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Ex 7.7

Ex 7.7, 1

Ex 7.7, 2 Important

Ex 7.7, 3

Ex 7.7, 4

Ex 7.7, 5 Important

Ex 7.7, 6

Ex 7.7, 7 Important

Ex 7.7, 8 Important

Ex 7.7, 9

Ex 7.7, 10

Ex 7.7, 11 Important

Ex 7.7, 12 (Supplementary NCERT) Important Deleted for CBSE Board 2023 Exams

Ex 7.7, 13 (Supplementary NCERT) Deleted for CBSE Board 2023 Exams

Ex 7.7, 14 (Supplementary NCERT) Important Deleted for CBSE Board 2023 Exams You are here

Chapter 7 Class 12 Integrals

Serial order wise

Last updated at March 16, 2023 by Teachoo

Ex 7.7, 14 (Supplementary NCERT) (𝑥+3) √(3−4𝑥〖−𝑥〗^2 ) (𝑥+3) √(3−4𝑥〖−𝑥〗^2 ) We can write it as:- x + 3 = A [𝑑/𝑑𝑥 (3−4𝑥−𝑥^2 )]+ B x + 3 = A [0−4−2𝑥]+ B x + 3 = A [−4−2𝑥]+ B x + 3 = −4"A"−2"A" 𝑥+ B x + 3 = −2"A" 𝑥+(−4"A"+𝐵) Comparing x and constant term Thus, we can write x + 3 = A [−4−2𝑥] + B x + 3 = (−1)/2 [−4−2𝑥] + 1 x = (−2A)x 𝑥/𝑥 = −2A 1 = −2A A = (−1)/2 3 = −4A + B 3 = −4((−1)/2) + B 3 = 2 + B B = 3 − 2 B = 1 Integrating the function w.r.t.x ∫1▒〖(𝑥+3) √(3−4𝑥−𝑥^2 ) 〗 𝑑𝑥 = ∫1▒〖[−1/2 [−4 −2𝑥]+1] 〗 √(3−4𝑥−𝑥^2 ) 𝑑𝑥 = ∫1▒〖[−1/2 [−4 −2𝑥] √(3−4𝑥−𝑥^2 )+1√(3−4𝑥−𝑥^2 )] 〗 𝑑𝑥 = ∫1▒〖−1/2 [−4 −2𝑥] √(3−4𝑥−𝑥^2 ) 𝑑𝑥+〗 ∫1▒√(3−4𝑥−𝑥^2 ) 𝑑𝑥 = −1/2 ∫1▒〖[−4 −2𝑥] √(3−4𝑥−𝑥^2 ) 𝑑𝑥+〗 ∫1▒√(3−4𝑥−𝑥^2 ) 𝑑𝑥 Solving 𝑰_𝟏 I_1 = (−1)/2 ∫1▒〖[−4−2𝑥] √(3−4𝑥−𝑥^2 )〗 𝑑𝑥 Let 3 − 4𝑥 − 𝑥^2 = t Diff. both sides w.r.t.x 0 − 4 −2x = 𝑑𝑡/𝑑𝑥 − 4 − 2x = 𝑑𝑡/𝑑𝑥 dx = 𝑑𝑡/(−4 − 2𝑥) Thus, our equation becomes I_1 = (−1)/2 ∫1▒〖[−4−2𝑥] √(3−4𝑥−𝑥^2 )〗 𝑑𝑥 Putting the value if (3−4𝑥−𝑥^2) and dx, we get I_1 = (−1)/2 ∫1▒〖[−4−2𝑥] √𝑡〗. 𝑑𝑥 I_1 = (−1)/2 ∫1▒〖[−4−2𝑥] √𝑡〗. 𝑑𝑡/[−4−2𝑥] ("Using t = " 3−4𝑥 −𝑥^2 ) Solving 𝑰_𝟐 I_2 = ∫1▒√(3−4𝑥−𝑥^2 ) 𝑑𝑥 I_2 = ∫1▒√(3−(4𝑥+𝑥^2)) 𝑑𝑥 I_2 = ∫1▒√(3−(𝑥^2+4𝑥)) 𝑑𝑥 I_2 = ∫1▒〖√(3−[𝑥^2+2(2) (𝑥))] 〗 𝑑𝑥 I_2 = ∫1▒〖√(3−[𝑥^2+2(2)+(2)^2 −〖(2)〗^2 )] 〗 𝑑𝑥 I_2 = ∫1▒〖√(3−[(〖𝑥+2)〗^2− 〖(2)〗^2 )] 〗 𝑑𝑥 I_2 = ∫1▒〖√(3−(𝑥+2)^2+ 〖(2)〗^2 ) 〗 𝑑𝑥 I_2 = ∫1▒〖√(3−(𝑥+2)^2+4) 〗 𝑑𝑥 I_2 = ∫1▒〖√(7−(𝑥+2)^2 ) 〗 𝑑𝑥 I_2 = ∫1▒〖√(〖(√7)〗^2−(𝑥+2)^2 ) 〗 𝑑𝑥 I_2 = (𝑥 + 2)/2 √((√7)^2−(𝑥+2)^2 )+〖(7)〗^2/2 〖𝑠𝑖𝑛〗^(−1) ((𝑥 + 2)/√7)+ C_2 I_2 = (𝑥 + 2)/2 √(7−〖(𝑥〗^2 + 4𝑥 +4)) +7/2 〖𝑠𝑖𝑛〗^(−1) ((𝑥 + 2)/√7)+ C_2 I_2 = (𝑥 + 2)/2 √(7−𝑥^2 + 4𝑥 +4) +7/2 〖𝑠𝑖𝑛〗^(−1) ((𝑥 + 2)/√7)+ C_2 I_2 = (𝑥 + 2)/2 √(3− 4𝑥−𝑥^2 ) +7/2 〖𝑠𝑖𝑛〗^(−1) ((𝑥 + 2)/√7)+ C_2 It is of form √(𝑎^2−𝑥^2 ) 𝑑𝑥=1/2 𝑥√(𝑎^2−𝑥^2 )+𝑎^2/2 〖𝑠𝑖𝑛〗^(−1) (𝑥/𝑎)+ 𝐶_2 Replacing x by (x + 2) a by √7 , we get Putting the value of I_1 and I_2 in (1) ∫1▒(𝑥+3) √(3− 4𝑥−𝑥^2 ) d𝑥 = (−1)/2 ∫1▒〖[−4−2𝑥] √(3−4𝑥−𝑥^2 )〗 𝑑𝑥+∫1▒√(3−4𝑥−𝑥^2 ) 𝑑𝑥 = (−1)/3 〖(3−4𝑥−𝑥^2)〗^(3/2) + C_1+ ((𝑥 +2) √(3 − 4𝑥 − 𝑥^2 ))/2+7/2 〖𝑠𝑖𝑛〗^(−1) ((𝑥 + 2)/√7)+ C_3 = (−𝟏)/𝟑 〖(𝟑−𝟒𝒙−𝒙^𝟐)〗^(𝟑/𝟐) +𝟕/𝟐 〖𝒔𝒊𝒏〗^(−𝟏) ((𝒙 + 𝟐)/√𝟕)+ ((𝒙 +𝟐) √(𝟑 − 𝟒𝒙 − 𝒙^𝟐 ))/𝟐+ 𝐂