Class 12 Maths NCERT solutions of Exercise 7.4 of chapter 7-Integrals - Future Study Point

Class 12 Maths NCERT solutions of Exercise 7.4 of chapter 7-Integrals

Class 12 Maths NCERT solutions of Exercise 7.4 of chapter 7-Integrals

Class 12 Maths NCERT solutions of Exercise 7.4-Integrals

Class 12 Maths NCERT solutions of Exercise 7.4 of chapter 7-Integrals are created by an expert teacher of maths. These NCERT solutions are the solutions to the unsolved questions of class 12 maths exercise 7.4 of chapter 7-Integrals of the NCERT textbook prescribed by the CBSE School board of India.NCERT solutions of exercise 7.4 will help in your preparation of the board exam of CBSE and in clearing the ways of methods of integration used in different kinds of functions. By studying these NCERT Solutions you will become capable to solve all problems related to the integration.

Class 12 Maths NCERT solutions of Exercise 7.4 of chapter 7-Integrals

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In solving unsolved questions of exercise 7.4 of chapter 7-Integrals you have to understand solutions of algebraic expressions so you can revise the maths chapter of previous classes like quadratic equations. Click here to revise complete square of polynomial

 

Class 12 Maths NCERT solutions of Chapter 7 Integrals

Exercise 7.1- Integrals

Exercise 7.2-Integral

Exercise 7.3-Integrals

Exercise 7.4 -Integral

 

Class 12 Maths NCERT solutions of Exercise 7.4 of chapter 7-Integrals

Exercise 7.4-Integrals

Integrate the functions in exercises 1 t0 23.

Ans. Integrating the functions

Let t = x³

dt = 3x²dx

Substituting the value dt = 3x²dx and t = x³

Substituting back the value of t = x³

Ans.We are given the function

Integrating it

Let t = 2x

Substituting the value 2x=t and dx=dt/2

Using the formula for integration

Therefore

Substituting back the value of t=2x

Ans. We are given

Let t= 2-x

dx= -dt

Putting value of dx =-dt and 2-x= t in the integral

Applying the integration formula

We have

Substituting back the value of t= 2-x, we get

Ans.

We are given

Integrating it

Let t= 5x

Substituting the value of dx =dt/5 and 5x = t

Applying the integration formula

We have

Substituting back the value of t=5x

Class 12 Maths NCERT solutions of Exercise 7.4 of chapter 7-Integrals

Ans. We are given

Integrating it

Let t= √2x²

Substituting  value of xdx ,we have

Applying the integration formula

We get

Substituting back the value of t=√2 x²

We are given

Integrating the given function

Let t = x³

Substituting the value x²dx = dt/3 and t= x³, we get

Applying the integration formula

Substituting back the value of t = x³

.

Ans.We are given the function

Integrating the function

Solving integration of both fraction separately

Solution of first part

Let t = x² – 1

Substituting the value of xdx = dt/2 and x² – 1 = t

Substituting the value of  t = x² -1

We have

Integration of first part =

Putting this value as the integration of  first part and applying the following  integration formula for solving the second fraction

We have

Ans. We are given

Integrating the function

Let t= x³

3x²dx =dt

Substituting the value of x²dx = dt/3 and x³ = t

Applying the integration formula

 

Substituting back the value of t = x³

Class 12 Maths NCERT solutions of Exercise 7.4 of chapter 7-Integrals

Ans. We are given

Integrating the given function

Let t = tanx

dt = sec²x.dx

Substituting the value of sec²x.dx = dt and tanx=t

Applying the integration formula

Substituting back the value of t=tanx

Ans. We are given

Integrating the given function

Let t = x+1 ,we can rewrite the denominator x² +2x+2= (x+1)² +1

dt = dx

 

Applying the integration formula

Substituting back the value of t = x+1

Ans.We are given the function

Integrating it

Rewriting the denominator of the given function

9x² +6x + 1 +4 = (3x+1)² +2²

Let t = 3x +1

Substituting the value 3x+1=t and dx= dt/3

Applying the integration formula

 

We have

Substituting back the value of t=3x+1

Ans. We are given

Converting the terms of denominator into a complete square form

7 -6x -x² = -(x² +6x -7+3² -3²)= -(x+3)² +16= 4² -(x+3)²

So,we have

Now,integrating the function

Let t= x+3

dt = dx

Substituting the value of dx=dt and x+3= t

Applying the formula of integration

 

Substituting back the value of t = x +3

Class 12 Maths NCERT solutions of Exercise 7.4 of chapter 7-Integrals

Ans. We are given the function

Integrating the given function

Converting the terms of denominator into complete square form

(x-1)(x-2)=x²-3x +2 +(3/2)² -(3/2)²

= (x-3/2)²-(1/2)²

So,we have

Now, integrating it

dt =dx

Substituting x-3/2= t and dx= dt in the integral

Using the integration formula

 

So, we have

Substituting back the value of t = x -3/2, we have

 

Converting the denominator 8 + 3x -x² into the terms of complete square

8 + 3x -x²

 

 

Now, we have

Now,integrating the function

Let t= x -3/2

dt =dx

Substituting the value dx=dt and t= x -3/2

Applying the integration formula

So, we have

 

Substituting back the value of t = x-3/2

Ans. We are given

Converting the terms of denominator into the form of complete square

(x -a)(x-b)  = x² – (a +b)x + ab=x² – (a +b)x + ab +[(a+b)/2]² -[(a+b)/2]²

Integrating it

dx =dt

Substituting the value of   

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