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BITS Pilani, K K Birla Goa Campus
BITS PilaniK K Birla Goa Campus
Engineering Mathematics – I
(MATH ZC 161)
Dr. Amit Setia (Assistant Professor)Department of Mathematics
BITS PilaniK K Birla Goa Campus
BITS Pilani, K K Birla Goa Campus
Chapter 1 & chapter 3Pratap Singh, Review of Elementary Calculus, WILPD- Notes
Summary of lecture 8
• L’Hopital’s Rule
• Rationalization method to find limit
• How to check continuity of a function
• Derivative
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• Derivative
• Rules of derivatives
• Higher order derivatives
• Derivative by Chain Rule
• Derivative by Substitution31/08/2012 3Engineering Mathematics – I (MATH ZC 161)
• Integration
• Fundamental Integration Formulae
• Some properties of Integration
• Various methods for integration
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• Various methods for integration
31/08/2012 4Engineering Mathematics – I (MATH ZC 161)
Outline of lecture 9
• Integration by parts
• Integration of some standard forms
• Definite Integral
• Properties of definite integral
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• Properties of definite integral
• Some examples based on the properties of
definite integral
31/08/2012 5Engineering Mathematics – I (MATH ZC 161)
Don’t forget
You need to
of "Pratap Singh, Review of Elementary Calculus,
remember all the formulae given in Chapter 2
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of "Pratap Singh, Review of Elementary Calculus,
WILPD Notes" uploaded on Taxila in a folder
as MATH_ZC161_Lecture_Notes
−
24/08/2012 6Engineering Mathematics – I (MATH ZC 161)
• If u and v are functions of x then the formula
for integration by parts is
duuvdx u vdx vdx dx
= −∫ ∫ ∫ ∫
Article: 4.10
Integration by Parts
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u is called I functions and v is called II function.
duuvdx u vdx vdx dx
dx = −
∫ ∫ ∫ ∫
31/08/2012 7Engineering Mathematics – I (MATH ZC 161)
• Rule 1:
If both u and v are integrable functions,
then take that function as the first function
which can be finished by repeated
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which can be finished by repeated
differentiation.
31/08/2012 8Engineering Mathematics – I (MATH ZC 161)
Example
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31/08/2012 9Engineering Mathematics – I (MATH ZC 161)
• Rule 2:
If the integral contains logarithmic or inverse
trigonometric function,
then take logarithmic or inverse trigometric
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then take logarithmic or inverse trigometric
function as the first function and other
function as the II function.
If there is no other function then take 1 as the
second function.
31/08/2012 10Engineering Mathematics – I (MATH ZC 161)
Example
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31/08/2012 11Engineering Mathematics – I (MATH ZC 161)
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31/08/2012 12Engineering Mathematics – I (MATH ZC 161)
Example
( )
( ) ( ) ( )( )( )
2
2 2
log 1
log 1 log 1
Integrating by parts, we get
III
x dx
dx dx x dx dx
dx = −
∫
∫ ∫ ∫
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31/08/2012 13Engineering Mathematics – I (MATH ZC 161)
( ) ( )( )( ) ( ) ( )
( ) ( ) ( )
( ) ( )
2
2
2
2loglog
log 2 log 1
log 2 log
III
dx
xx x x dx
x
x x x dx
x x x x x C
= −
= −
= − − +
∫ ∫ ∫
∫
∫
• Rule 3:
If both functions are integrable and none can
be finished by repeated differentiation, then
any function can be taken as the I function
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any function can be taken as the I function
and other as the II function. Repeat the rule of
integration by parts.
31/08/2012 14Engineering Mathematics – I (MATH ZC 161)
3
2
cosec
cosec cosec
Integrating by parts, we get
III
x dx
x x dx=
∫
∫
Example
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( ) ( )( )( ) ( )( )
( )
2 2
2
cosec ccosec
cosec
cosec c
c o
ot cosec cot
sec
cot co
osec
cosec co tt
x dx x dx
x x
x dx
x dx
x x x x dx
dx
dx
x x
= −
= −
=
−
−
−
−
−
∫ ∫∫
∫
∫31/08/2012 15Engineering Mathematics – I (MATH ZC 161)
( )( )( )
2
3
3
cosec cot cosec cosec 1
cosec cot cosec cosec
cosec cot cosec cosec
x x x x dx
x x x x dx
x x x dx x dx
= − − −
= − − −
= − − +
∫
∫
∫ ∫
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3
3
3
2 cosec cosec cot cosec
2 cosec cosec cot log cosec cot
1cosec cosec co
2
x dx x x x dx
x dx x x x x C
x dx x
⇒ = − +
⇒ = − + − +
⇒ = −
∫ ∫
∫ ∫
∫
∫1
t log cosec cot2 2
Cx x x+ − +
31/08/2012 16Engineering Mathematics – I (MATH ZC 161)
• Rule 4:
That function is taken as the first function
which comes first in the word ILATE
where I = inverse circular function,
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where I = inverse circular function,
L = Logarithmic function,
A = Algebraic function,
T = Trigonometric function,
E = Exponential function.
31/08/2012 17Engineering Mathematics – I (MATH ZC 161)
Example
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putting sin cosx a dx a dθ θ θ= ⇒ =
Article: 4.11
Some standard forms
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31/08/2012 19Engineering Mathematics – I (MATH ZC 161)
putting sin cosx a dx a dθ θ θ= ⇒ =
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31/08/2012 20Engineering Mathematics – I (MATH ZC 161)
Please try !
use sec sec tanHint: x a dx aθ θ θ= ⇒ =
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2
Please try !
use tanHint: secx a dx a dθ θ θ= ⇒ =
Please try !
do integration by parts,Hint:
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31/08/2012 22Engineering Mathematics – I (MATH ZC 161)
2 2
do integration by parts,
take as I function and 1 as II f
H
u
in
nc
t:
tiona x−
Similarly,
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31/08/2012 23Engineering Mathematics – I (MATH ZC 161)
The integrals of the above form can be evaluated
Article: 4.12
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The integrals of the above form can be evaluated
by writing in any one of the 6 forms just described.
Example
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Example
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Example
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Article: 4.13
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31/08/2012 30Engineering Mathematics – I (MATH ZC 161)
To solve the integral of above form, we write
Example
( )1
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( )
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( )1 becomes
( )2
Putting in first integral 2 5
2&
in 2 integral 1
2
nd
x x t
x dx t
x u
dx du
d
+ =
+ +
=
⇒
⇒ + =
=
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31/08/2012 33Engineering Mathematics – I (MATH ZC 161)
dx du⇒ =
( )( )2Evaluate: 3 45x x x dx+ − −∫
Problem
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31/08/2012 34Engineering Mathematics – I (MATH ZC 161)
Please try !
Article: 4.14
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31/08/2012 35Engineering Mathematics – I (MATH ZC 161)
22
1 2tan
To solve the integral of above fo
sec2 2 2 1
rm, we take
x x dtt dx dt dx
t⇒ ⇒
= = = +
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31/08/2012 36Engineering Mathematics – I (MATH ZC 161)
Example
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2
2tan
2 1|
x dtt dx
t = =⇒ +
∵
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31/08/2012 38Engineering Mathematics – I (MATH ZC 161)
Article: 4.15
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31/08/2012 39Engineering Mathematics – I (MATH ZC 161)
( )1 2
To solve the integral of above form, we take
sin cos Denominator Denominatord
a x b x K Kdx + = +
( )
To solve the integral of above form, we take
2sin 3cos 3sin 4cos 3si
2sin 3cos
3sin 4cos
n 4cosd
x x K x x K x
x xd
xx
x
x + = + + +
+
+∫
Example
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31/08/2012 40Engineering Mathematics – I (MATH ZC 161)
( )1 22sin 3cos 3sin 4cos 3sin 4cosd
x x K x x K xdx
x + = + + +
1 2
1 2
Equating coefficient of sin and cos , we get
3 4 2
4 3 3
18 1,
x x
K K
K K
K K
∴− =
+ =
⇒ = =
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31/08/2012 41Engineering Mathematics – I (MATH ZC 161)
1 2
18 1,
25 25K K⇒ = =
18 1log | 3sin 4cos |
25 25x x x C= + + +
Article: 4.16
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31/08/2012 42Engineering Mathematics – I (MATH ZC 161)
( )sin cos Denominator Denominator
where p,q,r
To solve the integral of above form, we
can be evaluated by comparing coefficients of
cos , sin and const
ta
ant terms
ke
da x b x
x
dx
x
C p q r + + = + +
Given integral
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31/08/2012 43Engineering Mathematics – I (MATH ZC 161)
The last integral part can be evaluated by using method
described in article 4.14
( ) [ ]( ) ( )
If is a continuous function on ,
Fundamental Theorem of
, and
,
Calculus
f x a b
f x dx F x C= +∫
Article: 4.17 Definite Integral
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31/08/2012 44Engineering Mathematics – I (MATH ZC 161)
( ) ( )
( ) ( ) ( )
,
then b
a
f x dx F x C
f x dx F b F a
= +
= −
∫
∫
Example
/22
0 0sin cos cos
I IIx x dx x x x dx
π π = + ∫ ∫
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31/08/2012 45Engineering Mathematics – I (MATH ZC 161)
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31/08/2012 46Engineering Mathematics – I (MATH ZC 161)
( )To evaluate
if we are taking a substitution ( ),
then find the values of t corresponding to
and .
b
af x dx
x g t
x a x b
=
= =
∫
Remark
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31/08/2012 47Engineering Mathematics – I (MATH ZC 161)
and .
If and
when x = a and x = b respectively, then
x a x b
t A t B
= == =
Example
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Properties of definite integral
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31/08/2012 50Engineering Mathematics – I (MATH ZC 161)
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31/08/2012 51Engineering Mathematics – I (MATH ZC 161)
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31/08/2012 52Engineering Mathematics – I (MATH ZC 161)
Example
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( )of property 7 , as∵
Example
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( )of property 4∵
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If sin and cos are replaced
by tan and cot or
by cosec and sec respectively
x x
x x
x x
Remark
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31/08/2012 56Engineering Mathematics – I (MATH ZC 161)
in the integral I,
then similarly we can evaluate the integral.
Example
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( )| of property 3∵
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Thanks
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Thanks