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Example 27 (iv) - Chapter 7 Class 12 Integrals
Last updated at Dec. 16, 2024 by Teachoo
Definite Integration - By Substitution
Chapter 7 Class 12 Integrals
Concept wise
- Using Formulaes
- Using Trignometric Formulaes
- Integration by substitution - x^n
- Integration by substitution - lnx
- Integration by substitution - e^x
- Integration by substitution - Trignometric - Normal
- Integration by substitution - Trignometric - Inverse
- Integration using trigo identities - sin^2,cos^2 etc formulae
- Integration using trigo identities - a-b formulae
- Integration using trigo identities - 2x formulae
- Integration using trigo identities - 3x formulae
- Integration using trigo identities - CD and CD inv formulae
- Integration using trigo identities - Inv Trigo formulae
- Integration by parts
- Integration by parts - e^x integration
- Integration by specific formulaes - Formula 1
- Integration by specific formulaes - Formula 2
- Integration by specific formulaes - Formula 3
- Integration by specific formulaes - Formula 4
- Integration by specific formulaes - Formula 5
- Integration by specific formulaes - Formula 6
- Integration by specific formulaes - Formula 7
- Integration by specific formulaes - Formula 8
- Integration by specific formulaes - Method 9
- Integration by specific formulaes - Method 10
- Integration by partial fraction - Type 1
- Integration by partial fraction - Type 2
- Integration by partial fraction - Type 3
- Integration by partial fraction - Type 4
- Integration by partial fraction - Type 5
- Definite Integral as a limit of a sum
- Definite Integration - By Formulae
- Definite Integration - By Partial Fraction
- Definite Integration - By e formula
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Definite Integration - By Substitution
- Definite Integration by properties - P2
- Definite Integration by properties - P3
- Definite Integration by properties - P4
- Definite Integration by properties - P6
- Definite Integration by properties - P7
Transcript
= 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
