NCERT Solutions for Class 12 Maths Chapter 7 Exercise 7.2

Physics Wallah Academic Expert
November 14, 2024

Class 12 Maths Chapter 7 Exercise 7.2:- Chapter 7 in Class 12 Maths covers "Integrals." Exercise 7.2 focuses on the evaluation of integrals using various methods, such as substitution, integration by parts, and partial fractions. This exercise helps students understand how to find the integral of different types of functions and improve their problem-solving skills in calculus. The problems in this exercise are designed to provide a comprehensive understanding of the integration techniques and their applications. Check out the NCERT Solutions for Class 12 Maths Chapter 7 Exercise 7.2  below.

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NCERT Solutions for Class 12 Maths Chapter 7 Exercise 7.2 

Get the NCERT Solutions for Class 12 Maths Chapter 7 Exercise 7.2 below:-

Integrate the functions in Exercise 1 to 8.

Question 1.chapter 7-Integrals Exercise 7.2/image001.png

Solution :

Let 1 + x2 = t

∴2x dx = dt
NCERT Solutions class 12 Maths Integrals

Question 2. chapter 7-Integrals Exercise 7.2

Solution :

Let log |x| = t

∴ 1/x dx = dt
chapter 7-Integrals Exercise 7.2
NCERT Solutions class 12 Maths Integrals

Check out: NCERT Solutions for Class 12 Maths Chapter 6 Exercise 6.1

Question 3. chapter 7-Integrals Exercise 7.2/image019.png

Solution :
chapter 7-Integrals Exercise 7.2/image020.png

Check out: NCERT Solutions for Class 12 Maths Chapter 6 Exercise 6.2

Question 4. sin x ⋅ sin (cos x)

Solution :

sin x ⋅ sin (cos x)

Let cos x = t

∴ −sin x dx = dt
NCERT Solutions class 12 Maths Integrals

Question 5. sin(ax + b) cos(ax + b)

Solution :
chapter 7-Integrals Exercise 7.2/image034.png

Questionc 6. √ax + b

Solution :

Let ax + b = t

⇒ adx = dt

chapter 7-Integrals Exercise 7.2/image042.png

Question 7. x√x + 2

Solution :

Let  (x + 2) = t

∴ dx = dt

chapter 7-Integrals Exercise 7.2/image048.png

Question 8. x√1 + 2x2
Solution :

Let 1 + 2x2 = t

∴ 4xdx = dt

chapter 7-Integrals Exercise 7.2/image059.png

Integrate the functions in Exercise 9 to 17.

Question 9.chapter 7-Integrals Exercise 7.2/image067.png

Solution :
chapter 7-Integrals Exercise 7.2/image068.png

Question 10.chapter 7-Integrals Exercise 7.2/image079.png

Solution :
chapter 7-Integrals Exercise 7.2

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Question 11.chapter 7-Integrals Exercise 7.2/image090.png

Solution :
chapter 7-Integrals Exercise 7.2/image091.png

Question 12.chapter 7-Integrals Exercise 7.2/image101.png

Solution :

Let x3 – 1 = t

∴ 3x2 dx = dt

chapter 7-Integrals Exercise 7.2/image103.png

Question 13.chapter 7-Integrals Exercise 7.2/image116.png

Solution :

Let 2 + 3x3 = t

∴ 9x2 dx = dt

chapter 7-Integrals Exercise 7.2/image117.png

Question 14. chapter 7-Integrals Exercise 7.2/image126.png
Solution :

Let log x = t

∴ 1/x dx = dt

chapter 7-Integrals Exercise 7.2/image128.png
Question 15. x/9 – 4x2

Solution :

Let 9 – 4x2 =  t

∴ −8x dx = dt

chapter 7-Integrals Exercise 7.2/image134.png

Question 16. chapter 7-Integrals Exercise 7.2/image143.png

Solution :

Let 2x + 3 = t

∴ 2dx = dt

chapter 7-Integrals Exercise 7.2/image144.png

Question 17. chapter 7-Integrals Exercise 7.2/image148.png
Solution :

Let x2 = t

∴ 2xdx = dt

chapter 7-Integrals Exercise 7.2/image150.png

Integrate the functions in Exercise 18 to 26.

Question 18. chapter 7-Integrals Exercise 7.2/image157.png

Solution :
chapter 7-Integrals Exercise 7.2/image158.png

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Question 19.chapter 7-Integrals Exercise 7.2/image165.png

Solution :
chapter 7-Integrals Exercise 7.2/image165.png

Dividing numerator and denominator by ex, we obtain

chapter 7-Integrals Exercise 7.2/image168.png

Question 20. chapter 7-Integrals Exercise 7.2/image177.png
Solution :

chapter 7-Integrals Exercise 7.2/image178.png

Question 21. tan2 (2x – 3)

Solution :
chapter 7-Integrals Exercise 7.2/image189.png

Question 22. sec2 (7 – 4x)

Solution :

Let 7 − 4x = t

∴ −4dx = dt

chapter 7-Integrals Exercise 7.2/image196.png

Question 23. chapter 7-Integrals Exercise 7.2/image188.png

Solution :

chapter 7-Integrals Exercise 7.2
Question 24. chapter 7-Integrals Exercise 7.2/image194.png

Solution :

chapter 7-Integrals Exercise 7.2/image207.png

Question 25. chapter 7-Integrals Exercise 7.2/image198.png

Solution :
chapter 7-Integrals Exercise 7.2/image215.png

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Question 26. chapter 7-Integrals Exercise 7.2/image206.png

Solution :

Let √x = t

chapter 7-Integrals Exercise 7.2/image227.png

Integrate the functions in Exercise 27 to 37.

Question 27.chapter 7-Integrals Exercise 7.2/image214.png

Solution :

Let sin 2x = t

chapter 7-Integrals Exercise 7.2/image235.png

Question 28.chapter 7-Integrals Exercise 7.2/image243.png

Solution :

Let 1 + sin x = t

∴ cos x dx = dt

chapter 7-Integrals Exercise 7.2/image245.png

Question 29. cot x log sin x

Solution :

Let log sin x = t

chapter 7-Integrals Exercise 7.2/image256.png
Question 30. sin x/1 + cos x

Solution :

Let 1 + cos x = t

∴ −sin x dx = dt

chapter 7-Integrals Exercise 7.2/image262.png

Question 31.  sin x/(1 + cos x)2

Solution :

Let 1 + cos x = t

∴ −sin x dx = dt

chapter 7-Integrals Exercise 7.2/image269.png

Question 32. 1/1 + cot x
Solution :
chapter 7-Integrals Exercise 7.2/image274.png

Question 33. 1/1 – tan x

Solution :

chapter 7-Integrals Exercise 7.2/image295.png

Question 34. chapter 7-Integrals Exercise 7.2/image308.png

Solution :
chapter 7-Integrals Exercise 7.2/image309.png
Question 35. chapter 7-Integrals Exercise 7.2/image319.png

Solution :

Let 1 + log x = t

∴ 1/x dx = dt
chapter 7-Integrals Exercise 7.2/image320.png

Question 36. chapter 7-Integrals Exercise 7.2/image323.png

Solution :

chapter 7-Integrals Exercise 7.2/image324.png

Question 37.chapter 7-Integrals Exercise 7.2/image330.png

Solution :

Let x4 = t

∴ 4x3 dx = dt

chapter 7-Integrals Exercise 7.2

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Choose the correct answer in Exercise 38 and 39.

Question 38. chapter 7-Integrals Exercise 7.2/image339.png equals

(A) 10x – x10 + C
(B) 10x + x10 + C
(C) (10x – x10)-1 + C
(D) log(10x + x10) + C

Solution :

chapter 7-Integrals Exercise 7.2/image344.png

Therefore, option (D) is correct.
Question 39.chapter 7-Integrals Exercise 7.2/image348.pngequals

(A) tan x + cot x + C
(B) tan x – cot x + C
(C) tan x cot x + C
(D) tan x – cot 2x + C

Solution :

chapter 7-Integrals Exercise 7.2/image353.png

Therefore, option (B) is correct.

Class 12 Maths Chapter 7 Exercise 7.2 Summary

Chapter 7 of Class 12 Maths deals with "Integrals," and Exercise 7.2 specifically focuses on integrating functions using different techniques. Here is a summary of Exercise 7.2:

  1. Integration by Substitution: This technique involves substituting a part of the integral with a new variable to simplify the integral into a more manageable form. The goal is to make the integral easier to evaluate.

  2. Integration by Parts: This method is used when the integral is a product of two functions. It is based on the formula:
    ∫u dv=uv−∫v du\int u \, dv = uv - \int v \, du∫udv=uv−∫vdu
    where uuu and dvdvdv are parts of the original integral.

  3. Partial Fractions: This approach is used to integrate rational functions by expressing them as a sum of simpler fractions, making it easier to integrate each term individually.

  4. Special Integrals: Exercise 7.2 also includes problems that involve integrals of specific forms that have standard solutions, such as ∫1a2+x2 dx\int \frac{1}{a^2 + x^2} \, dx∫a2+x21​dx and ∫1a2−x2 dx\int \frac{1}{\sqrt{a^2 - x^2}} \, dx∫a2−x2​1​dx.

Class 12 Maths Chapter 7 Exercise 7.2 FAQs

Q1. What is the primary focus of Exercise 7.2 in Chapter 7?

Ans. Exercise 7.2 primarily focuses on the evaluation of integrals using techniques like substitution, integration by parts, and partial fractions.

Q2. How do you determine when to use substitution in an integral?

Ans. Substitution is useful when the integral contains a function and its derivative. By substituting the inner function with a new variable, the integral can be simplified.

Q3. What is the formula for integration by parts?

Ans. The formula is ∫u dv=uv−∫v du\int u \, dv = uv - \int v \, du∫udv=uv−∫vdu, where uuu and dvdvdv are parts of the original integral.

Q4. When should you use integration by parts?

Ans. Integration by parts is typically used when integrating the product of two functions, especially if one function is easily differentiable and the other is easily integrable.

Q5. What are partial fractions and when are they used in integration?

Ans. Partial fractions involve breaking down a complex rational function into simpler fractions. This method is used when integrating rational functions to simplify the integration process.

 

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