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  1. Chapter 7 Class 12 Integrals
  2. Serial order wise

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 is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.