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Ex 7.7

Ex 7.7, 1

Ex 7.7, 2 Important You are here

Ex 7.7, 3

Ex 7.7, 4

Ex 7.7, 5 Important

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Ex 7.7, 7 Important

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Ex 7.7, 11 Important

Ex 7.7, 12 (Supplementary NCERT) Important Deleted for CBSE Board 2024 Exams

Ex 7.7, 13 (Supplementary NCERT) Deleted for CBSE Board 2024 Exams

Ex 7.7, 14 (Supplementary NCERT) Important Deleted for CBSE Board 2024 Exams

Last updated at May 29, 2023 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