# Example 27

Last updated at May 29, 2018 by Teachoo

Last updated at May 29, 2018 by Teachoo

Transcript

Example 27 Evaluate the following integrals: (i) 2 3 2 Step 1 :- 2 = 2 + 1 2 + 1 = 3 3 Hence F = 3 3 Step 2 :- 2 3 2 = 3 2 = 3 3 3 2 3 3 = 27 3 8 3 = 19 3 Example 27 Evaluate the following integrals: (ii) 4 9 30 3 2 2 Step 1 :- 30 3 2 2 Let 30 3 2 = Differentiating w.r.t. both sides 30 3 2 = 3 2 3 2 1 = 3 2 1 2 = = 3 2 1 2 = 2 3 Therefore, our equation becomes 30 3 2 2 = 2 2 3 = 2 3 2 = 2 3 2 = 2 3 2 + 1 2 + 1 = 2 3 1 1 = 2 3 1 = 2 3 Putting = 30 3 2 = 2 3 30 3 2 Hence F = 2 3 30 3 2 Step 2 :- 4 9 30 3 2 = 9 4 = 2 3 30 9 3 2 2 3 30 4 3 2 = 2 3 30 3 2 2 3 2 3 30 2 2 2 3 = 2 3 1 30 3 3 1 30 2 3 = 2 3 1 30 27 1 30 8 = 2 3 1 3 1 22 = 2 3 22 3 3 22 = 2 3 19 66 = 19 3 (33) = Example 27 Evaluate the following integrals: (iii) 1 2 + 1 + 2 Step 1 :- = + 1 + 2 We can write the integrate as : + 1 + 2 = A + 1 + B + 2 + 1 + 2 = A + 2 + B + 1 + 1 + 2 By canceling denominators =A +2 +B +1 Therefore + 1 + 2 = 1 + 1 + 2 + 2 Integrating w.r.t. +1 +2 = 1 +1 + 2 +2 = +1 +2 +2 = +1 + +2 2 = +2 2 +1 = + 2 2 + 1 Hence = + 2 2 + 1 Step 2 :- 1 2 + 1 + 2 = 2 1 1 2 + 1 + 2 = 2 + 2 2 2 + 1 1 + 2 2 1 + 1 = 4 2 3 3 2 2 = 4 2 3 3 2 2 = 4 2 3 2 3 2 = 16 3 2 9 = 32 27 = Example 27 Evaluate the following integrals: (iv) 0 4 sin 3 2 cos 2 0 4 3 2 2 Step 1 :- F = 3 2 2 Let s 2 = Differentiating w.r.t. ( sin 2 ) = 2c 2 = = 2 2 Hence the integrate 3 2 2 = 3 2 2 2 = 1 2 3 = 1 2 3+1 3+1 = 1 2 4 4 = 4 8 Putting back = 2 = 1 8 4 2 Hence F = 1 8 4 2 Step 2 :- 0 4 3 2 2 = 4 0 = 1 8 4 2 4 1 8 4 2 0 = 1 8 4 2 1 8 4 0 = 1 8 1 4 1 8 0 4 = 1 8 1 0 = 1 8

Ex 7.1, 10
Important

Ex 7.1, 18 Important

Ex 7.1, 20 Important

Ex 7.2, 20 Important

Ex 7.2, 26 Important

Ex 7.2, 35 Important

Ex 7.2, 36 Important

Ex 7.3, 6 Important

Ex 7.3, 13 Important

Ex 7.3, 18 Important

Ex 7.3, 22 Important

Ex 7.3, 24 Important

Example 9 Important

Example 10 Important

Ex 7.4, 8 Important

Ex 7.4, 15 Important

Ex 7.4, 21 Important

Ex 7.4, 22 Important

Ex 7.4, 25 Important

Example 15 Important

Ex 7.5, 9 Important

Ex 7.5, 11 Important

Ex 7.5, 17 Important

Ex 7.5, 18 Important

Ex 7.5, 21 Important

Example 20 Important

Example 22 Important

Ex 7.6, 13 Important

Ex 7.6, 14 Important

Ex 7.6, 18 Important

Ex 7.6, 19 Important

Ex 7.6, 24 Important

Ex 7.7, 5 Important

Ex 7.7, 10 Important

Ex 7.7, 11 Important

Example 25 Important

Ex 7.8, 4 Important

Ex 7.8, 6 Important

Example 27 Important You are here

Ex 7.9, 15 Important

Ex 7.9, 16 Important

Ex 7.9, 20 Important

Ex 7.9, 22 Important

Ex 7.10, 4 Important

Ex 7.10, 7 Important

Ex 7.10, 8 Important

Ex 7.10, 9 Important

Example 30 Important

Example 34 Important

Example 36 Important

Ex 7.11,8 Important

Ex 7.11, 18 Important

Example 40 Important

Example 41 Important

Example 44 Important

Misc 18 Important

Misc 8 Important

Misc 19 Important

Misc 24 Important

Misc 30 Important

Misc 32 Important

Misc 41 Important

Misc 44 Important

Integration Formula Sheet - Chapter 7 Class 12 Formulas Important

Class 12

Important Question for exams Class 12

- Chapter 1 Class 12 Relation and Functions
- Chapter 2 Class 12 Inverse Trigonometric Functions
- Chapter 3 Class 12 Matrices
- Chapter 4 Class 12 Determinants
- Chapter 5 Class 12 Continuity and Differentiability
- Chapter 6 Class 12 Application of Derivatives
- Chapter 7 Class 12 Integrals
- Chapter 8 Class 12 Application of Integrals
- Chapter 9 Class 12 Differential Equations
- Chapter 10 Class 12 Vector Algebra
- Chapter 11 Class 12 Three Dimensional Geometry
- Chapter 12 Class 12 Linear Programming
- Chapter 13 Class 12 Probability

About the Author

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.