Ex 7.8, 10 - Chapter 7 Class 12 Integrals
Last updated at Dec. 16, 2024 by Teachoo
Definite Integration - By Formulae
Ex 7.8, 1
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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
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Definite Integration - By Formulae
- Definite Integration - By Partial Fraction
- Definite Integration - By e formula
- 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
Ex 7.8, 10 β«_0^1βππ₯/(1 + π₯2) Let F(π₯)=β«1βππ₯/(1 + π₯^2 ) =β«1β1/(1^2 + π₯^2 ) ππ₯ =1/1 .tan^(β1)β‘(π₯/1) =tan^(β1) π₯ Hence F(π₯)=tan^(β1) π₯ Now, β«_0^1βγππ₯/(1 + π₯^2 )=πΉ(1)βπΉ(0) γ =tan^(β1)β‘γ(1)βtan^(β1)β‘(0) γ =π/4β0 =π /π
