ = = −+ = = −+ = = −
− −− −− −− −
∴ = =∴ = =∴ = =∴ = =
− −− −− −− −

118. Mathematics For Engineering
118
squareunits
2 2 2 2
2 2
2 2 2
2 2
2
24 4
2
4 4
4
2 1
4
4(16 )
1 1
4(16 ) 4(16 )
1 4(16 ) 1
1 16 64 3
2 44(16 )
2 1 64 3
2
3 3 4 3
64 3 32sin 8 1
2 8 92 3
x
dy x x x
dx x x
dy x x
y x x
dx x
dy
S y dx x dx
dx
x x
x
ππππ
ππππ
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 3
cos , sinx a y aθ θθ θθ θθ θ= == == == = .
Solution:
We shall use
2 2
2
b
x
a
dx dy
S y d
d d
π θπ θπ θπ θ
θ θθ θθ θθ θ
            
= += += += +            
            
∫∫∫∫ Now
3 3
2 2
2 2
2 4 2 2 4 2 2 2 2
2 2
3 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
2
b
x
a
x a y a
dx dy
a a
d d
dx dy
a a a
d d
dx dy
y a a a
d d
dx dy
S y
d d
θ θθ θθ θθ θ
θ θ θ θθ θ θ θθ θ θ θθ θ 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 θ
θ θθ θθ θθ θ
θ θ θ θ θθ θ θ θ θθ θ θ θ θθ θ θ θ θ
θ θθ θθ θθ θ
ππππ
θ θθ θθ θθ θ
= == == == =
= − == − == − == − =
            
+ = + =+ = + =+ = + =+ = + =            
            
            
+ = =+ = =+ = =+ = =            
            
        
∴ = +∴ = +∴ = +∴ = +    
    
∫∫∫∫
[[[[ ]]]]
2 / 2
2 4
0
2 2
/ 2
0
2(2 ) 3 sin cos
12 12
sin5
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 π
θθθθ

====    
    
= == == == =
∫∫∫∫

119. 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 2
2 2 2 2
2 2 2
0
2
2 3 2
0
( sin ), (1 cos )
(1 cos ), sin
(1 cos ) sin 2 sin
2
2 4 (1 cos )sin
2
64
8 sin
2 3
b
x
a
x a t t y a t
dx dy
a t a t
dt dt
dx dy t
dL dt a t a t a dt
dt dt
dx dy t
S y dt a a t dt
dt dt
t
a 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 . .]
1
0 3; [ . 982 82 1)/ 9 . .]
3
1
0 2; [ . 9 82 ln( 82 9) . .]
2
1
8 (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) . .]
128
2cos cos2 , 2sin sin2 [ . . .]
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
ππππ
ππππ
θ θ θ θθ θ θ θθ θ θ θθ θ θ θ
−−−−
− = − −− = − −− = − −− = − −
− +− +− +− +
= − = −= − = −= − = −= − = −