# Ex 7.7, 2 - Chapter 7 Class 12 Integrals (Term 2)

Last updated at Dec. 20, 2019 by Teachoo

Ex 7.7

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Ex 7.7, 2 Important You are here

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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

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Last updated at Dec. 20, 2019 by Teachoo

Ex 7.7, 2 (Method 1) √(1−4𝑥2) ∫1▒〖√(1−4𝑥^2 ).𝑑𝑥〗 =∫1▒〖√(4(1/4−𝑥^2 ) ).𝑑𝑥〗 =∫1▒〖√4 √(1/4−𝑥^2 ).𝑑𝑥〗 =2∫1▒〖√((1/2)^2−𝑥^2 ).𝑑𝑥" " 〗 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+𝐶1〗 Replacing a with 2 and 1/2 , we get =2[1/2 𝑥√((1/2)^2−𝑥^2 ) +(1/2)^2.1/2 𝑠𝑖𝑛^(−1) 𝑥/(1/2)+𝐶1] =2[1/2 𝑥√(1/4−𝑥^2 ) +1/4 . 1/2 𝑠𝑖𝑛^(−1) 2𝑥+𝐶1] =𝑥√(1/4−𝑥^2 )+1/4 𝑠𝑖𝑛^(−1) 2𝑥+2𝐶1 =1/4 𝑠𝑖𝑛^(−1) 2𝑥+𝑥√((1 − 4𝑥^2)/4)+𝐶 =𝟏/𝟒 𝒔𝒊𝒏^(−𝟏) 𝟐𝒙+𝟏/𝟐 𝒙√(𝟏 − 𝟒𝒙^𝟐 )++𝑪 [where 𝐶=2𝐶1] Ex 7.7, 2 (Method 2) √(1−4𝑥2) Let 2𝑥=𝑡 Differentiating both sides w.r.t. 𝑥 2=𝑑𝑡/𝑑𝑥 𝑑𝑥=𝑑𝑡/2 Integrating the function ∫1▒〖√(1−4𝑥^2 ) 𝑑𝑥〗 =∫1▒〖√(1−(2𝑥)^2 ) 𝑑𝑥〗 Putting value of t = 2𝑥 and 𝑑𝑥 = 𝑑𝑡/2 =∫1▒〖√(1−𝑡^2 ) .𝑑𝑡/2〗 =1/2 ∫1▒〖√(1−𝑡^2 ) 𝑑𝑡〗 =1/2 ∫1▒〖√((1)^2−(𝑡)^2 ) 𝑑𝑡〗 =1/2 [𝑡/2 √((1)^2−𝑡^2 )+(1)^2/2 𝑠𝑖𝑛^(−1) 𝑡/((1) )+𝐶1] =𝑡/4 √(1−𝑡^2 )+1/4 𝑠𝑖𝑛^(−1) (𝑡)+𝐶1" " /2 =𝑡/4 √(1−𝑡^2 )+1/4 𝑠𝑖𝑛^(−1) (𝑡)+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 +𝐶1〗 Replacing a by 1 and 𝑥 by 𝑡 , we get Putting back t = 2x =2𝑥/4 √(1−(2𝑥)^2 )+1/4 𝑠𝑖𝑛^(−1) 2𝑥+𝐶 =𝟏/𝟒 𝒔𝒊𝒏^(−𝟏) 𝟐𝒙+𝒙/𝟐 √(𝟏−𝟒𝒙^𝟐 )+𝑪