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Edited By Ramraj Saini | Updated on Sep 14, 2023 08:18 PM IST | #CBSE Class 12th

**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.

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- NCERT Integrals Class 12 Questions And Answers
- NCERT Integrals Class 12 Questions And Answers PDF Free Download
- Class 12 maths chapter 7 NCERT Solutions - Important Formulae
- NCERT Integrals Class 12 Questions And Answers (Intext Questions and Exercise)
- NCERT Solutions for Class 12 Maths Chapter 7 Integrals - Topics
- NCERT solutions for class 12 maths - Chapter wise
- Key Features of NCERT Solutions for Class 12 Maths Chapter 7 Integrals
- NCERT solutions for class 12 - subject wise
- NCERT Books and NCERT Syllabus

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.

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**Also read:**

- Class 12 Maths Chapter 7 Integrals Notes
- Ncert Exemplar Solutions For Class 12 Maths Chapter 7 Integrals

>> 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:

∫x^{n} dx = x^{n+1}/(n+1) + C (n ≠ -1)

∫cos x dx = sin x + C

∫sin x dx = -cos x + C

∫sec^{2} x dx = tan x + C

∫cosec^{2} x dx = -cot x + C

∫sec x tan x dx = sec x + C

∫cosec x cot x dx = -cosec x + C

∫e^{x} dx = e^{x} + C

∫a^{x} dx = (a^{x})/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. **

The anti derivative of equals

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 ** .

** C****lass 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. **

C**lass 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,

this denominator can be written as

Now,

......................................by using the form

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'. **

** Put x = t ^{ 2 } . **

Differentiating the equation wrt x, we get

Thus

** or **

Using integration by parts, we get :

** **

or

We know that

Thus it becomes :

So I come to be :-

Given,

** = I (let) **

Let

using the above substitution we can write the integral as

Given to evaluate

now the integral becomes

Let tan x = f(x)

Given,

** using partial fraction we can simplify the integral as **

Let

Equating the coefficients of x, x ^{ 2 } and constant value, we get:

A + C = 1

3A + B + 2C = 1

2A+2B+C =1

Solving these:

A= -2, B=1 and C=3