Integral calculus

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

Mathematics For Engineering
2
Chapter 1
Indefinite Integration
If ( )F x is a function such that ( ) ( )F x f x′′′′ ==== on the interval [ , ]a b Then
( )F x is called an anti-derivative or indefinite of ( )f x . The indefinite
integral of the given function is not unique for example
2 2 2
, 3, 5x x x+ ++ ++ ++ + are indefinite integral of ( ) 2f x x==== since
2 2 2
( ) ( 3) ( 5) 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
(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
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
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
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
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
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
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
(((( ))))
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
(((( ))))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
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
(((( )))) (((( ))))
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
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
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
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
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
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
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
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
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
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
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
(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
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
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
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
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
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
+ + ++ + ++ + ++ + +
∴ = + + +∴ = + + +∴ = + + +∴ = + + +∫∫∫∫
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
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
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
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
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
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
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
(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
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
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
− −− −− −− −
= = −= = −= = −= = −
= → == → == → == → =
∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫
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
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∫∫∫∫
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
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∫∫∫∫
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
− −− −− −− −
= = −= = −= = −= = −
= → == → == → == → =
∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫
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
−−−−
−−−−
= = += = += = += = +
− −− −− −− −
∫∫∫∫
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∫∫∫∫

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
− −− −− −− −
= = += = += = += = +
= → = −= → = −= → = −= → = −
∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫
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
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
= == == == =∫ ∫∫ ∫∫ ∫∫ ∫
if m and n are both an even we use
[[[[ ]]]] [[[[ ]]]]2 21 1
2 2
x x x x= − = += − = += − = += − = +
sin cosx xdx∫∫∫∫
Solution:
(((( ))))
2
22 2 1
(12) sin cos sin cos sin2
2
x x dx x x dx x dx
    
= == == == =     
    
∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫

41
21 1 1 1 1
sin 2 (1 cos4 ) ( sin4 )
4 4 2 8 4
x dx x dx x x c= = − = − += = − = − += = − = − += = − = − +∫ ∫∫ ∫∫ ∫∫ ∫
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
====
= −= −= −= −
= −= −= −= −
= −= −= −= −
−−−−
= + += + += + += + +
∫ ∫∫ ∫∫ ∫∫ ∫
∫∫∫∫
∫∫∫∫
∫∫∫∫
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
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
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
( 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
m n
m n m n
x x
m n x x m x x
++++
++++
= − + + + += − + + + += − + + + += − + + + +
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
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
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
− +− +− +− +
− + −− + −− + −− + −
− +− +− +− +
    
    
= + − + −= + − + −= + − + −= + − + −
= − + + −= − + + −= − + + −= − + + −
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
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
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
+ −+ −+ −+ −
+ − −+ − −+ − −+ − −
+ −+ −+ −+ −
    
    
= − − + += − − + += − − + += − − + +
= − − + += − − + += − − + += − − + +
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
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
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
= == == == =∫ ∫∫ ∫∫ ∫∫ ∫
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
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
= ∴ == ∴ == ∴ == ∴ =
            
∴ = − = − = − +∴ = − = − = − +∴ = − = − = − +∴ = − = − = − +            
            
∫∫∫∫
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
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
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
−−−−
====
====
∫ ∫∫ ∫∫ ∫∫ ∫
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
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
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
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
+ = = =+ = = =+ = = =+ = = =
∴ + = + = + =∴ + = + = + =∴ + = + = + =∴ + = + = + =
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
− = = =− = = =− = = =− = = =
∴ − = − = − =∴ − = − = − =∴ − = − = − =∴ − = − = − =
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
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 θθ θθ θθ θ
θθθθ
= ∴ == ∴ == ∴ == ∴ =
− = − = = − = −− = − = = − = −− = − = = − = −− = − = = − = −
∴ = =∴ = =∴ = =∴ = =
−−−−
= + += + += + += + +
            
= − + + − = − + + − += − + + − = − + + − += − + + − = − + + − += − + + − = − + + − +            
            
∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫∫ ∫ ∫

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
θθθθ
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 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
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
(((( ))))
(((( ))))
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
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
    = − + + − += − + + − += − + + − += − + + − +
        −−−−
− = − − − +− = − − − +− = − − − +− = − − − +
∫∫∫∫
∫∫∫∫

Integral calculus

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