



Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class
Ex 7.7
Ex 7.7, 2 Important You are here
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 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