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Transcript

Ex 7.9, 7 Evaluate the integrals using substitution โˆซ_(โˆ’1)^(1 )โ–’ใ€– ๐‘‘๐‘ฅ/(๐‘ฅ^2 + 2๐‘ฅ + 5)ใ€— we can write โˆซ_(โˆ’1)^1โ–’ใ€–๐‘‘๐‘ฅ/(๐‘ฅ^2 + 2๐‘ฅ + 5)=โˆซ_(โˆ’1)^1โ–’๐‘‘๐‘ฅ/((๐‘ฅ + 2๐‘ฅ + 1) + 4)ใ€— =โˆซ_(โˆ’1)^1โ–’๐‘‘๐‘ฅ/((๐‘ฅ + 1)^2 +ใ€– 2ใ€—^2 ) Putting ๐‘ฅ+1=๐‘ก Differentiating w.r.t.๐‘ฅ ๐‘‘/๐‘‘๐‘ฅ (๐‘ฅ+1)=๐‘‘๐‘ก/๐‘‘๐‘ฅ 1=๐‘‘๐‘ก/๐‘‘๐‘ฅ ๐‘‘๐‘ฅ=๐‘‘๐‘ก Hence when ๐‘ฅ varies from โ€“ 1 to 1 then ๐‘ก varies from 0 to 2 Therefore, โˆซ_(โˆ’1)^1โ–’ใ€–๐‘‘๐‘ฅ/((๐‘ฅ+1)^2 + 2^2 )=โˆซ_0^2โ–’๐‘‘๐‘ก/(๐‘ก^2 + 2^2 )ใ€— =[1/2 tan^(โˆ’1)โกใ€–๐‘ก/2ใ€— ]_0^2 =1/2 tan^(โˆ’1)โกใ€–2/2โˆ’1/2 tan^(โˆ’1)โกใ€–0/2ใ€— ใ€— =1/2 tan^(โˆ’1)โกใ€–1โˆ’1/2 tan^(โˆ’1)โก0 ใ€— =1/2 ร— ๐œ‹/4โˆ’0 =๐…/๐Ÿ–

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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 14 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.