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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


Transcript

Ex 7.7, 2 (Method 1) โˆš(1โˆ’4๐‘ฅ2) โˆซ1โ–’ใ€–โˆš(1โˆ’4๐‘ฅ^2 ).๐‘‘๐‘ฅใ€— =โˆซ1โ–’ใ€–โˆš(4(1/4โˆ’๐‘ฅ^2 ) ).๐‘‘๐‘ฅใ€— =โˆซ1โ–’ใ€–โˆš4 โˆš(1/4โˆ’๐‘ฅ^2 ).๐‘‘๐‘ฅใ€— =2โˆซ1โ–’ใ€–โˆš((1/2)^2โˆ’๐‘ฅ^2 ).๐‘‘๐‘ฅ" " ใ€— It is of the form โˆซ1โ–’ใ€–โˆš(๐‘Ž^2โˆ’๐‘ฅ^2 ) .๐‘‘๐‘ฅ=1/2 ๐‘ฅโˆš(๐‘Ž^2โˆ’๐‘ฅ^2 )+๐‘Ž^2/2 ๐‘ ๐‘–๐‘›^(โˆ’1) ๐‘ฅ/๐‘Ž+๐ถ1ใ€— Replacing a with 2 and 1/2 , we get =2[1/2 ๐‘ฅโˆš((1/2)^2โˆ’๐‘ฅ^2 ) +(1/2)^2.1/2 ๐‘ ๐‘–๐‘›^(โˆ’1) ๐‘ฅ/(1/2)+๐ถ1] =2[1/2 ๐‘ฅโˆš(1/4โˆ’๐‘ฅ^2 ) +1/4 . 1/2 ๐‘ ๐‘–๐‘›^(โˆ’1) 2๐‘ฅ+๐ถ1] =๐‘ฅโˆš(1/4โˆ’๐‘ฅ^2 )+1/4 ๐‘ ๐‘–๐‘›^(โˆ’1) 2๐‘ฅ+2๐ถ1 =1/4 ๐‘ ๐‘–๐‘›^(โˆ’1) 2๐‘ฅ+๐‘ฅโˆš((1 โˆ’ 4๐‘ฅ^2)/4)+๐ถ =๐Ÿ/๐Ÿ’ ๐’”๐’Š๐’^(โˆ’๐Ÿ) ๐Ÿ๐’™+๐Ÿ/๐Ÿ ๐’™โˆš(๐Ÿ โˆ’ ๐Ÿ’๐’™^๐Ÿ )++๐‘ช [where ๐ถ=2๐ถ1] Ex 7.7, 2 (Method 2) โˆš(1โˆ’4๐‘ฅ2) Let 2๐‘ฅ=๐‘ก Differentiating both sides w.r.t. ๐‘ฅ 2=๐‘‘๐‘ก/๐‘‘๐‘ฅ ๐‘‘๐‘ฅ=๐‘‘๐‘ก/2 Integrating the function โˆซ1โ–’ใ€–โˆš(1โˆ’4๐‘ฅ^2 ) ๐‘‘๐‘ฅใ€— =โˆซ1โ–’ใ€–โˆš(1โˆ’(2๐‘ฅ)^2 ) ๐‘‘๐‘ฅใ€— Putting value of t = 2๐‘ฅ and ๐‘‘๐‘ฅ = ๐‘‘๐‘ก/2 =โˆซ1โ–’ใ€–โˆš(1โˆ’๐‘ก^2 ) .๐‘‘๐‘ก/2ใ€— =1/2 โˆซ1โ–’ใ€–โˆš(1โˆ’๐‘ก^2 ) ๐‘‘๐‘กใ€— =1/2 โˆซ1โ–’ใ€–โˆš((1)^2โˆ’(๐‘ก)^2 ) ๐‘‘๐‘กใ€— =1/2 [๐‘ก/2 โˆš((1)^2โˆ’๐‘ก^2 )+(1)^2/2 ๐‘ ๐‘–๐‘›^(โˆ’1) ๐‘ก/((1) )+๐ถ1] =๐‘ก/4 โˆš(1โˆ’๐‘ก^2 )+1/4 ๐‘ ๐‘–๐‘›^(โˆ’1) (๐‘ก)+๐ถ1" " /2 =๐‘ก/4 โˆš(1โˆ’๐‘ก^2 )+1/4 ๐‘ ๐‘–๐‘›^(โˆ’1) (๐‘ก)+๐ถ It is of the form โˆซ1โ–’ใ€–โˆš(๐‘Ž^2โˆ’๐‘ฅ^2 ) ๐‘‘๐‘ฅ=1/2 ๐‘ฅโˆš(๐‘Ž^2โˆ’๐‘ฅ^2 )+๐‘Ž^2/2 ๐‘ ๐‘–๐‘›^(โˆ’1) ๐‘ฅ/๐‘Ž +๐ถ1ใ€— Replacing a by 1 and ๐‘ฅ by ๐‘ก , we get Putting back t = 2x =2๐‘ฅ/4 โˆš(1โˆ’(2๐‘ฅ)^2 )+1/4 ๐‘ ๐‘–๐‘›^(โˆ’1) 2๐‘ฅ+๐ถ =๐Ÿ/๐Ÿ’ ๐’”๐’Š๐’^(โˆ’๐Ÿ) ๐Ÿ๐’™+๐’™/๐Ÿ โˆš(๐Ÿโˆ’๐Ÿ’๐’™^๐Ÿ )+๐‘ช

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.