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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 You are here

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

Chapter 7 Class 12 Integrals

Serial order wise

Last updated at March 16, 2023 by Teachoo

Ex 7.7, 7 √(1+3𝑥−𝑥2) ∫1▒〖√(1+3𝑥−𝑥^2 ) 𝑑𝑥〗 =∫1▒〖√(1−(−3𝑥+𝑥^2 ) ) 𝑑𝑥〗 =∫1▒〖√(1−(𝑥^2−3𝑥) ) 𝑑𝑥〗 =∫1▒〖√(1−[𝑥^2−2(3/2)(𝑥)] ) 𝑑𝑥〗 =∫1▒〖√(1−[𝑥^2−2(3/2)(𝑥)+(3/2)^2−(3/2)^2 ] ) 𝑑𝑥〗 =∫1▒〖√(1−[(𝑥−3/2)^2−9/4] ) 𝑑𝑥〗 =∫1▒〖√(1−(𝑥−3/2)^2+9/4) 𝑑𝑥〗 =∫1▒〖√(1+〖9/4−(𝑥−3/2)〗^2 ) 𝑑𝑥〗 =∫1▒〖√(〖13/4−(𝑥−3/2)〗^2 ) 𝑑𝑥〗 =∫1▒〖√(〖(√13/2)^2−(𝑥−3/2)〗^2 ) 𝑑𝑥〗 =(𝑥 − 3/2)/2 √((√13/2)^2−(𝑥−3/2)^2 )+(√13/2)^2/2 𝑠𝑖𝑛^(−1) ((𝑥 − 3/2))/(√13/2)+𝐶 It is of the form ∫1▒〖√(𝑎^2−𝑥^2 ) 𝑑𝑥=1/2 𝑥√(𝑎^2−𝑥^2 )+𝑎^2/2 𝑠𝑖𝑛^(−1) 𝑥/𝑎+𝐶〗 ∴ Replacing 𝑥 by 𝑥−3/2 and a by √13/2 , we get =((2𝑥 − 3)/2)/2 √(13/4−(𝑥−3/2)^2 )+(13/4)/2 𝑠𝑖𝑛^(−1) (((2𝑥 − 3)/2))/(√13/2)+𝐶 =(2𝑥 − 3)/4 √(13/4−(𝑥−3/2)^2 )+13/8 𝑠𝑖𝑛^(−1) (2𝑥 − 3)/√13+𝐶 =(2𝑥 − 3)/4 √(13/4−[𝑥^2+9/4−2(𝑥)(3/2)] )+13/8 𝑠𝑖𝑛^(−1) (2𝑥 − 3)/√13+𝐶 =(2𝑥 − 3)/4 √(13/4−𝑥^2−9/4+3𝑥)+13/8 𝑠𝑖𝑛^(−1) (2𝑥 − 3)/√13+𝐶 =(2𝑥 − 3)/4 √(13/4−9/4+3𝑥−𝑥^2 )+13/8 𝑠𝑖𝑛^(−1) (2𝑥 − 3)/√13+𝐶 =(𝟐𝒙 − 𝟑)/𝟒 √(𝟏+𝟑𝒙−𝒙^𝟐 )+𝟏𝟑/𝟖 𝒔𝒊𝒏^(−𝟏) ((𝟐𝒙 − 𝟑)/√𝟏𝟑)+𝑪