# Ex 7.4, 8 - Chapter 7 Class 12 Integrals (Term 2)

Last updated at Dec. 20, 2019 by Teachoo

Ex 7.4

Ex 7.4, 1

Ex 7.4, 2 Important

Ex 7.4, 3

Ex 7.4, 4

Ex 7.4, 5 Important

Ex 7.4, 6

Ex 7.4, 7

Ex 7.4, 8 Important You are here

Ex 7.4, 9

Ex 7.4, 10

Ex 7.4, 11 Important

Ex 7.4, 12

Ex 7.4, 13 Important

Ex 7.4, 14

Ex 7.4, 15 Important

Ex 7.4, 16

Ex 7.4, 17 Important

Ex 7.4, 18

Ex 7.4, 19 Important

Ex 7.4, 20

Ex 7.4, 21 Important

Ex 7.4, 22

Ex 7.4, 23 Important

Ex 7.4, 24 (MCQ)

Ex 7.4, 25 (MCQ) Important

Chapter 7 Class 12 Integrals

Serial order wise

Last updated at Dec. 20, 2019 by Teachoo

Ex 7.4, 8 Integrate 𝑥^2/√(𝑥^6 + 𝑎^6 ) Let 𝑥^3=𝑡 Differentiating both sides w.r.t. x 3𝑥^2=𝑑𝑡/𝑑𝑥 𝑑𝑥=𝑑𝑡/(3𝑥^2 ) Integrating the function ∫1▒𝑥^2/√(𝑥^6 + 𝑎^6 ) 𝑑𝑥=∫1▒𝑥^2/√((𝑥^3 )^2 + (𝑎^3 )^2 ) 𝑑𝑥 Putting values of 𝑥^3=𝑡 and 𝑑𝑥=𝑑𝑡/(3𝑥^2 ) , we get =∫1▒𝑥^2/√(𝑡^2 + (𝑎^3 )^2 ) 𝑑𝑥 =∫1▒𝑥^2/√(𝑡^2 + (𝑎^3 )^2 ) . 𝑑𝑡/(3𝑥^2 ) =∫1▒1/√((𝑡^2 + (𝑎^3 )^2 ) ) . 𝑑𝑡/3 =1/3 ∫1▒𝑑𝑡/√(𝑡^2 + (𝑎^3 )^2 ) =1/3 [log|𝑡+√(𝑡^2 + (𝑎^3 )^2 )|+𝐶1] It is of form ∫1▒𝑑𝑥/√(𝑥^2 + 𝑎^2 ) =log|𝑥+√(𝑥^2 + 𝑎^2 )|+𝐶1 ∴ Replacing 𝑥 by 𝑡 and a by 𝑎^3, we get =1/3 log|𝑡+√(𝑡^2 + 𝑎^6 ) |+𝐶 =1/3 log|𝑥^3+√((𝑥^3 )^2 + 𝑎^6 ) |+𝐶 =𝟏/𝟑 𝒍𝒐𝒈|𝒙^𝟑+√(𝒙^𝟔+ 𝒂^𝟔 ) |+𝑪 ("Using" 𝑡=𝑥^3 )