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Edited By Team Careers360 | Updated on Dec 04, 2023 01:21 PM IST

**NCERT Solutions for miscellaneous exercise chapter 7 class 12 Integrals **are discussed here. These NCERT solutions are created by subject matter expert at Careers360 considering the **latest syllabus and pattern of CBSE 2023-24. **NCERT solutions for class 12 maths chapter 7 miscellaneous exercise provides questions based on integration of square root functions, trigonometric functions etc. Class 12 maths chapter 7 miscellaneous exercise is last but not the least as you can find some of the questions in previous years from this exercise only. class 12 maths chapter 7 miscellaneous exercise can be found difficult for some students but after giving sufficient time, you will be able to deal with the questions. NCERT solutions for class 12 maths chapter 7 with all the other exercises can be found in NCERT Class 12th book.

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- NCERT Solutions For Class 12 Chapter 7 Miscellaneous Exercise
- NCERT solutions for class 12 maths chapter 7 Integrals-Miscellaneous Exercise
- More about NCERT solutions for class 12 maths chapter 7 miscellaneous exercise
- Benefits of ncert solutions for class 12 maths chapter 7 miscellaneous exercises
- Key Features Of NCERT Solutions For Class 12 Chapter 7 Miscellaneous Exercise
- Subject wise NCERT Exemplar solutions

**Miscellaneous exercise class 12 chapter 7**** **are designed as per the students demand covering comprehensive, step by step solutions of every problem. Practice these questions and answers to command the concepts, boost confidence and in depth understanding of concepts. Students can find all exercise enumerated in NCERT Book together using the link provided below.

- NCERT solutions for class 12 maths chapter 7 Integrals Exercise 7.1
- NCERT solutions for class 12 maths chapter 7 Integrals Exercise 7.2
- NCERT solutions for class 12 maths chapter 7 Integrals Exercise 7.3
- NCERT solutions for class 12 maths chapter 7 Integrals Exercise 7.4
- NCERT solutions for class 12 maths chapter 7 Integrals Exercise 7.5
- NCERT solutions for class 12 maths chapter 7 Integrals Exercise 7.6
- NCERT solutions for class 12 maths chapter 7 Integrals Exercise 7.7
- NCERT solutions for class 12 maths chapter 7 Integrals Exercise 7.8
- NCERT solutions for class 12 maths chapter 7 Integrals Exercise 7.9
- NCERT solutions for class 12 maths chapter 7 Integrals Exercise 7.10
- NCERT solutions for class 12 maths chapter 7 Integrals Exercise 7.11

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 **

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

We have

Let us assume that :

Differentiating wrt x,

Substituting this in the original equation, we get

or

or

** Using integration by parts ** , we get

or

or

Putting all the assumed values back in the expression,

or

Here let's first reduce the log function.

Now, let

So our function in terms if new variable t is :

now let's solve this By using integration by parts

Since, we have multiplied by some function, let's try to make that function in any function and its derivative.Basically we want to use the property,

So,

** Here let's use the property **

so,

First, let's convert sin and cos into tan and sec. (because we have a good relation in tan and square of sec,)

Let' divide both numerator and denominator by

** Now lets change the variable **

the limits will also change since the variable is changing

So, the integration becomes:

Lets first simplify the function.

As we have a good relation in between squares of the tan and square of sec lets try to take our equation there,

AS we can write square of sec in term of tan,

Now let's calculate the integral of the second function, (we already have calculated the first function)

let

here we are changing the variable so we have to calculate the limits of the new variable

when x = 0, t = 2tanx = 2tan(0)=0

when

our function in terms of t is

Hence our total solution of the function is

Here first let convert sin2x as the angle of x ( sinx, and cosx)

Now let's remove the square root form function by making a perfect square inside the square root

Now let

,

since we are changing the variable, limit of integration will change

our function in terms of t :

First, let's get rid of the square roots from the denominator,

First let's assume t = cosx - sin x so that (sinx +cosx)dx=dt

So,

Now since we are changing the variable, the new limit of the integration will be,

when x = 0, t = cos0-sin0=1-0=1

when

Now,

Hence our function in terms of t becomes,

Let I =

Here, we can see that if we put sinx = t, then the whole function will convert in term of t with dx being changed to dt.so

Now the important step here is to change the limit of the integration as we are changing the variable.so,

So our function becomes,

Now, let's integrate this by using integration by parts method,

Let I = -(i)

Replacing x with ( -x),

- (ii)

Adding (i) and (ii)

Given integral

So, we split it in according to intervals they are positive or negative.

Now,

as is positive in the given x -range

Therefore,

as is in the given x -range and in the range

Therefore,

as is in the given x -range and in the range

Therefore,

So, We have the sum

** . **

L.H.S =

We can write the numerator as [(x+1) -x]

= RHS

Hence proved.

Integrating I by parts

Applying Limits from 0 to 1

Hence proved I = 1

The Integrand g(x) therefore is an odd function and therefore

For I _{ 2 } let cosx=t, -sinxdx=dt

The limits change to 0 and 1

I _{ 1 } -I _{ 2 } =2/3

Hence proved.

The integral is written as

Hence Proved

Integrating by parts we get

For I _{ 2 } take 1-x ^{ 2 } = t ^{ 2 } , -xdx=tdt

Hence Proved

As we know

where b-a=hn

In the given problem b=1, a=0 and

the above integral can be re arranged as

let e ^{ x } =t, e ^{ x } dx=dt

(A) is correct

cos2x=cos ^{ 2 } x-sin ^{ 2 } x

let sinx+cosx=t,(cosx-sinx)dx=dt

hence the given integral can be written as

B is correct

As we know

Using the above property we can write the integral as

Answer (D) is correct

as

Now the integral can be written as

(B) is correct.

The NCERT class 12 maths chapter integrals is most important in the whole maths syllabus and it has applications in Physics and chemistry also. NCERT solutions for class 12 maths chapter 7 miscellaneous exercise deals with quite tricky and interesting questions which can be realised while solving the questions. NCERT solutions for class 12 maths chapter 7 miscellaneous exercise will take some longer time to complete but one should remain focussed while solving it as it is quite important. class 12 maths chapter 7 miscellaneous solutions is a good source to practice well.

- The class 12th maths chapter 7 exercise has great importance and it should be done in step by step manner.
- Class 12 maths chapter 7 miscellaneous exercises Will take more effort than other exercises, Hence should be done patiently.
- These class 12 maths chapter 7 miscellaneous exercises along with previous exercises are good to go for the exam.

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Download EBook**Comprehensive Coverage:**The solutions encompass all the topics covered in miscellaneous exercise class 12 chapter 7, ensuring a thorough understanding of the concepts.**Step-by-Step Solutions:**In this class 12 chapter 7 maths miscellaneous solutions, each problem is solved systematically, providing a stepwise approach to aid in better comprehension for students.**Accuracy and Clarity:**Solutions for class 12 maths miscellaneous exercise chapter 7 are presented accurately and concisely, using simple language to help students grasp the concepts easily.**Conceptual Clarity:**In this class 12 maths ch 7 miscellaneous exercise solutions, emphasis is placed on conceptual clarity, providing explanations that assist students in understanding the underlying principles behind each problem.**Inclusive Approach:**Solutions for class 12 chapter 7 miscellaneous exercise cater to different learning styles and abilities, ensuring that students of various levels can grasp the concepts effectively.**Relevance to Curriculum:**The solutions for miscellaneous exercise class 12 chapter 7 align closely with the NCERT curriculum, ensuring that students are prepared in line with the prescribed syllabus.

**NCERT Solutions Subject Wise**

- NCERT solutions class 12 chemistry
- NCERT solutions for class 12 physics
- NCERT solutions for class 12 biology
- NCERT solutions for class 12 mathematics

- NCERT Exemplar Class 12th Maths
- NCERT Exemplar Class 12th Physics
- NCERT Exemplar Class 12th Chemistry
- NCERT Exemplar Class 12th Biology

Happy learning!!!

1. What is the level of questions in Miscellaneous exercise ?

Advanced level problems are dealt with in this exercise.

2. How much time will it take topcomlete miscellaneous exercise for the first time ?

It can take 3 to 4 hours for the first time.

3. Can shortcuts be used while solving the questions.?

In boards, step by step method is used but shortcuts can be used in JEE and NEET.

4. Is it mandatory to solve every question of Miscellaneous exercise ?

No, but questions on different concepts must be solved.

5. What is the weightage from miscellaneous exercise in the examination ?

Around 5 marks of questions are asked in the examination.

6. Can this exercise be done with self study only ?

Yes, but for that basic concepts must be clear.

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