Properties of Indefinite Integrals Video Lecture | Mathematics (Maths) Class 12 - JEE

solving indefinite integrals can be
quickened by some simple properties
let's learn about these properties one
by one the first property states at the
process of differentiation and
integration are inverses of each other
that is the derivative of integral of f
of X with respect to X is equal to f of
X and the integral of the derivative of
f of X with respect to X is equal to f
of X plus C where C is any arbitrary
constant let's see the proof for this
property let G be any antiderivative of
F that is d by DX of G of X is equal to
f of X let's take this as equation 1
then integral of f of X with respect to
X is equal to G of X plus C where C is a
constant of integration let's take this
as equation 2 now on differentiating
both the sides with respect to X we get
derivative of integral of f of X DX is
equal to derivative of G of X plus C
since the derivative of a constant is
equal to zero so we get D by DX of G of
X which is equal to f of X from equation
1 therefore innovative of integral of f
of X DX is equal to f of X now let's
prove that integral of derivative of f
of X with respect to X is equal to f of
X plus C we know that the derivative of
f of
eggs can be written as f dash of eggs so
the integral of derivative of f of X
with respect to X can also be written as
integral of F dash of X with respect to
eggs and integral of F dash of X into DX
is equal to f of X plus C here C is
called the constant of integration let's
evaluate the derivative of integral of 1
by 1 plus X square DX by using this
property we know that integral of 1 by 1
plus X square into DX is equal to tan
inverse X plus C where C is the
integration constant therefore the
derivative of integral of 1 by 1 plus X
square into DX is equal to derivative of
tan inverse X plus C this can be written
as d by DX of tan inverse X plus D by DX
of the constant on the Francey engine we
get 1 by 1 plus X square therefore the
derivative of integral of 1 by 1 plus X
square into DX is equal to 1 by 1 plus X
square now let's evaluate the integral
of derivative of sine X DX
we know that derivative of sine X with
respect to X is equal to cos x therefore
integral of derivative of sine X into DX
is equal to integral of cos x DX which
is equal to sine X plus C where C is the
constant of integration
raishin hence the integral of derivative
of sine X into DX is equal to sine X
plus C now let's move on to the second
property of indefinite integration it
thinks that two indefinite integrals
with the same derivative lead to the
same family of curves and so they are
equivalent let's see the proof for this
property let F and G be two functions
such that the derivative of the integral
of f of X with respect to X is equal to
the derivative of the integral of G of X
with respect to X now on transposing the
terms from the right hand side to the
left hand side we get derivative of
integral of f of X into DX minus
derivative of integral G of X into DX is
equal to zero this can be written as
shown here since the derivative of a
constant is equal to zero the integral
of f of X into DX minus integral of G of
X into DX is equal to C which is an
arbitrary constant now on transposing
the terms to the right hand side we get
integral of f of X DX is equal to
integral of G of X DX plus C now let's
see B equal to C 2 minus C 1 this
implies that integral of f of X DX plus
C 1 is equal to integral of G of X DX
plus C 2 where C 1 and C 2 are constants
of integration
so the family of curves integral of f of
X DX plus c1 and integral of G of X DX
plus C 2 where c1 and c2 belong to the
set of real numbers are identical thus
we can say that integral of f of X DX
and integral of G of X DX are equivalent
thus the boundary of curves that is
shown here can be expressed as integral
of f of X DX and integral of G of X DX
without mentioning the parameter let's
move on to the next property
of indefinite integration link the
parameter let's move on to the next
property of indefinite integration the
third property states that the integral
of sum of two functions is the sum of
the integrals of the functions
thus the boundary of curves that is
shown here can be expressed as integral
of f of X DX and integral of G of X DX
without mentioning the parameter let's
move on to the next property of
indefinite integration the third
property states that the integral of sum
of two functions is the sum of the
integrals of the functions that is if F
and G are two functions then integral of
f of X plus G of X into DX is equal to
integral of f of X DX plus integral of G
of X DX
before we proceed with the proof let's
look at property 1 of indefinite
integrals once again
it states that the derivative of
integral of f of X into DX is equal to f
of X in this property let's replace f of
X by f of X plus G of X which gives
derivative of integral of f of X plus G
of X into DX is equal to f of X plus G
of X let's take this as equation 1 also
the derivative of integral of f of X
into DX plus integral of G of X into DX
can be written as derivative of integral
of f of X into DX plus derivative of
integral of G of X into DX this is
further equal to f of X plus G of X
let's think this as equation 2 now from
equations 1 & 2 it can be observed that
the derivative of integral of f of X
plus G of X into DX is equal to the
derivative of integral of f of X into DX
plus integral of G of X into DX let's
take this as equation 3 but by property
2 on indefinite integration that is two
indefinite integrals having the same
derivative lead to the same family of
curves and they are equivalent so
equation 3 can be reduced as integral of
f of X plus G of X into DX is equal to
integral of f of X DX plus integral of G
of X DX and
integral of the sum of functions is
equal to the sum of the integral of the
functions let's evaluate integral of e
power X plus X into DX based on this
property by using property 3 this can be
written as integral of e power X into DX
plus integral of X into DX on
integrating we get e power X plus X
square by 2 plus C where C is the
constant of integration
hence integral of e power X plus X into
DX is equal to e raised to the power X
plus X square by 2 plus C X property of
indefinite integration states that the
integral of a constant multiple of a
function is a constant multiple of the
integral of the function that is the
integral of K into f of X DX is equal to
K into integral of f of X into DX where
K is any real number let's see the proof
of this property property 1 of
indefinite integration is derivative of
integral of f of X into DX is equal to f
of X using this property we have D by DX
of integral of K into f of X DX is equal
to K into f of X let's take this as
equation 1 where K is any real number we
know that the derivative of a constant
multiple of X is equal to constant
multiple into derivative of X where K is
any real number in a similar way D by DX
of
into integral of f of X into DX is equal
to K into D by DX of integral of f of X
DX this can be further simplified as K
into f of X since we know that the
derivative of integral of G of X into DX
is equal to G of X therefore derivative
of K into integral of f of X into DX is
equal to K into f of X let's take this
as equation 2 from equations 1 and 2 we
have d by DX of integral of K into f of
X DX is equal to D by DX of K into
integral of f of X into DX on
transposing the terms on the right hand
side to the left hand side we get D by
DX of integral of K into f of X DX minus
D by DX of K into integral of f of X
into DX is equal to 0 this can be
written as d by DX of integral K into f
of X DX minus K into integral of f of X
into DX is equal to 0 since the
derivative of a constant is equal to 0
we have integral of K into f of X DX
minus K into integral of f of X DX is
equal to C where C is an arbitrary
constant therefore integral of K into f
of X DX is equal to K into integral of f
of X DX plus C where C is an arbitrary
constant
hence integral of K into f of X DX is
equal to K into integral of f of X DX
let's now evaluate integral of ten into
secant square X DX based on this
property this can be written as ten into
integral of secant square X DX on
integrating we get ten into tan X plus C
where C is the constant of integration
hence integral of ten into secant square
X into DX is equal to 10 into tan X plus
C let's move on to the next property of
indefinite integration it states that
the integral of the sum of the constant
multiples of n functions with respect to
X is equal to the sum of the constant
multiples of integrals of n functions
with respect to X that is integral of K
1 into F 1 of X plus K 2 into F 2 of X
plus and so on plus KN into f n of X
into DX is equal to K 1 into integral of
F 1 of X DX plus K 2 into integral of F
2 of X DX plus K 3 into integral of F 3
of X DX plus and so on plus KN into
integral of F n of X into DX to prove
this property we combine properties 3 &
4 for a finite number of functions such
as f 1 f 2
f 3 and so on up to FN and for real
numbers such as K 1 K 2 K 3 and so on up
to K n this gives integral of K 1 into F
1 of X plus K 2 into F 2 of X plus and
so on plus KN into f n of X into DX is
equal to K 1 into integral of F 1 DX
plus K 2 into integral F 2 DX plus K 3
into integral of F 3 DX plus and so on
plus KN into integral of F n into DX
let's see an example based on this
property let's evaluate integral of 10 X
plus 2 sine X plus 15 divided by X plus
12 divided by 1 plus X square into DX by
using this property this can be written
as 10 into integral of X DX plus 2 into
integral of sine X DX plus 15 into
integral of 1 by X DX plus 12 into
integral of 1 by 1 plus X square DX this
is further equal to 10 into X square by
2 plus 2 into minus cos x plus 15 into
log modulus of x plus 12 into tan
inverse X plus C hence on simplification
we get the answer as shown