# NCERT Solutions for Miscellaneous Exercise Chapter 7 Class 12 - Integrals

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. Apart from class 12 maths chapter 7 miscellaneous exercise solutions, you can also refer below exercises for wide understanding of the topic.

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

**NCERT solutions for class 12 maths chapter 7 Integrals-Miscellaneous Exercise **

### ** Question:1 ** Integrate the functions in Exercises 1 to 24.

** **

### ** Answer: **

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

### ** Question:2 ** Integrate the functions in Exercises 1 to 24.

### ** Answer: **

At first we will simplify the given expression,

** or **

Now taking its integral we get,

** or **

** or **

### ** Question:3** Integrate the functions in Exercises 1 to 24.

[Hint: Put ]

### ** Answer: **

Let

Using the above substitution we can write the integral is

** or **

** or **

** or **

** or **

### ** Question:4 ** Integrate the functions in Exercises 1 to 24.

** . **

### ** Answer: **

For the simplifying the expression, we will multiply and dividing it by x ^{ -3 } .

We then have,

Now, let

Thus,

** or **

### ** Question:5 ** Integrate the functions in Exercises 1 to 24.

[Hint: , put ]

### ** Answer: **

Put

We get,

** or **

** or **

** or **

Now put in the above result :

### ** Question:6 ** Integrate the functions in Exercises 1 to 24.

### ** Answer: **

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

### ** Question:7 ** Integrate the functions in Exercises 1 to 24.

### ** Answer: **

We have,

Assume :-

Putting this in above integral :

or

or

or

or

### ** Question:** Integrate the functions in Exercises 1 to 24.

** **

### ** Answer: **

We have the given integral

Assume

So, this substitution gives,

or

### ** Question:10 ** Integrate the functions in Exercises 1 to 24.

** **

### ** Answer: **

We have

Simplifying the given expression, we get :

** or **

** or **

** or **

** Thus, **

** and **

### ** Question:11 ** Integrate the functions in Exercises 1 to 24.

** **

### ** Answer: **

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

### ** Question:12** Integrate the functions in Exercises 1 to 24.

### ** Answer: **

Given that to integrate

Let

the required solution is

### ** Question:13 ** Integrate the functions in Exercises 1 to 24.

** ** ** **

### ** Answer: **

we have to integrate the following function

** Let **

** using this we can write the integral as **

### ** Question:14 ** Integrate the functions in Exercises 1 to 24.

** **

### ** Answer: **

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

### ** Question:15** Integrate the functions in Exercises 1 to 24.

** **

### ** Answer: **

Given,

** (let) **

Let

using the above substitution the integral is written as

### ** Question:16** Integrate the functions in Exercises 1 to 24.

### ** Answer: **

Given the function to be integrated as

Let

Let

### ** Question:17 ** Integrate the functions in Exercises 1 to 24.

** **

### ** Answer: **

Given,

Let

Let f(ax +b) = t ⇒ a .f ' (ax + b)dx = dt

Now we can write the ntegral as

### ** Question:18 ** Integrate the functions in Exercises 1 to 24.

** . **

### ** Answer: **

Given,

** Let **

We know the identity that

sin (A+B) = sin A cos B + cos A sin B

### ** Question:19 ** Integrate the functions in Exercises 1 to 24.

** . **

### ** Answer: **

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 :-

### ** Question:20 ** Integrate the functions in Exercises 1 to 24.

** **

### ** Answer: **

Given,

** = I (let) **

Let

using the above substitution we can write the integral as

### ** Question:21** Integrate the functions in Exercises 1 to 24.

** **

### ** Answer: **

Given to evaluate

now the integral becomes

Let tan x = f(x)

### ** Question:22 ** Integrate the functions in Exercises 1 to 24.

### ** Answer: **

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

### ** Question:23 ** Integrate the functions in Exercises 1 to 24.

** **

### ** Answer: **

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

### ** Question:24 ** Integrate the functions in Exercises 1 to 24.

### ** Answer: **

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

### ** Question:25 ** Evaluate the definite integrals in Exercises 25 to 33.

### ** Answer: **

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,

### ** Question:26 ** Evaluate the definite integrals in Exercises 25 to 33.

** **

### ** Answer: **

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:

### ** Question:27 ** Evaluate the definite integrals in Exercises 25 to 33.

** **

### ** Answer: **

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

### ** Question:28 ** Evaluate the definite integrals in Exercises 25 to 33.

** **

### ** Answer: **

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 :

### ** Question:29 ** Evaluate the definite integrals in Exercises 25 to 33.

** **

### ** Answer: **

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

### ** Question:30 ** Evaluate the definite integrals in Exercises 25 to 33.

** **

### ** Answer: **

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,

### ** Question:31 ** Evaluate the definite integrals in Exercises 25 to 33.

** **

### ** Answer: **

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,

### ** Question:32 ** Evaluate the definite integrals in Exercises 25 to 33.

** **

### ** Answer: **

Let I = -(i)

Replacing x with ( -x),

- (ii)

Adding (i) and (ii)

### ** Question:33 ** Evaluate the definite integrals in Exercises 25 to 33.

** **

### ** Answer: **

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

### ** Question:34 ** Prove the following (Exercises 34 to 39)

** . **

### ** Answer: **

L.H.S =

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

= RHS

Hence proved.

### ** Question:35 ** Prove the following (Exercises 34 to 39)

### ** Answer: **

Integrating I by parts

Applying Limits from 0 to 1

Hence proved I = 1

### ** Question:36 ** Prove the following (Exercises 34 to 39)

** **

### ** Answer: **

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

### ** Question:37 ** Prove the following (Exercises 34 to 39)

** **

### ** Answer: **

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

The limits change to 0 and 1

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

Hence proved.

### ** Question:38 ** Prove the following (Exercises 34 to 39)

** **

### ** Answer: **

The integral is written as

Hence Proved

### ** Question:39 ** Prove the following (Exercises 34 to 39)\

** **

### ** Answer: **

Integrating by parts we get

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

Hence Proved

### ** Question:40 ** Evaluate as a limit of a sum.

### ** Answer: **

As we know

where b-a=hn

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

### ** Question:41 ** Choose the correct answers in Exercises 41 to 44.

** . ** is equal to

(A)

(B)

(C)

(D)

### ** Answer: **

the above integral can be re arranged as

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

(A) is correct

### ** Question:42** Choose the correct answers in Exercises 41 to 44.

** . ** is equal to

(A)

(B)

(C)

(D)

### ** Answer: **

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

### ** Question:43** Choose the correct answers in Exercises 41 to 44.

If , then is equal to

(A)

(B)

(C)

(D)

### ** Answer: **

As we know

Using the above property we can write the integral as

Answer (D) is correct

### ** Question: 44 ** Choose the correct answers in Exercises 41 to 44.

** ** The value of is

(A) 1

(B) 0

(C) -1

(D)

### ** Answer: **

as

Now the integral can be written as

(B) is correct.

## More about NCERT solutions for class 12 maths chapter 7 miscellaneous exercise

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.

## Benefits of ncert solutions for class 12 maths chapter 7 miscellaneous exercises

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.

### Also see-

NCERT exemplar solutions class 12 maths chapter 7

NCERT solutions for class 12 maths chapter 7

**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

## Subject wise NCERT Exemplar solutions

NCERT Exemplar Class 12th Maths

NCERT Exemplar Class 12th Physics

NCERT Exemplar Class 12th Chemistry

NCERT Exemplar Class 12th Biology

Happy learning!!!

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