INTEGRATION Syllabus in IIT JEE : Integration as the inverse process of differentiation, indefinite integrals of standard functions, definite integrals and their properties, Fundamental Fund amental Theorem of Integral Calculus. Integration Integr ation by parts, integration by the methods of substitution and partial fractions,
INTRODUCTION: The word integration literally means summation. summation. It is, in fact, the process of finding the limit of a sum of a certain number of elements, as the number of elements tends to infinity and each element becomes infinitesimally small. The integral calculus calcu lus has its origin in an attemp to find a general method for the determination determination of an area of a region with straight or curved boundaries, by supposing the given region to be divided into an infinite number of infinitesimal infinitesimal elements; the sum of these elements being the area required. It can be shown that when once such a method of summation is discovered, it may be applied to a wide variety of geometrical and physical problems such as the finding of arc length, volume, centroid, moment of inertia, work and fluid pressure, etc. Accordingly it has played a very important role in development of all the branches of science. The subject of integration may also be viewed from the point of converse of differentiation. The calculation of derivative or expression (discussed below) sug suggests gests the opposite problem of passing from the derivative or differential expression to its original, called the primitive, or indefinite integral. We shall, first of all, confine our attention for the next few chapters to this inverse problem of diffrentiation diffrentiation and then define a definite integral as the limit of sum. Finally, Finally, we shall discuss the conditions under which it may be true to assert that the operations of derivation and of integration are inverse operations. That means we shall establish the validity of the process of the employment of indefinite indefinite integration for the purpose of evaluating definite integrals. integrals . INTEGRATION AS ANTI - DERIVA DERIVATIVE : DEFINITIONS : THEOREMS. DIFFERENTIALS: Up to this point in our work, for y = f (x) we have regarded regarded dy/ dx as a composite symbol for the derivative f !(x) , whose component parts, dy and dx , had no meaning by themselves. It is now convenient convenient to modify this point of view and attach meaning to dy and dx, so that thereafter we can treat dy/ dy/ dx as though it were a fraction in fact as well as in appearance. We We shall not however enter into any discussions on it. We We shall only state that, for a function of a single variable variable y = f (x) , the the diffrential of y denoted by dy is is the product of the derivative derivative of y (with respect to x) and the diffrential diffrential of x denoted by dx. Thus,
Diffrential of y = f (x) is dy = f (x)dx. Fo r
4
3
y = x , dy = 4x dx , or simply d(x 4) = 4x3 dx . Thus d (sinx) = cosx dx , d (y 2) = 2y dy , d (tanu) = sec 2u du.
INTEGRATION AS ANTI-DERIVATIVE Integration defined as an anti-derivative, i.e., the inverse of a derivative or the inverse of a diffrential expression may be put in the form of a question – What is the function, function, which on being differentiated differentiated produces a given function ?
Bansal Classes
Page # 1
For example, what is the function which on being differentiated with respect to x produces 2x? In this case, we know from Differential Calculus that x 2 is the function required. We, therefore, conclude that x 2 is an integral of 2x with respect to x. Similarly, we may say that sinx is an integral of cosx, e x is an integral of e x, since the derivative of sinx is cosx and the derivative of e x is ex, both being with respect to x. In general , if we consider
d
f ( x ) # " ( x ) dx or, using differentials d f(x) = " (x) dx ; then an integral of " (x) with respect to x or an integral of " (x) dx is f (x) and symbolically, we write,
$ "(x ) dx # f (x ) where the symbol $ which is an elongated S (the first letter of the word sum, or, of the Latin word Summa) is known as the sign of integration. Now we come to some formal definitions: The actual process of finding the function, when its derivative or its differential is known, is called Integration as anti-derivative ; the function to which the integration is applied is called Integrand and the function obtained as a result of integration is said to be Integral. In the above case, " (x) is the integrand and f (x) is the integral. The process of integrating many ordinary functions is simple, but in general, integration is more involved than differentiation, as will be evident from future discussions. Summary:
d
If
dx
%F( x ) ' C& = f (x) then F(x) + C is called an antiderivative of f (x) on [a, b] and is
written as
$ f (x ) dx = F(x) + C.
In this case we say that the function f (x) is integrable on [a, b]. Note that every function is not integrable.
0
e.g.
* f (x) = ( ) 1
if x , Q if x + Q
is not integrable in [0, 1]. Every function which is continuous
on a closed and bounded interval is integrable. However for integrability function f (x) may only be piece wise continuous in (a, b)
Bansal Classes
Page # 2
ILLUSTRATIONS FOR TEACHERS FIRST LECTURE ON INTEGRATION XII (ANTIDERIVATIVE AS PROCESS OF SUMMATION OR REVERSE PHENOMENON OF DIFFRENTIATING)
d
i.e
dx
-F(x) ' C. # f (x) / $ f (x) dx # F(x) ' C
Function (Loving integrands)
$
1.(a )
x n dx
#
$
x
n '1
n '1
$
' C, n 4 1
( b) (ax ' b) dx # n
Illustration
(ax ' b) n '1 a (n ' 1)
$e
'C
n 4 11, n , R
$
1
$
3.(a ) e x dx # e x
' C; $ eax ' b dx #
(b) a x dx; a px ' q dx #
$
$
a
px ' q
p ln a
'b
a
a30
$ 1 sin( ax b ) dx cos(ax ' b) ' C ' # 1 $ a $ cos x dx # sin x ' C 1 $ cos(ax ' b) dx # a sin(ax ' b) ' C
4. sin x dx # 1 cos x ' C
$ 1 $ sec (ax ' b) dx # a tan(ax ' b) ' C $ cosec x dx # 1 cot x ' C 1 $ cosec x dx # 1 a cot(ax ' b) ' C
$
2
2 2
# sec x ' C $ $ cosec x cot x dx # 1 cosec x ' C
6. sec x tan x dx
7.
dx
$ 1' x ; $ a dx $ 11 x ; $ 2
2
$x
dx x2
dx 2
'x
#
dx a2 1 x2 dx
$ 11 x ;
2
1 a
tan
x2 1 a2
#
a
dx #
x 2 x '3
'3x 110
dx;
dx;
m '1 '1
x ln 2
'C
ln 2 ' 1
dx
x dx
x dx
$ 3 1 2x ; $ a ' bx ; $ x
2
(1 ' x)
3
dx
$
' C, m 4 1;
2x'3 '
$ 2x13 ;
x
' 2x ' 1 dx;
x 2
1 5x ' 6 2 x '1 1 5 x 11 dx ; a x · e x dx ; (2 x ' 3x ) 2 dx; x 10 3x e ' e5x dx; a mx · b nx dx 1x x e 'e 1 ' sin x ; cos 2x cos 3x dx; sin 4 x dx;
$
$
2
$
2
cos 3 x ' sin 3 x
3
4
$
dx; sin x 2 dx cos x ' sin x dx cos 2x ' 2 sin 2 x 1 1 cos x x ; dx; dx # tan 2 2 1 ' cos x 1 ' cos x 2 cos x 2 2 2 2 2 cot x dx; sec x cosec x dx; cot x cos x dx
$ $ $ $ $ $ $ $ (cot x 1 cos x)dx; $ tan x sin x dx # $ (tan x 1 sin x) dx a sin x ' b cos x $ sin x cos x ; $ sin x cot x dx [Ans. C 1 (sin x ' cosec x)]; dx sin x ; $ 1 1 sin 3x $ cos x (1 1 3 sin x) dx 2
2
2
2
2
3
2
3
3
2
3
$
x 2 ' cos 2 x x2 '1 x4
$ 1' x
a sec
'2
x m'1
2
a 11 x
1
14
3
2
m ln x
# $ x ln 2 dx #
2
11 x
# sin
$
: 1 7 dx; dx ; 5 9 e x 6 $ 2 x
$
dx; ln8
$ $ $ $ $ cos x 1 cos 2 x $ 1 1 cos x dx [Ans. (2 sin x ' x)] $ sin 2x · cos x dx; $ sin x dx; $ cos x dx; $ cos x dx; cos 5x ' cos 4 x 1 $ 1 1 2 cos 3x dx [Ans. (1 sin x 1 2 sin 2x)];
2
5. sec x dx # tan x ' C
1ln x 2
dx; e
x4 ' x
$x 'C
dx ; e
2ln x dx
$x e ax
$
x
ln 2 ' ln x
$
$ x dx # ln x ' C ln (ax ' b) dx # 'C ( b) $ ax ' b a
2.(a )
e ln
11 x
a
2
dx;
$ ( 2 x 1 7)
· cosec 2 x dx;
$
x2
$ 1' x
2
dx;
dx 9 1 4x 2
dx ( x 1 3)(x 1 4)
;
$ (x
dx 2
1 4 x ' 4)(x 2 1 4x ' 5)
General: In each case, find a function f satisfying the given conditions. (a) f '(x 2) = 1/x for x > 0, f (1) = 1; (b) f '(sin2x) = cos2x for all x, f (1) = 1 (c) f '(sin x) = cos 2x for all x, f (1) = 1 1
1
1
1
2
2
3
3
[Ans.(a) f (t)=2 t – 1 if t > 0, (b) f (t)= t – t2+ if 0 0 t 0 1, (c) f (t) = t – t3 +
Bansal Classes
if | t | 0 1]
Page # 3
Notes on indefinite integration: 1. Geometrical interpretation
/
x
2
y=
$ 2x dx #
'C
y=
$ f ( x) dx = F(x) + C
2
F ' (x) = f (x) ; F ' (x1) = f (x 1)
$
Hence y = f ( x ) dx denotes a family of curves such that the slope of the tangent at x = x1 on every member is same. i.e. F ' (x1) = f (x) (when x1 lies in the domain of f (x)) hence antiderivative of a function is not unique. If g 1(x) and g2(x) are two antiderivatives of a function f (x) on [a, b] then they differ only by a constant i.e. g1(x) – g2(x) = C Antiderivative of a continuous function is differentiable
(2)
i.e. If f (x) is continuous then
$ f ( x) dx = F (x) + C
/
F ' (x) = f (x) / F ' (x) is
always exists / F (x) is differentiable If integrand is discontinuous at x = x 1 then its antiderivative at x = x 1 need not be discontinuous.
(3)
i.e.
$ x 11 3dx . Here x – 1/3 is discontinuous at x = 0.
e.g.
$
but x
11 3
dx #
d
3 2
x 2 3 ' C is continuous at x = 0
-F( x ) ' C. = f (x) / f ( x ) dx # F(x ) ' C then only we say that f (x) is integrable. dx Antiderivative of a periodic function need not be a periodic function
(4)
If
(5)
e.g.
$
$
f (x) = cos x + 1 is periodic but (cos x ' 1) dx = sin x + x + C is aperiodic.
ELEMENTARY INTEGRATION HOME-WORK AFTER THE FIRST LECTURE Find the antiderivative/primitive/integrals of the following by simple manipulation/simplifying and converting them into loving integrands . Q.1
$
x
2 . e dx
$
Q.9
$
Q.14
e 3 !n x
4 cos
$
x 2
· cos x · sin
Q.6
$
x dx
Q.10
$
sin x
' cos x
1 ' sin 2x
11 dx 2 x '1
x6
Bansal Classes
3
Q.15
$
Q.12
11 dx sec 2x ' 1 sec 2x
Q.16
dx (cosx + sinx > 0)
Q.21
$
Q.3
1 1 tan 2 x
$ 1 ' tan
sin 3 x ' cos 3 x sin 2 x cos 2 x
Q.19
dx
Q.22
2
x
dx
cos 2 x
$
(ea ln x + ex ln a)dx(a > 0)Q.7
(3 sin x cos x 1 sin x) dx
$ 2 x ' 1 dx
$
$ 1 ' cos 2x dx
21 2
1 ' cos 2 x
Q.2
2
x
Q.18
Q.20
1 e 4 !n x dx 1 e 2 !n x
e 5 !n x
Q.5
Q.11
x
cos 2 x sin 2 x
cos x
1 ' tan 2 x
$ 1 ' cot
Q.4
dx
2
x
dx
1 ' 2x2
$x
Q.8
2
-1 ' x . 2
dx
1 sin x
$ cos x ' sin x (2 + 2 sin 2x) dx
$ $
cos x° dx
11 dx x12
2x
Q.13
Q.17
(1 ' x)
$x $
2
-1 ' x . 2
11
e 2x e
x
dx
dx
1 cos 2; dx cos x 1 cos ;
$
cos 2x
$
x
' x2 ' 1 dx Q.23 2 2 -1 ' x . 4
$
1 1 sin 2x dx
Page # 4
Q.24
$
' cos 6 x
6
sin x 2
2
sin x . cos x
dx
* 2 : 9 < x 7 (sin 89 8 ' 4 65 1 )
: 7 < x 7 = ' dx 9 8 4 65 ?>
Q.25
$
Q.27
$
-
' 1. - x 2 1
x
x x cos 4 x
x
' x' '1
.
x
dx
Q.26
$
Q.28
A function g defined for all positive real numbers, satisfies g'(x2) = x3 for all x>0 and g (1) = 1. Compute g (4).
Q.29
Q.31
$
* 7= 2 : x (sin ; sin (x 1 ;) ' sin 89 2 1 ; 65 ? dx ) >
Q.30
$
* cot 2 2x 1 1 = ( 2 cot 2x 1 cos 8x cot 4x? dx ) >
Q.32
Q.33
$
Q.35
$
Q.38
sin 2 8
' 3x2 ' 4 x ' 5 dx 2x ' 1
2x3
dx
dx
$
$
$ sin x 1 sin k ' cos x 1 cos k dx
Q.44
$ x ln(ex)dx
25 ' 4 x
$
sin 2x 1 sin 2k
Q.41
dx
$
Q.39
1 ' sin x
$
Q.34
Q.36
9 1 16 x 2
$
2
1 ' 2 cos 5x
$
x
' sin 5x 1 sin 3x dx 2 cos x ' 1 1 2 sin 2x
sin 2x
1 sin 4 x
cos 4 x
1 ' cos 4x
-x
2
dx (cos2x>0)
' sin 2 x. sec2 x 1 ' x2
Q.40
$
tan
2
2
dx
11 : 1 ' cos x 7
8 9
sin x
2 ' 3x2
$
'3 dx Q.43 ( x ' 1)
x 6
dx
dx
1 tan x
Q.37
cos 8x 1 cos 7 x
Q.42
cot x
-
x 2 1 ' x2
$
.
5 dx 6
dx
sin x cos x cos 2x cos 4x dx
x
ANSWER SHEET 2x . e x
Q.1
1 ' !n 2 x3
Q.5
3
1
Q.8
x
+ tan 11 x + C Q.9
Q.10 sin 2x + C Q.14
* ()x 1
1 2
2
Q.11
a
1
?> + C
Q.17 e x + e1x + C Q.20
x5 5
x3
1
3
+ x 1 2
tan 11 x
'1
'
ax !n a
+ c
2
Q.25 Q.28 Q.31 Q.33
Q.36
2
cos3x 3
Q.12
+ C
180
<
sin x° + C
1 x + C
tan x 1 x + C
Q.16
2x + 3 ln (x 1 2) +C
Q.18
x+C
Q.19
2 (sin x + x cos ;) +C
Q.26 Q.29
5 cos8x
x3 3
1 10
8
'
x2 2
tan 11
Q.32
+ C
'
3x 2
2x 5
'
7 4
+ C
Bansal Classes
ln(2x+1)
Q.13 ln x + 2 tan11 x +C
Q.15
+ C Q.21
67
1
tan x 1 x + C
Q.4
1 (cot x + tan x) + C
Q.7
sec x 1 cosec x + C
+ C
Q.27
1
(x 1 sin x) + C
Q.30
1 2 cos x + C
Q.34
Q.37
2 cos
2
2
11 = ( 3 ' tan x? +C ) >
tan x 1 cot x 1 3x + C
1 1
Q.22
1 * x3
Q.24
Q.23 (sin x + cos x) sgn (cos x - sin x) + C x2
sin 2x + C
1 1 1 1 1 *( cos 9 x ' cos10 x ' cos11x ' cos12x =? + C 10 11 12 )9 >
!n (2x ' 1) =
2
1
(tan x + x) + C Q.3
xa '1
Q.6
+ C
1
1
Q.2
+ C
x 2
x 2
8
+ C
+ C
tan x 1 tan11 x + c
cos4 x
4
Q.35
Q.38
1 4
sin11
4 3
x+ C
tan x 1 sec x + C
Page # 5
Q.39
sin 3x
sin 2x
1
'C
Q.40
Q.41
3 2 (sin x – cos x) + (sin k + cos k)x + C
Q.42
C –
2 x
+
2 1 3x
@
3 1
–
5x
– 2tan – 1x
5
1
2
+ tan 11 x + C
x
Q.43
1
1
cos 8x + C
64
SECOND & THIRD LECTURE
Substitution
By part (product rule)
xx + C
Q.44
(T. O. I.)
Partial (fraction)
Kuturputur & Misc.
SUBSTITUTION: Theory :
$
I = dI dx dI
/
dz
f (x) dx
dx
= f (x) ; =
dz
d I dx
= " ! (z) dI
= f (x) . " ! (z) or
·
d x dz
Hence I #
and let x = " (z)
$ f -"(z). "' (z) dz
= f (" (z))
dz
" ! (z)
....(1)
Substitution is said to be appropriate if the integrand in (1) is a loving one . f ! (x ) or dx $ [ f (x) ]n f ! (x) d x $ n
If
[f (x)]
Start with
$ -tan x .dx # ln sec x ' C # 1 ln(cos x) ' C; $ (cot x) dx # ln(sin x) (loving integrals)
Example:
$
1.
f (x) = t
tan(!n x ) x
dx ;
2.
$ cos(x1a ) dx ;
4.
dx
5.
$ cos( x1a )cos( x1b) ;
7.
$ tan 3x tan 2x tan x dx
8.
$ (sec
10.
11.
: $ 889
$
2
6.
x · 5 ' tan x ) dx ; 9.
cos x x
1
$ 1 1 cos x
x cos x
$ (x sin x ' cos x) sin 2x dx
$ sin5x.sin3x
$
2
2
];
dx
(use 2x = 5x 1 3x)
ln : 8 x ' 1 ' x 2 75
9
1' x
x
2
6 dx [Ans.
ln 2 : 8 x ' 1 ' x 2 75
9
2
6 ' C ]
7
· sin x 5 dx [Ans. 2 x cos x + C] 5
6
sec x dx cos(2 x ' ;) ' cos ;
Bansal Classes
dx [Ans. ln (1 1 cos x ) 1 cot
x cos x
cosx
3.
1 ' sin x
[Ans C 1
2(cot ; 1 tan x ) ] sin ;
Page # 6
x2
$ 1' x
12.
15.
$
6
1
dx [Ans.
tan x sec 2 x
18. (tan x – x tan x)dx
$ 2 ' tan
19.
21.
23.
2
x
(tan x 1 x ) 2
[Ans.
2
dx
20. $
cot x !n
sin 2 x
dx ; 14.
17.
$
1' x 6
dx ;
tan x
$ sin 2 x dx ;
+C]
( x 2 ' 2e x sin x 1 cos 2 x ) 2
2
x
x
1
dx
[Ans. C –
' 1) 1 sin 2( x 2 ' 1) $ x · 2 sin(x 2 ' 1) ' sin 2(x 2 ' 1) dx 2 sin( x
dx ;
'1 dx x (x ' 1)
22. $
x ' e x (sin x ' cos x ) ' sin x cos x
24.
12
x 2 tan 11 x 3
2
! n (sec x)
$ (a sin 2 x ' b cos 2 x)2 dx ;
$
$ 1'x
tan ( x ) ' C ]; 13.
2
sec x tan x
x5
3
dx ; 16. $ sec x ln (sec x+tan x)dx;
x 3
3
11
2( x 2 ' 2e x sin x 1 cos 2 x )
2
x 2 '1
[Ans. ln sec
]
'C]
2
ASKING:
$
1.
3.
2
x sin x
x1 2
$ 1' x
34
32
dx
1 1 dx ;
[Ans.
4
-x 3
34
cos
$
2.
B
B sin 2 B
dB
1 ln -1 ' x 3 4 ..]
: < ' x 7 ' C;F n (sec x tan x ) C or n tan l l # ' ' 8 5 C $ 4 2 9 6 E (loving integrands) x $ cosec x dx # ln(cosec x 1 cot x) or ln tan 2 ' C CD sec x dx
Lighter faces of the integrands:
dx
Example: 3. 6.
$ $
1.
$ cos x(1 1 4 sin
cos ec(tan 11 x ) 1' x 2 dx 3sin x ' cosx dx
dx ; 4. 7.
;
$
2
#$
x)
cos2x sin x
dx cos 3x ; 2.
dx ; 5.
tan x 'secx 11
$ tanx1secx'1 dx ; dx
$ sinx cos x 2
dx
$ asinx'bcosx ; dx
$ tan x ' cot x ' sec x ' cosec x
$ secx 'cos ecx
1
(sin x 1 cos x 1 x ) ' C ] 2 Other examples (may be taken as recap of 2nd lecture before the 3rd lecture commences) 8.
1.
4.
e
x 2 ' !nx
dx ;
ex (1 ' x)
$ sin (xe ) dx ; 2
x
Bansal Classes
2.
$e
dx x
'1
5.
;
3.
$ (27e
9x
[Ans.
11 $ ex ' 1 dx e
x
'e
12 x 1 / 3
)
3x 4 3
dx [Ans.
( 27 ' e ) 4
'C] ; Page # 7
6.
$ (1 ' x
2
) ' (1 ' x )
e
x
$
11
1.
$ sin
dx
1
[Ans.
x 'e Trigonometric Functions: e
n
: 81 ' 9
[Ans. 2 tan 11 e x
x e11 ' e x 11
8.
1' x2
[Ans. ln
2 3
dx
$
7.
x dx
x
$ sin
m
x . cos x dx ; 2.
4
e
' ex ) ]
2
x cos x dx [Ans.
: 11 H cos8 2 cot 9 a 1x
$
x2' a2
Loving Integrals:
$
dx
-x
' 4.
2
6
a
2
'
1;
(9 1 x 2 ) 3
$
1
2
5 [Ans. x 1 ' x 6 2 dx ;
x6
];
1 2 x 2 ' 3x ' 5
x dx
$
a
3
1 x3
2
1x
dx
6
$
note that
$
a 2 ' x 2 dx J
x
2
1 a 2 dx
to be executed by parts.
x 2 ' a 2 75 &
$x
;
1 x2
dx
$
6
2
4x
x
$
dx a
11 x 7
x
= a 1 x ; x = a sin B ? ? ? ? a 2 ' x 2 ; x = a tan B ? ? ? ? x 2 1 a 2 ; x = a sec B ? ? a 2 1 x2 ?> 2 2 H IB x = a cos a 2 ' x2
# ln: 8x ' 9
: x 1 1 sin 4 x 7 1 1 sin 3 2x ' C 8 5 ] 16 9 4 6 48 1
Examples :
2
dx
# t2 ]
6
General Substitution : 2
with 1 ' x 2
1 1 G ex 11 # t 2 ]
ln ( x
e
1 ' x 2 75
'C
x2 1 a2
dx 2
;
' a2
# ln: 8x ' 9
$x
x 2 1 a 2 75 (loving integrals)
6
dx x
2
;
1 a2
$
dx x
2
1 a2
;
$
dx x
2
' a2
Examples: 1.
5.
7.
$
sin 2x 9 1 sin x 4
$a e
dx
2 x
$
dx ; 2.
[Ans.
' b 2e1 x
tan x 2
a ' b tan x
e x dx
$
e
2x
11
ex
$ 4'e
; 3.
11 : 8 ae
7 5 ]; tan 8 5 ab 9 b 6 1
2x
dx ; 4.
$
dx 2ax 1 x 2
;
2
x
6.
: x 1 1 7 dx [Ans. tan 11 x ' 1 $ 89 1 ' x 2 65 1' x2
]
dx (b > a)
Some More Examples : 1.
2.
$
e x dx 5 1 4e x
$ 3x
' e2x
4x ' 3 2
' 3x ' 1
Bansal Classes
)
* [Ans. ln ( e
dx ; 3.
$
x
1 2. '
e2x
2x 1 1 4x 2 ' 4x ' 2
1 4e x ' 5 =? ]; >
dx ;
4.
$x
x dx 4
' x2 '1 Page # 8
H
Note for integral of the type write ax + b= A
d dx
$ px
ax ' b
' qx ' r 2 ( px ' qx ' r ) ' B 2
ax ' b
$
dx or
px
2
dx
' qx ' r
Home Work: after 2nd & 3rd lecture + Berman : Q.No 1703 to 1831. FOURTH & FIFTH LECTURE (INTEGRATION) (May consume two period) Theory:
If f (x) and g (x) are derivable functions then
[f ( x ) . g ( x ) ] # f ( x ) . g ' ( x ) ' g( x ) . f ' ( x ) dx f ( x ) . g ' ( x ) dx = f (x) . g (x) – g ( x ) . f ' ( x ) dx
$
K
I
$
II
$ f (x) · g(x) dx
I=
"
"
I
II
= 1st function L integral of 2 nd
1
$
(diff. co-eff. of 1 st) L (integral of 2 nd) d x
Remember ILATE for deciding the choice of the first and second function which is not arbitrary. Start with
$ x cos x dx .
EXAMPLE:
$
1.
x tan
11
x dx ;
sin 11 x
$ (11x
4.
2 3 / 2
cos 11
)
2.
dx ; 5.(a)
$ cosec
$e
x
2
x · ln (sec x ) dx ;
(1 ' x ) · ln ( xe ) dx ; (b) x
1
3.
$
sin x·ln(sec x + tan x) dx ;
$ln(1' x)
1'x
$ x dx ; 7. $ sin(ln x)dx ; 8. $ x 1a dx $ 9. $ sec x dx $ cosec x dx ; 10. $ x sin x cos x dx ; x dx ; 11. $ 12. $ x cos x cos 2 x dx ; 13. $ x e 1 ' sin x 14. $ ( x ' 3x ' 1)e dx $ e sin 3x dx 6.
2
2
3
3
x 2 'a 2 dx ;
2
2 3x
3
3x
$ x ln x dx ; 3
2
(a)
$
dx ;
2x
$
cos 11 x
cos 1 x 1
dx [Ans. C – ' 2 2x x3 Two Classic Integrands (HO NA HO WO NA HO): 15.
dx ;
16.
11 x
2
2x
]
$ -f (x) ' xf !(x).dx = x f (x) + C (i) $ e (tan x 1 ln cos x ) dx ; (ii) $ e (sin x 1 cos x ) dx (iii) $ e [ln( sec x ' tan x ) ' sec x ]dx [Ans. e ln (secx + tanx)]
e -f (x) ' f !(x) .dx x
# e x f (x) ' C
& (b)
x
e.g.
x
x
x
EXAMPLE FOR ILLUSTRATION:
xe x
$ (1'x)
1.
3.
$
2
dx [Ans.
e
x
1' x
];
2.
$
ex
1 ' sin x 1 ' cos x
dx ;
[ sin(ln x) + cos(ln x)dx [Ans. x sin(ln x) ]; 4.
Bansal Classes
$
e 2x (sin 4x 1 2) 11cos 4x
dx [Ans.
1 2
e 2 x cot 2 x ];
Page # 9
5.
6.
x 'sin x
$ 1'cosx
$e
7.
cos 11 x
$
e
x
·
dx
[Ans. x tan
( x ' 1) ' 1 1 x 2 ( x ' 1)
11 x
2
2
x
];
2
dx
: x 4 ' 2 7 8 5 8 (1 ' x 2 )5 2 5 dx [Ans. 9 6
[Ans. C – x
2
e (1 ' x ' x ) 2 32
(1 ' x )
e
cos 11 x
]
1' x
'C]
EXAMPLE (FOR ASKING): (1)
(3)
(5)
(7)
(9)
(11)
ex
$x
(1 ' x · ln x )dx [Ans. e ln x + C] ; (2) x
x 1 sin x
$ 1 1 cos x
dx [Ans.
!n x
$ (1'!n x) x 2 ex
$ (x'2)
e x ( x 1 1)
$
e x (x 2
d x [Ans.
2
dx [Ans. e
2
$ (x ' 1)
1 x cot
3
dx
' 1)
( x ' 1)
2
[Ans.
x12 x'2
ex (1 ' x )
(6)
];
1 ' lnx x
(4)
] ;
2
x
2
(8)
];
$
$ $
(10)
] ;
1 1 cos x e tan
11 x
x dx [Ans. – e cot
(1' x ' x 2 )
1' x 2 e x ( x 2 '5x '7) ( x '3) 2 e x (1' x ' x 3 )
(1' x 2 )3 / 2
tan dx [Ans. x e
dx [Ans.
x 2
]
11 x
e x ( x ' 2) x'3
]
]
ex x dx [Ans.
1 ' sin 2 x
$ (1 ' cos 2x )e1
x 2 '1
]
x
x
dx [Ans.
e sec x 2
]
x
dx
[Ans.
: !n(!n x )' 1 7 dx ; 5 $8 !n 2 x
(12)
x
$
x
e (1 1 sin x )
e ( x 1 1) x '1
]
x [Ans. x ln (ln x) –
ln x
];
TOUGHER PROBLEMS
$
1.
sin 4 x · e
tan 2 x
tan dx [Ans. C – 2 e
2
x
cos 4 x ]
GENERAL CONCEPT:
$ eg( x ) -f (x )g' ( x ) ' f ' (x ).dx Proof: 1.
2.
[Ans. eg(x) · f (x)]
Explain
: x 2 cos2 x 1 ( x sin x ' cos x ) 7 8 5 dx $e 2 8 5 x 9 6 x sin x ' cos x 7 ( x sin x ' cos x ) : 2 1 e cos x 8 5 dx $ x2 9 6 : cos x 7 : cos x 7' 75 : ( x sin x ' cos x ) 8 x cos x 8 5'8 5 dx $e e(x sin x + cos x) · 8 x 6 9 x 6 5 ; 9 9 6 $ etan x -sin x 1 sec x .dx = $ e tan x sin x dx 1 $ etan x sec x dx ( x sin x ' cos x )
cos x x
+ C
= – e tan x · cos x + $ e tan x sec 2 x cos x dx 1 $ e tan x sec x dx = – e tan x · cos x
Bansal Classes
Page # 10
$
Integrals of (*) e
ax
sin( bx ' c) dx or
$e 2. Assume $ e
*
1. Integrate
ax
2x
$e
ax
cos( bx ' c) dx (* Discuss three different ways)
sin bx dx
cos 3x dx = e2x(Acos 3x + B sin 3x) and then differentiate both sides.
3. Use Euler's equation Let hence P + iQ = 1
P + iQ =
P=
$e
$e
·e
ax
ax
ibx
e(a+ib)x =
a ' ib
cos bx dx dx =
$e
( a ' ib ) x
a 1 ib a
'b
2
and Q =
$e
ax
sin bx dx ,
dx
eax (cosbx + i sin bx)
2
(ae ax cos bx ' be ax sin bx ) 1 i(ae ax sin bx 1 be ax cos bx )
=
a
– P =
2
' b2
e ax (a cos bx ' b sin bx )
ax
e (a sin bx 1 b cos bx )
Q= a 2 ' b2 a 2 ' b2 Note: Discuss the reduction formulae of (sin x)n , (cos x)n and (tan x) n and give the remaining 3 for home work: GENERAL PROBLEM:
x 2011
If the primitive of the function f (x) = then find (m + n) (where m, n
(1 ' x )
2 1007
w.r.t. x is equal to
, N).
m
1 : x 2
7 8 5 'C 2 5 8 n 9 1 ' x
[Ans. 3018]
Home Work after 4th & 5th lecture : Berman : Q.No 1832 to 2011. SIXTH LECTURE (INTEGRATION) PARTIAL FRACTION & MISC. & KUTURPUTUR Techniques of Partial Fraction (Discuss the rules of splitting a given fraction) Dergee of Nr is < then Dr Degree of Nr Dr
Loving Integrands
$a
dx 2
1 x2
#
1 2a
: a ' x 7 5 9 a 1 x 6
ln 8
&
$x
dx 2
1 a2
#
1 2a
: x 1 a 7 5 9 x ' a 6
ln 8
Examples: 1.
2
'41x 191 $ (x11)( x'3)(x14) dx ; 2x
4(a)
$x
x 3dx 4
' 3x 2 ' 2
2
(x =t), (b)
2.
$ (x
x 3 12x 2 ' 4
$x
2
( x12) 2
( x 2 1 x ) dx 2
' 2) ( x 2 ' 1)
dx;
3.
x
$ (x11)(x
2
'4)
dx;
dx
dx : x 2 1( x 2 11) 7 x 2 1 ( x 2 1 1)dx 5 5.(a) $ 3 = $ 8 8 x 3 '1 5 dx , (b) $ x 3 1 1 = $ x '1 x3 11 9 6 dx
6.
dx
$ sin x(3 ' 2 cos x) (cos x = t)
General: (1)
dx
$ sin 2x12sinx
Bansal Classes
; (2)
sin x
$ sin 4x
dx ; (3)
1 ' x cos x
$ x(1 1 x e
2 2 sin x
)
dx
Page # 11
Type 1: Manipulating Integrands: (Kuturputur) 1.
4.
7.
dx
$ x(x
$x
9.(a)
$ x(x
' 1)
n
dx 4
3
( x '1)
$x
8. if
dx
2
12
2
12
$ (x $ (x
10.
6.
( x 4 11) dx
$x
2
x
4
2
' x '1
$
6
4
'x 'x
2
) · (2 x
1' x ' x
4
2
' 3x ' 6)
dx [Ans.
2
12
$ (11x
dx
-
2
x 1' x x4
[Ans.
[Ans.
n
n 11 n
.
' x2 ' 1 x
x x 2 ' 2 x cos x ' 1
dx [Ans.
(2x
6
4 3 / 2
)
;
;
]
+ C]
' 3x 4 ' 6 x 2 ) 3 2 18
]
x 4 ' x 3 ' x 2 + C]
: x ' 8 l n [Ans. 8 9
dx 2
)
x dx
3.
3
' 2x cos ; ' 1)3 / 2
2
$2
$x
2 5
dx ;
( x cos ; ' 1) dx
2 ' 3x ' 4 x 2
(b)
$ (11x
11 $ 4x 3 1x dx ; x
dx =
' x '1
2.
;
' 1)
5.
x 2 1x x
2
x7
: 8 x ' 1' x 75 9 6 2
1 ' x 2 75
5 6
x
1
x2 '1 x
'C]
Type 2: Forcing integrand by part (1)
$ (x
dx 4
1 1)
2
#$
4x 3 4
1
. 2
( x 1 1) 4x & #%# $
(3)
$
1' x
2
(2)
"
dx =
$
$x
dx 3
(1 ' x)
3
; 1 + x = t2 !
$ (t
I
II
x4
3
dx ;
x 1' x & #%# $ 2
2t 2
. 3
1
1 1) t 3 & #%# $
dt
"
I
II
· x 3 dx "
I
II
I= x
3
1' x
2
1 3$ x
2
I
1 ' x dx # x
3
1' x
2
1 3$
x 2 (1 ' x 2 ) 1' x2
dx
GENERAL PROBLEM: Suppose f (x) is a quadratic function such that f (0) = 1 and f ( – 1) = 4. If 1. is a rational function, find the value of f (10). [Ans. 521] Type 3 :
$
5x 2 1 12
-x
2
1 6x ' 13.
2
f ( x ) dx
$x
2
( x ' 1) 2
dx
put 5x2 – 12 = A(x 2 – 6x + 13) + B(2x – 6) + C (2)
If g(x) =
$ (x
2
2
1 12) 1 6x ' k ) 2
(x
dx (where k
, N)
is a rational function find the sum of all
possible values of k.
[Ans. 21]
Home Work : Selected problems from Berman from 2012 to 2067 (say 15)
Bansal Classes
Page # 12
SEVENTH LECTURE (INTEGRATION) Integrals of the Type of : (Trigonometric functions) Type
dx
$ a 'bsin
1
dx
2
$ a 'bcos x
$ asin
2
x
dx 2
x ' bcos 2 x 'csinx cosx
Multiply Nr and Dr by sec2x and proceed/enjoy
dx
Type
dx
$ a'bsinx
2
dx
$ a 'bcosx
$ a'bsinx'ccosx
Convert sin x and cos x into their corresponding tangent to h alf the angles.
dx
Examples on T 1 & T 2: 1.
dx
$ 5'4sinx ;
4.
7.(a)
(b)
8.
$ 5 ' 4 cos x
$ cos x(5 ' 3 cos x) $ 513cosx dx
$ 3 ' 5 cos x
; 9.
$ (a ' b cos x)
General :(i)
$ sin
(ii)
3
2
$ sin
x
2
; 3.
dx
$ 3 ' cos
2
x
$ 3'2sinx 'cosx ;
;
(5 1 3 cos x ) dx
$ 4 ' 5 sin x
;
dx # 1
x 3
'
5
$
dx
3 (5 1 3 cos x )
'C;
dx
$ 1 1 cos x ' sin x ;
10.
a 7 dx [Ans. – 2 : ln( a ' b cos x ) ' 8 5 +C] a ' b cos x 6 b 2 9
sin 2x 4
x ' cos x 4
dx 4
$
2
$ (3sinx 14cosx)
; 2.
dx
; 6.
1 5 1 3 cos x 1 5
dx =
sin 2 x
11.
dx
5.
dx
cosx
$ 415sin
dx
x ' cos 4 x
dx
[Ans. c – tan – 1(cot2x)]
;
Type 3: a sin x ' b cos x ' c
$ ! sin x ' m cos x ' n
(a)
dx;
Nr
d 7 # A( D r ) ' B: 8 Dr 5 ' C 9 dx 6
' 5e 1 d x 1x x – x =A(4ex – 5e – x)+ B dx 1 ( 4 e 5 e ) ; 3e +5e $ 4e x 1 5e1x dx 3e
(b)
[Ans. 3x 2 1 2 x ' 4
$ (x
(c)
2
1 6x ' 13) 2
Example : (a)
7 8
ln(4ex – 5e – x) –
x 8
+ c]
dx
6 ' 3 sin x ' 14 cos x
$ 3 ' 4 sin x ' 5 cos x
(b)
$
11 cos x 1 16 sin x 2 cos x ' 5 sin x
(c)
$ cos 2x · ln (1 ' tan x) dx
(a) [ A = 2 ; B = 1 ; C = 0] (d)
$e
sin x x
1 sin x 1 cos x
Bansal Classes
dx (mixed profile)
Page # 13
$x
Type 4:
$
e.g.
x
x (x
x 2 '1 4
'k x 2 '1
4.
$ 1' x
8.
$
'k x 2 '1
x2
$x
4
dx;
'1
$x
dx 4
$
6.
'a 4
cosx
$
cot B M tan B dB 9.
dx Take example with k = 7 & k = 1 23.
* : x x 1 1 7= 8 x 5? 1 ( 11 x 5? ] ( tan 8 [Ans. 8 2( 2 5? 8 5? () 9 6 > 2 dx ( x '1) 2. $ 4 3. $ 4 2 dx x '1 x ' x '1
' 1)(ln x ' 1) dx ; 4x x '1
dx [x6 = t] 5.
24
4
2x
Other examples: 1.
x17
$x
dx or
x 2 11
81sin 2x
tan B dB
$
7.
dx
$ cos ecx 'cosx
dx 10.
cot B dB
11.
cosx
$ 10'sin 2x dx
Note: cos x + sin x or cos x - sin x is loving in the numerator if denominator contains a + sin 2x or
' sin2 x . However if cos x + sin x or cos x - sin x appears in denominator
b
and sin 2x in numerator, then manipulate differently.
dx
e.g.
1
2 sin x cos x
1
1 1 1 (cos x 1 sin x ) 2
sin 2 x
$ cos ecx 1secx = 2 $ cos x 1 sin x dx = 2 $ cos x 1 sin x dx = 2 $
cos x 1 sin x
dx
INTEGRATION OF IRRATIONAL ALGEBRAIC FUNCTION Type 1 (a)
(b)
(c)
Type Type
Type
2 3
4
Note : Type Case-I:
dx
$ (x1;) $ $
( x 1;)(N1x )
dx
x12
dx
$ (a x
2
'bx 'c)
px ' q
this reduces to 2 5
$ (ax
;
' qx ' r
dx 2
( x 1 ; ) ( x 1 N)
$ (2x ' 1) O
dx px
dx
[Ans. 2 ln
$t
11
-
x12 3
x 1N ' x 1;
.' C ]
]
dx
; e.g.
px 'q
$ (ax ' b)
(Start: x = ; cos 2B + N sin 2B)
[Ans. ( x 1 2)(5 1 x ) ' 3 sin
dx
$ (a x ' b )
;)
$ (x 1 ;)
( x 1; )( x 1N)
51 x
(N >
;
$ (x ' 1) O
e.g.
e.g.
$ -x
4x ' 3 x 1' x 1 x
2
dx
2
' 5x ' 2 .
x12
dt 4
' 9t 2 ' 16
dx
px 2 ' qx ' r When (ax2 + bx + c) breaks up into two linear factors, e.g. I =
Bansal Classes
2
' bx ' c)
$ (x
dx 2
1 x 1 2)
x2 ' x '1
then
Page # 14
: A ' B 7 $ 89 x 1 2 x ' 1 65
=
1 x
2
' x '1
dx = A
$ (x 1 2)
dx
x2 ' x '1 &## # %## # $
'
B
put x 1 2#1 / t
$ (x ' 1)
dx
x2 ' x '1 &## # %## # $ put x '1#1 / t
If ax2 + bx + c is a perfect square say ( lx + m)2 then put lx + m = 1/t
Case-II: Case-III:
$ (ax
If b = 0; q = 0 e.g.
dx
' b)
2
px
2
'r
then put x =
1 t
or the trigonometric
substitution are also helpful.
$ (x
e.g.
dx 2
' 4) 4x 2 ' 1
Special Integrands: 1.
x2
$ ( x sin x ' cos x)
2
[Ans. tan(x – tan – 1x) + c]
dx
dx
2.
3.
4.
6.
$ (a'bcosx) $ (sin e
2
(a > b)
e x dx x
'e
1x
cos e )
dx
$ (sin x ' 2 sec x)
$
2
5.
$e
dx 4
[Ans. tan (ex – tan – 1 ex) + C]
x 2
( x 1 1) 3 ( x ' 2)5
sin x
: xcos3 x 1sin x 7 8 5 8 cos 2 x 5 dx 9 6 14
=
4 : x 1 1 7
8
5
3 9 x ' 1 6
Home Work : Complete Berman and T/S 1 st Exercise only) by the end of 8 th period. (Leaving Q. No. 2132 to 2150)
Bansal Classes
Page # 15
