Pearson | PTE
Trusted by 3,500+ universities and colleges globally | Accepted for migration visa applications to AUS, CAN, New Zealand , and the UK
NCERT Solutions for Class 12 Maths Chapter 7 Integrals are discussed here. This chapter deals with definite and indefinite integrals. Integration class 12 also includes elementary properties of integration including basic techniques of integration. NCERT Class 12 maths chapter 7 solutions will be very helpful when you are solving the questions from NCERT books for Class 12 Maths. These NCERT Class 12 Maths solutions chapter 7 are prepared by subject matter experts that are very easy to understand. students can practice integrals class 12 ncert solutions to get good hold on the concepts.
NCERT solutions for class 12 math chapter 7 integrals are important for board exams as well as for competitive examinations like JEE Main, VITEEE, BITSAT, etc but without command, on the concepts of integrals ncert solutions meritorious marks cant be scoured. Therefore chapter 7 maths class 12 is recommended to students. Also, you can check the NCERT solutions for other Classes here.
Also read:
>> Integration as Inverse of Differentiation: Integration is the inverse process of differentiation.
In differential calculus, we find the derivative of a given function, while in integral calculus, we find a function whose derivative is given.
Indefinite Integrals:
∫f(x) dx = F(x) + C
These integrals are called indefinite integrals or general integrals.
C is an arbitrary constant that leads to different anti-derivatives of the given function.
Multiple Anti-Derivatives:
A derivative of a function is unique, but a function can have infinite anti-derivatives or integrals.
Properties of Indefinite Integral:
∫[f(x) + g(x)] dx = ∫f(x) dx + ∫g(x) dx
For any real number k, ∫k f(x) dx = k∫f(x)dx.
In general, if f1, f2, …, fn are functions and k1, k2, …, kn are real numbers, then ∫[k1f1(x) + k2f2(x) + … + knfn(x)] dx = k1 ∫f1(x) dx + k2 ∫f2(x) dx + … + kn ∫fn(x) dx
First Fundamental Theorem of Integral Calculus:
Define the area function A(x) = ∫[a, x]f(t)dt for x ≥ a, where f is continuous on [a, b].
Then A'(x) = f(x) for every x ∈ [a, b].
Second Fundamental Theorem of Integral Calculus:
If f is a continuous function on [a, b], then ∫[a, b]f(x)dx = F(b) - F(a), where F(x) is an antiderivative of f(x).
Standard Integral Formulas:
∫xn dx = xn+1/(n+1) + C (n ≠ -1)
∫cos x dx = sin x + C
∫sin x dx = -cos x + C
∫sec2 x dx = tan x + C
∫cosec2 x dx = -cot x + C
∫sec x tan x dx = sec x + C
∫cosec x cot x dx = -cosec x + C
∫ex dx = ex + C
∫ax dx = (ax)/ln(a) + C
∫(1/x) dx = ln|x| + C
Other Integral Formulas:
∫tan x dx = ln|sec x| + C
∫cot x dx = ln|sin x| + C
∫sec x dx = ln|sec x + tan x| + C
∫cosec x dx = ln|cosec x - cot x| + C
Free download Class 12 Maths Chapter 7 Question Answer for CBSE Exam.
Class 12 Integrals NCERT solutions Exercise: 7.1
Question:1 Find an anti derivative (or integral) of the following functions by the method of inspection.
GIven ;
So, the anti derivative of is a function of x whose derivative is
.
Therefore, we have
Or, antiderivative of is
.
GIven ;
So, the antiderivative of is a function of x whose derivative is
.
Therefore, we have the anti derivative of is
.
GIven ;
So, the anti derivative of is a function of x whose derivative is
.
Therefore, we have the anti derivative of is
.
GIven ;
So, the anti derivative of is a function of x whose derivative is
.
Therefore, we have the anti derivative of is
.
GIven ;
So, the anti derivative of is a function of x whose derivative is
.
Therefore, we have the anti derivative of is
.
Given intergral ;
or , where C is any constant value.
Given intergral ;
or , where C is any constant value.
Given intergral ;
or , where C is any constant value.
Given intergral ;
or , where C is any constant value.
Given integral ;
or
, where C is any constant value.
Given intergral ;
or
Or, , where C is any constant value.
Given intergral ;
or
Or, , where C is any constant value.
Given integral
It can be written as
Taking common out
Now, cancelling out the term from both numerator and denominator.
Splitting the terms inside the brackets
Given intergral ;
or
, where C is any constant value.
Given intergral ;
or
, where C is any constant value.
Given integral ;
splitting the integral as the sum of three integrals
, where C is any constant value.
Given integral ;
Using the integral of trigonometric functions
, where C is any constant value.
Given integral ;
, where C is any constant value.
Given integral ;
Using antiderivative of trigonometric functions
, where C is any constant value.
Given to find the anti derivative or integral of ;
, where C is any constant value.
Hence the correct option is (C).
If such that f (2) = 0. Then f (x) is
Given that the anti derivative of
So,
Now, to find the constant C;
we will put the condition given, f (2) = 0
or
Therefore the correct answer is A .
Class 12 Integrals NCERT solutions Exercise: 7.2
Given to integrate function,
Let us assume
we get,
now back substituting the value of
as is positive we can write
Given to integrate function,
Let us assume
we get,
Given to integrate function,
Let us assume
we get,
Back substituting the value of t we get,
Given to integrate function,
Let us assume
we get,
Now, by back substituting the value of t,
Given to integrate function,
Let us assume
we get,
Now, by back substituting the value of t,
Given function ,
Assume the 19634
Back substituting the value of t in the above equation.
or, , where C is any constant value.
Given function ,
Assume the
Or
, where C is any constant value.
Given function ,
Assume the
Now, back substituting the value of t in the above equation,
, where C is any constant value.
Given function ,
Can be written in the form:
Assume the
, where C is any constant value.
Given function ,
Assume the so,
, where C is any constant value.
Given function ,
Assume the
, where C is any constant value.
Given function ,
Assume the
, where C is any constant value.
Given function ,
Assume the
, where C is any constant value.
Given function ,
Assume the
Now back substituting the value of t ;
, where C is any constant value.
Given function ,
Assume the
Now back substituting the value of t ;
, where C is any constant value.
Given function ,
Assume the
, where C is any constant value.
Given function ,
Simplifying it by dividing both numerator and denominator by , we obtain
Assume the
Now, back substituting the value of t,
, where C is any constant value.
Given function ,
Assume the
Now, back substituting the value of t,
, where C is any constant value.
Given function ,
Assume the
Now, back substituting the value of t,
or , where C is any constant value.
Given function ,
Assume the
Now, back substituted the value of t.
, where C is any constant value.
Given function ,
Assume the
Now, back substituted the value of t.
, where C is any constant value.
Given function ,
or simplified as
Assume the
Now, back substituted the value of t.
, where C is any constant value.
Given function ,
or simplified as
Assume the
Now, back substituted the value of t.
where C is any constant value.
Given function ,
Assume the
Now, back substituted the value of t.
, where C is any constant value.
Given function ,
Assume the
Now, back substituted the value of t.
, where C is any constant value.
Given function ,
Assume the
Now, back substituted the value of t.
, where C is any constant value.
Given function ,
Assume the
Now, back substituted the value of t.
, where C is any constant value.
Given function ,
Assume the
Now, back substituted the value of t.
, where C is any constant value.
Given function ,
Assume the
Now, back substituted the value of t.
, where C is any constant value.
Given function
Assume that
Now solving the assumed integral;
Now, to solve further we will assume
Or,
Now, back substituting the value of t,
Given function
Assume that
Now solving the assumed integral;
Now, to solve further we will assume
Or,
Now, back substituting the value of t,
Given function
Assume that
Now solving the assumed integral;
Multiplying numerator and denominator by ;
Now, to solve further we will assume
Or,
Now, back substituting the value of t,
Given function
Assume that
Now, back substituting the value of t,
Given function
Simplifying to solve easier;
Assume that
Now, back substituting the value of t,
Given function
Assume that
......................(1)
Now to solve further we take
So, from the equation (1), we will get
Now back substitute the value of u,
and then back substituting the value of t,
Given integral
Taking the denominator
Now differentiating both sides we get
Back substituting the value of t,
Therefore the correct answer is D.
Class 12 Integrals NCERT Solutions Exercise: 7.3
using the trigonometric identity
we can write the given question as
=
Using identity
, therefore the given integral can be written as
Using identity
Again use the same identity mentioned in the first line
The integral can be written as
Let
Now, replace the value of t, we get;
rewrite the integral as follows
Let
......(replace the value of t as
)
Using the formula
we can write the integral as follows
Using identity
we can write the following integral as
=
We know the identities
Using the above relations we can write
The integral is rewritten using trigonometric identities
can be written as follows using trigonometric identities
Therefore,
now using the identity
now using the below two identities
the value
.
the integral of the given function can be written as
Using trigonometric identities we can write the given integral as follows.
We know that,
Using this identity we can rewrite the given integral as
Therefore integration of =
.....................(i)
Let assume
So, that
Now, the equation (i) becomes,
the given question can be rearranged using trigonometric identities
Therefore, the integration of =
...................(i)
Considering only
let
now the final solution is,
now splitting the terms we can write
Therefore, the integration of
The integral of the above equation is
Thus after evaluation, the value of integral is tanx+ c
Let
We can write 1 =
Then, the equation can be written as
put the value of tan = t
So, that
we know that
therefore,
let
Now the given integral can be written as
using the trigonometric identities we can evaluate the following integral as follows
Using the trigonometric identities following integrals can be simplified as follows
The correct option is (A)
On reducing the above integral becomes
NCERT solutions for maths chapter 7 class 12 Integrals Exercise: 7.4
The given integral can be calculated as follows
Let
, therefore,
let suppose 2x = t
therefore 2dx = dt
.................using formula
Let assume 5x =t,
then 5dx = dt
The above result is obtained using the identity
let
then
using the special identities we can simplify the integral as follows
We can write above eq as
............................................(i)
for let
Now, by using eq (i)
The integration can be down as follows
let
........................using
the above equation can be also written as,
let 1+x = t
then dx = dt
therefore,
the denominator can be also written as,
therefore
Let x+3 = t
then dx =dt
......................................using formula
(x-1)(x-2) can be also written as
=
=
therefore
let suppose
Now,
.............by using formula
let
By equating the coefficient of x and constant term on each side, we get
A = 1 and B=0
Let
let
By comparing the coefficients and constant term on both sides, we get;
A=1/2 and B=2
then
let
By comparing the coefficients and constants we get the value of A and B
A = and B =
NOW,
...........................(i)
put
Thus
let
By comparing the coefficients and constants on both sides, we get
A =3 and B =34
....................................(i)
Considering
let
Now consider
here the denominator can be also written as
Dr =
Now put the values of and
in eq (i)
let
By equating the coefficients and constant term on both sides we get
A = -1/2 and B = 4
(x+2) = -1/2(4-2x)+4
....................(i)
Considering
let
now,
put the value of and
...........(i)
take
let
considering
putting the values in equation (i)
Let
By comparing the coefficients and constant term, we get;
A = 1/2 and B =4
..............(i)
put
let
On comparing, we get
A =5/2 and B = -7
...........................................(i)
put
The correct option is (B)
the denominator can be written as
now,
The following integration can be done as
The correct option is (B)
NCERT solutions for maths chapter 7 class 12 Integrals Exercise: 7.5
Question:1 Integrate the rational functions
Answer:
Given function
Partial function of this function:
Now, equating the coefficients of x and constant term, we obtain
On solving, we get
Given function
The partial function of this function:
Now, equating the coefficients of x and constant term, we obtain
On solving, we get
Given function
Partial function of this function:
.(1)
Now, substituting respectively in equation (1), we get
That implies
Given function
Partial function of this function:
.....(1)
Now, substituting respectively in equation (1), we get
That implies
Given function
Partial function of this function:
...........(1)
Now, substituting respectively in equation (1), we get
That implies
Given function
Integral is not a proper fraction so,
Therefore, on dividing by
, we get
Partial function of this function:
...........(1)
Now, substituting respectively in equation (1), we get
No, substituting in equation (1) we get
Given function
Partial function of this function:
Now, equating the coefficients of and the constant term, we get
and
On solving these equations, we get
From equation (1), we get
Now, consider ,
and we will assume
So,
or
Given function
Partial function of this function:
Now, putting in the above equation, we get
By equating the coefficients of and constant term, we get
then after solving, we get
Therefore,
Given function
can be rewritten as
Partial function of this function:
................(1)
Now, putting in the above equation, we get
By equating the coefficients of and
, we get
then after solving, we get
Therefore,
Given function
can be rewritten as
The partial function of this function:
Equating the coefficients of , we get
Therefore,
Given function
can be rewritten as
The partial function of this function:
Now, substituting the value of respectively in the equation above, we get
Therefore,
Given function
As the given integral is not a proper fraction.
So, we divide by
, we get
can be rewritten as
....................(1)
Now, substituting in equation (1), we get
Therefore,
Given function
can be rewritten as
....................(1)
Now, equating the coefficient of and constant term, we get
,
, and
Solving these equations, we get
Therefore,
Given function
can be rewritten as
Now, equating the coefficient of and constant term, we get
and
,
Solving these equations, we get
Therefore,
Given function
can be rewritten as
The partial fraction of above equation,
Now, equating the coefficient of and constant term, we get
and
and
Solving these equations, we get
Therefore,
[Hint: multiply numerator and denominator by and put
]
Given function
Applying Hint multiplying numerator and denominator by and putting
Putting
can be rewritten as
Partial fraction of above equation,
................(1)
Now, substituting in equation (1), we get
Given function
Applying the given hint: putting
We get,
Partial fraction of above equation,
................(1)
Now, substituting in equation (1), we get
Back substituting the value of t in the above equation, we get
Given function
We can rewrite it as:
Partial fraction of above equation,
Now, equating the coefficients of and constant term, we get
,
,
,
After solving these equations, we get
Given function
Taking
The partial fraction of above equation,
..............(1)
Now, substituting in equation (1), we get
Given function
So, we multiply numerator and denominator by , to obtain
Now, putting
we get,
Taking
Partial fraction of above equation,
..............(1)
Now, substituting in equation (1), we get
Back substituting the value of t,
Given function
So, applying the hint: Putting
Then
Partial fraction of above equation,
..............(1)
Now, substituting in equation (1), we get
Now, back substituting the value of t,
Given integral
Partial fraction of above equation,
..............(1)
Now, substituting in equation (1), we get
Therefore, the correct answer is B.
Given integral
Partial fraction of above equation,
Now, equating the coefficients of and the constant term, we get
,
,
We have the values,
Therefore, the correct answer is A.
NCERT solutions for maths chapter 7 class 12 Integrals Exercise: 7.6
Question:1 Integrate the functions
Answer:
Given function is
We will use integrate by parts method
Therefore, the answer is
Given function is
We will use integration by parts method
Therefore, the answer is
Given function is
We will use integration by parts method
Again use integration by parts in
Put this value in our equation
we will get,
Therefore, answer is
Given function is
We will use integration by parts method
Therefore, the answer is
Given function is
We will use integration by parts method
Therefore, the answer is
Given function is
We will use integration by parts method
Therefore, the answer is
Given function is
We will use integration by parts method
Now, we need to integrate
Put this value in our equation
Therefore, the answer is
Given function is
We will use integration by parts method
Put this value in our equation
Therefore, the answer is
Given function is
We will use integration by parts method
Now, we need to integrate
Put this value in our equation
Therefore, the answer is
Given function is
we will use integration by parts method
Therefore, answer is
Consider
So, we have then:
After taking as a first function and
as second function and integrating by parts, we get
Or,
Consider
So, we have then:
After taking as a first function and
as second function and integrating by parts, we get
Consider
So, we have then:
After taking as a first function and
as second function and integrating by parts, we get
Consider
So, we have then:
After taking as a first function and
as second function and integrating by parts, we get
Consider
So, we have then:
Let us take ....................(1)
Where, and
So,
After taking as a first function and
as second function and integrating by parts, we get
....................(2)
After taking as a first function and
as second function and integrating by parts, we get
................(3)
Now, using the two equations (2) and (3) in (1) we get,
Let suppose
we know that,
Thus, the solution of the given integral is given by
Let suppose
by rearranging the equation, we get
let
It is known that
therefore the solution of the given integral is
Let
substitute and
let
It is known that
Therefore the solution of the given integral is
It is known that
let
Therefore the required solution of the given above integral is
It is known that
So, By adjusting the given equation, we get
to let
Therefore the required solution of the given integral is
let
Taking as a first function and
as a second function, by using by parts method
we know that,
from above integral
let
thus, the solution of the above integral is
NCERT class 12 maths ch 7 question answer Exercise: 7.7
Given function ,
So, let us consider the function to be;
Then it is known that,
Therefore,
Given function to integrate
Now we can rewrite as
As we know the integration of this form is
Given function ,
So, let us consider the function to be;
And we know that,
Given function ,
So, let us consider the function to be;
And we know that,
Given function ,
So, let us consider the function to be;
And we know that,
Given function ,
So, let us consider the function to be;
a
And we know that,
Given function ,
So, let us consider the function to be;
And we know that,
Given function ,
So, let us consider the function to be;
And we know that,
Given function ,
So, let us consider the function to be;
And we know that,
(A)
(B)
(C)
(D)
As we know that,
So,
Therefore the correct answer is A.
(A)
(B)
(C)
(D)
Given integral
So, let us consider the function to be;
And we know that,
Therefore the correct answer is D.
NCERT class 12 maths ch 7 question answer - Exercise:7.8
We know that,
This is how the integral is evaluated using limit of a sum
We know that
let
Here a = 0, b = 5 and
therefore
We know that
here a = 2 and b = 3 , so h = 1/n
Let
for the second part, we already know the general solution of
So, here a = 1 and b = 4
therefore
So,
let
We know that
Here a =-1, b = 1 and
therefore h = 2/n
By using sum of n terms of GP ....where a = 1st term and r = ratio
.........using
It is known that,
..........................(
)
NCERT class 12 maths ch 7 question answer - Exercise:7.9
Given integral:
Consider the integral
So, we have the function of ,
Now, by Second fundamental theorem of calculus, we have
Given integral:
Consider the integral
So, we have the function of ,
Now, by Second fundamental theorem of calculus, we have
Given integral:
Consider the integral
So, we have the function of ,
Now, by Second fundamental theorem of calculus, we have
Given integral:
Consider the integral
So, we have the function of ,
Now, by Second fundamental theorem of calculus, we have
Given integral:
Consider the integral
So, we have the function of ,
Now, by Second fundamental theorem of calculus, we have
Given integral:
Consider the integral
So, we have the function of ,
Now, by Second fundamental theorem of calculus, we have
Given integral:
Consider the integral
So, we have the function of ,
Now, by Second fundamental theorem of calculus, we have
Given integral:
Consider the integral
So, we have the function of ,
Now, by Second fundamental theorem of calculus, we have
Given integral:
Consider the integral
So, we have the function of ,
Now, by Second fundamental theorem of calculus, we have
Given integral:
Consider the integral
So, we have the function of ,
Now, by Second fundamental theorem of calculus, we have
Given integral:
Consider the integral
So, we have the function of ,
Now, by Second fundamental theorem of calculus, we have
Given integral:
Consider the integral
So, we have the function of ,
Now, by Second fundamental theorem of calculus, we have
Given integral:
Consider the integral
So, we have the function of ,
Now, by Second fundamental theorem of calculus, we have
Given integral:
Consider the integral
Multiplying by 5 both in numerator and denominator:
So, we have the function of ,
Now, by Second fundamental theorem of calculus, we have
Given integral:
Consider the integral
Putting which gives,
As, and as
.
So, we have now:
So, we have the function of ,
Now, by Second fundamental theorem of calculus, we have
Given integral:
So, we can rewrite the integral as;
where
. ................(1)
Now, consider
Take numerator
We now equate the coefficients of x and constant term, we get
Now take denominator
Then we have
Then substituting the value of in equation (1), we get
Given integral:
Consider the integral
So, we have the function of ,
Now, by Second fundamental theorem of calculus, we have
Given integral:
Consider the integral
can be rewritten as:
So, we have the function of ,
Now, by Second fundamental theorem of calculus, we have
Given integral:
Consider the integral
can be rewritten as:
So, we have the function of ,
Now, by Second fundamental theorem of calculus, we have
or we have
Given integral:
Consider the integral
can be rewritten as:
So, we have the function of ,
Now, by Second fundamental theorem of calculus, we have
Given definite integral
Consider
we have then the function of x, as
By applying the second fundamental theorem of calculus, we will get
Therefore the correct answer is D.
(A)
(B)
(C)
(D)
Given definite integral
Consider
Now, putting
we get,
Therefore we have,
we have the function of x , as
So, by applying the second fundamental theorem of calculus, we get
Therefore the correct answer is C.
NCERT solutions for class 12 maths chapter 7 Integrals - Exercise:7.10
let
when x = 0 then t = 1 and when x =1 then t = 2
let
when and
using the above substitution we can evaluate the integral as
let
when x = 0 then and when x = 1 then
Taking as a first function and
as a second function, by using by parts method
Let
when x = 0 then t = and when x=2 then t = 2
let
when x=0 then t = 1 and when x= then t = 0
By adjusting, the denominator can also be written as
Now,
let
when x= 0 then t =-1/2 and when x =2 then t = 3/2
On rationalisation, we get
the Dr can be written as
and put x+1 = t then dx =dt
when x= -1 then t = 0 and when x = 1 then t = 2
let
when x = 1 then t = 2 and when x = 2 then t= 4
let
(A) 6
(B) 0
(C) 3
(D) 4
The value of integral is (A) = 6
let
now, when x = 1/3, t = 8 and when x = 1 , t = 0
therefore
(A)
(B)
(C)
(D)
The correct answer is (B) =
by using by parts method,
So,
NCERT solutions for class 12 maths chapter 7 Integrals - Exercise:7.11
We have ............................................................. (i)
By using
We get :-
or
................................................................ (ii)
Adding both (i) and (ii), we get :-
or
or
or
or
We have .......................................................................... (i)
By using ,
We get,
or .......................................................(ii)
Adding (i) and (ii), we get,
or
or
Thus
We have ..................................................................(i)
By using :
We get,
or . ............................................................(ii)
Adding (i) and (ii), we get :
or
or
Thus
We have ..................................................................(i)
By using :
We get,
or . ............................................................(ii)
Adding (i) and (ii), we get :
or
or
Thus
We have,
For opening the modulas we need to define the bracket :
If (x + 2) < 0 then x belongs to (-5, -2). And if (x + 2) > 0 then x belongs to (-2, 5).
So the integral becomes :-
or
This gives
We have,
For opening the modulas we need to define the bracket :
If (x - 5) < 0 then x belongs to (2, 5). And if (x - 5) > 0 then x belongs to (5, 8).
So the integral becomes:-
or
This gives
We have
U sing the property : -
We get : -
or
or
or
or
or
We have
By using the identity
We get,
or
or
or
or
or
or
We have
By using the identity
We get :
or
or
or
or
or
We have
or
or ..............................................................(i)
By using the identity :
We get :
or ....................................................................(ii)
Adding (i) and (ii) we get :-
or
or
We have
We know that sin 2 x is an even function. i.e., sin 2 (-x) = (-sinx) 2 = sin 2 x.
Also,
So,
or
or
We have ..........................................................................(i)
By using the identity :-
We get,
or ............................................................................(ii)
Adding both (i) and (ii) we get,
or
or
or
We have
We know that is an odd function.
So the following property holds here:-
Hence
We have
I t is known that :-
If f (2a - x) = f(x)
If f (2a - x) = - f(x)
Now, using the above property
Therefore,
We have ................................................................(i)
By using the property :-
We get ,
or ......................................................................(ii)
Adding both (i) and (ii), we get
Thus I = 0
We have .....................................................................................(i)
By using the property:-
We get,
or
....................................................................(ii)
Adding both (i) and (ii) we get,
or
or
or ........................................................................(iii)
or ........................................................................(iv)
or .....................................................................(v)
Adding (iv) and (v) we get,
We have ................................................................................(i)
By using, we get
We get,
.................................................................(ii)
Adding (i) and (ii) we get :
or
or
We have,
For opening the modulas we need to define the bracket :
If (x - 1) < 0 then x belongs to (0, 1). And if (x - 1) > 0 then x belongs to (1, 4).
So the integral becomes:-
or
This gives
Let ........................................................(i)
This can also be written as :
or ................................................................(ii)
Adding (i) and (ii), we get,
or
(A) 0
(B) 2
(C)
(D) 1
We have
This can be written as :
Also if a function is even function then
And if the function is an odd function then :
Using the above property I become:-
or
or
We have
.................................................................................(i)
By using :
We get,
or .............................................................................(ii)
Adding (i) and (ii), we get:
or
Thus
NCERT solutions for class 12 maths chapter 7 Integrals-Miscellaneous Exercise
Firstly we will simplify the given equation :-
Let
By solving the equation and equating the coefficients of x 2 , x and the constant term, we get
Thus the integral can be written as :
or
At first we will simplify the given expression,
or
Now taking its integral we get,
or
or
Let
Using the above substitution we can write the integral is
or
or
or
or
For the simplifying the expression, we will multiply and dividing it by x -3 .
We then have,
Now, let
Thus,
or
[Hint:
, put
]
Put
We get,
or
or
or
Now put in the above result :
Let us assume that :
Solving the equation and comparing coefficients of x 2 , x and the constant term.
We get,
Thus the equation becomes :
or
or
or
or
We have,
Assume :-
Putting this in above integral :
or
or
or
or
We have the given integral
Assume
So, this substitution gives,
or
We have
Simplifying the given expression, we get :
or
or
or
Thus,
and
For simplifying the given equation, we need to multiply and divide the expression by .
Thus we obtain :
or
or
or
Thus integral becomes :
or
or
Given that to integrate
Let
the required solution is
we have to integrate the following function
Let
using this we can write the integral as
Given,
Let
Now, Using partial differentiation,
Equating the coefficients of and constant value,
A + C = 0 C = -A
B + D = 0 B = -D
4A + C =0 4A = -C
4A = A
A = 0 = C
4B + D = 1 4B – B = 1
B = 1/3 = -D
Putting these values in equation, we have
Given,
(let)
Let
using the above substitution the integral is written as
Given the function to be integrated as
Let
Let
Given,
Let
Let f(ax +b) = t ⇒ a .f ' (ax + b)dx = dt
Now we can write the ntegral as
Given,
Let
We know the identity that
sin (A+B) = sin A cos B + cos A sin B
We have
or
or
or
or
or
Thus
Now we will solve I'.