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.
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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
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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