You are here
Misc 2 - Chapter 7 Class 12 Integrals
Last updated at April 16, 2024 by Teachoo
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
- 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
Misc 2 Integrate the function 1/(√(𝑥 + 𝑎) + √(𝑥 +𝑏) ) ∫1▒〖1/(√(𝑥 + 𝑎) + √(𝑥 + 𝑏) ) 𝑑𝑥〗 Rationalising = ∫1▒〖(1/(√(𝑥 + 𝑎) + √(𝑥 + 𝑏) ) × (√(𝑥 + 𝑎) − √(𝑥 + 𝑏))/(√(𝑥 + 𝑎) − √(𝑥 + 𝑏) )) 𝑑𝑥〗 = ∫1▒〖(√(𝑥 + 𝑎) − √(𝑥 + 𝑏))/((√(𝑥 + 𝑎) + √(𝑥 + 𝑏) )(√(𝑥 + 𝑎) − √(𝑥 + 𝑏) ) ) 𝑑𝑥〗 Using (𝑎−𝑏)(𝑎+𝑏)=𝑎^2−𝑏^2 =∫1▒〖(√(𝑥 + 𝑎) − √(𝑥 + 𝑏))/((√(𝑥 + 𝑎) )^2 − (√(𝑥 + 𝑏) )^2 ) 𝑑𝑥〗 = ∫1▒〖(√(𝑥 + 𝑎) − √(𝑥 + 𝑏))/(𝑥 + 𝑎 −(𝑥 + 𝑏) ) 𝑑𝑥〗 = ∫1▒〖(√(𝑥 + 𝑎) − √(𝑥 + 𝑏))/(𝑎 − 𝑏) 𝑑𝑥〗 = 1/(𝑎 − 𝑏) ∫1▒(√(𝑥+𝑎) −√(𝑥+𝑏) ) 𝑑𝑥 = 1/(𝑎 − 𝑏) ∫1▒((𝑥+𝑎)^(1/2) −(𝑥+𝑎)^(1/2) ) 𝑑𝑥 = 1/(𝑎 − 𝑏) [∫1▒(𝑥+𝑎)^(1/2) 𝑑𝑥−∫1▒(𝑥+𝑎)^(1/2) 𝑑𝑥] = 1/(𝑎 − 𝑏) [(𝑥 + 𝑎)^(1/2 + 1)/(1/2 + 1) − (𝑥 + 𝑏)^(1/2 + 1)/(1/2 + 1) ] + 𝐶 = 1/(𝑎 − 𝑏) [(𝑥 + 𝑎)^(3/2)/(3/2) − (𝑥 + 𝑏)^(3/2)/(3/2)] + 𝐶 = 1/(𝑎 − 𝑏) [〖2(𝑥 + 𝑎)〗^(3/2)/3 − 〖2(𝑥 + 𝑏)〗^(3/2)/3] + 𝐶 = 𝟐/𝟑(𝒂 − 𝒃) [(𝒙+𝒂)^(𝟑/𝟐)−(𝒙+𝒃)^(𝟑/𝟐) ] + 𝑪
