Ex 7.7, 2 - Integrate root 1 - 4x2 - Chapter 7 Integrals

Ex 7.7, 2 - Chapter 7 Class 12 Integrals - Part 2
Ex 7.7, 2 - Chapter 7 Class 12 Integrals - Part 3 Ex 7.7, 2 - Chapter 7 Class 12 Integrals - Part 4 Ex 7.7, 2 - Chapter 7 Class 12 Integrals - Part 5

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Ex 7.7, 2 (Method 1) √(1βˆ’4π‘₯2) ∫1β–’γ€–βˆš(1βˆ’4π‘₯^2 ).𝑑π‘₯γ€— =∫1β–’γ€–βˆš(4(1/4βˆ’π‘₯^2 ) ).𝑑π‘₯γ€— =∫1β–’γ€–βˆš4 √(1/4βˆ’π‘₯^2 ).𝑑π‘₯γ€— =2∫1β–’γ€–βˆš((1/2)^2βˆ’π‘₯^2 ).𝑑π‘₯" " γ€— It is of the form ∫1β–’γ€–βˆš(π‘Ž^2βˆ’π‘₯^2 ) .𝑑π‘₯=1/2 π‘₯√(π‘Ž^2βˆ’π‘₯^2 )+π‘Ž^2/2 𝑠𝑖𝑛^(βˆ’1) π‘₯/π‘Ž+𝐢1γ€— Replacing a with 2 and 1/2 , we get =2[1/2 π‘₯√((1/2)^2βˆ’π‘₯^2 ) +(1/2)^2.1/2 𝑠𝑖𝑛^(βˆ’1) π‘₯/(1/2)+𝐢1] =2[1/2 π‘₯√(1/4βˆ’π‘₯^2 ) +1/4 . 1/2 𝑠𝑖𝑛^(βˆ’1) 2π‘₯+𝐢1] =π‘₯√(1/4βˆ’π‘₯^2 )+1/4 𝑠𝑖𝑛^(βˆ’1) 2π‘₯+2𝐢1 =1/4 𝑠𝑖𝑛^(βˆ’1) 2π‘₯+π‘₯√((1 βˆ’ 4π‘₯^2)/4)+𝐢 =𝟏/πŸ’ π’”π’Šπ’^(βˆ’πŸ) πŸπ’™+𝟏/𝟐 π’™βˆš(𝟏 βˆ’ πŸ’π’™^𝟐 )++π‘ͺ [where 𝐢=2𝐢1] Ex 7.7, 2 (Method 2) √(1βˆ’4π‘₯2) Let 2π‘₯=𝑑 Differentiating both sides w.r.t. π‘₯ 2=𝑑𝑑/𝑑π‘₯ 𝑑π‘₯=𝑑𝑑/2 Integrating the function ∫1β–’γ€–βˆš(1βˆ’4π‘₯^2 ) 𝑑π‘₯γ€— =∫1β–’γ€–βˆš(1βˆ’(2π‘₯)^2 ) 𝑑π‘₯γ€— Putting value of t = 2π‘₯ and 𝑑π‘₯ = 𝑑𝑑/2 =∫1β–’γ€–βˆš(1βˆ’π‘‘^2 ) .𝑑𝑑/2γ€— =1/2 ∫1β–’γ€–βˆš(1βˆ’π‘‘^2 ) 𝑑𝑑〗 =1/2 ∫1β–’γ€–βˆš((1)^2βˆ’(𝑑)^2 ) 𝑑𝑑〗 =1/2 [𝑑/2 √((1)^2βˆ’π‘‘^2 )+(1)^2/2 𝑠𝑖𝑛^(βˆ’1) 𝑑/((1) )+𝐢1] =𝑑/4 √(1βˆ’π‘‘^2 )+1/4 𝑠𝑖𝑛^(βˆ’1) (𝑑)+𝐢1" " /2 =𝑑/4 √(1βˆ’π‘‘^2 )+1/4 𝑠𝑖𝑛^(βˆ’1) (𝑑)+𝐢 It is of the form ∫1β–’γ€–βˆš(π‘Ž^2βˆ’π‘₯^2 ) 𝑑π‘₯=1/2 π‘₯√(π‘Ž^2βˆ’π‘₯^2 )+π‘Ž^2/2 𝑠𝑖𝑛^(βˆ’1) π‘₯/π‘Ž +𝐢1γ€— Replacing a by 1 and π‘₯ by 𝑑 , we get Putting back t = 2x =2π‘₯/4 √(1βˆ’(2π‘₯)^2 )+1/4 𝑠𝑖𝑛^(βˆ’1) 2π‘₯+𝐢 =𝟏/πŸ’ π’”π’Šπ’^(βˆ’πŸ) πŸπ’™+𝒙/𝟐 √(πŸβˆ’πŸ’π’™^𝟐 )+π‘ͺ

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.