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Ex 7.8, 8 โˆซ_(๐œ‹/6)^(๐œ‹/4)โ–’ใ€–๐‘๐‘œ๐‘ ๐‘’๐‘ ๐‘ฅใ€—โก๐‘‘๐‘ฅ Let F(๐‘ฅ)=โˆซ1โ–’ใ€–๐‘๐‘œ๐‘ ๐‘’๐‘ ๐‘ฅ . ๐‘‘๐‘ฅใ€— Multiplying and Dividing by ๐‘๐‘œ๐‘ ๐‘’๐‘ ๐‘ฅ+๐‘๐‘œ๐‘ก ๐‘ฅ F(๐‘ฅ)=โˆซ1โ–’(๐‘๐‘œ๐‘ ๐‘’๐‘ ๐‘ฅ (๐‘๐‘œ๐‘ ๐‘’๐‘ ๐‘ฅ + ๐‘๐‘œ๐‘ก ๐‘ฅ))/(๐‘๐‘œ๐‘ ๐‘’๐‘ ๐‘ฅ + ๐‘๐‘œ๐‘ก ๐‘ฅ) ๐‘‘๐‘ฅ Let c๐‘œ๐‘ ๐‘’๐‘ ๐‘ฅ+๐‘๐‘œ๐‘ก ๐‘ฅ=๐‘ก Differentiating w.r.t. ๐‘ฅ ๐‘‘/๐‘‘๐‘ฅ (๐‘๐‘œ๐‘ ๐‘’๐‘ ๐‘ฅ+๐‘๐‘œ๐‘ก ๐‘ฅ)=๐‘‘๐‘ก/๐‘‘๐‘ฅ โˆ’๐‘๐‘œ๐‘ ๐‘’๐‘^2 ๐‘ฅโˆ’๐‘๐‘œ๐‘ ๐‘’๐‘ ๐‘ฅ ๐‘๐‘œ๐‘ก ๐‘ฅ=๐‘‘๐‘ก/๐‘‘๐‘ฅ โˆ’๐‘๐‘œ๐‘ ๐‘’๐‘ ๐‘ฅ(๐‘๐‘œ๐‘ ๐‘’๐‘ ๐‘ฅ+๐‘๐‘œ๐‘ก ๐‘ฅ)=๐‘‘๐‘ก/๐‘‘๐‘ฅ ๐‘‘๐‘ฅ=๐‘‘๐‘ก/(โˆ’๐‘๐‘œ๐‘ ๐‘’๐‘ ๐‘ฅ (๐‘๐‘œ๐‘ ๐‘’๐‘ ๐‘ฅ + ๐‘๐‘œ๐‘ก ๐‘ฅ) ) Therefore, โˆซ1โ–’(๐‘๐‘œ๐‘ ๐‘’๐‘ ๐‘ฅ(๐‘๐‘œ๐‘ ๐‘’๐‘ ๐‘ฅ + ๐‘๐‘œ๐‘ก ๐‘ฅ))/(๐‘๐‘œ๐‘ ๐‘’๐‘ ๐‘ฅ + ๐‘๐‘œ๐‘ก ๐‘ฅ) ๐‘‘๐‘ฅ =โˆซ1โ–’(๐‘๐‘œ๐‘ ๐‘’๐‘ ๐‘ฅ(๐‘๐‘œ๐‘ ๐‘’๐‘ ๐‘ฅ + ๐‘๐‘œ๐‘ก ๐‘ฅ))/๐‘ก. ๐‘‘๐‘ก/(โˆ’๐‘๐‘œ๐‘ ๐‘’๐‘ ๐‘ฅ(๐‘๐‘œ๐‘ ๐‘’๐‘ ๐‘ฅ + ๐‘๐‘œ๐‘ก ๐‘ฅ) ) =โˆ’โˆซ1โ–’๐‘‘๐‘ก/๐‘ก =โˆ’logโกใ€– |๐‘ก|ใ€— =โˆ’logโกใ€– |๐‘๐‘œ๐‘ ๐‘’๐‘ ๐‘ฅ+๐‘๐‘œ๐‘ก ๐‘ฅ|ใ€— Hence, F(๐‘ฅ)=โˆ’logโก|๐‘๐‘œ๐‘ ๐‘’๐‘ ๐‘ฅ+cotโก๐‘ฅ | Now, โˆซ_(๐œ‹/6)^(๐œ‹/4)โ–’ใ€–๐‘๐‘œ๐‘ ๐‘’๐‘ ๐‘ฅ=๐น(๐œ‹/4)โˆ’๐น(๐œ‹/6) ใ€— =โˆ’๐‘™๐‘œ๐‘”|๐‘๐‘œ๐‘ ๐‘’๐‘(๐œ‹/4)+๐‘๐‘œ๐‘ก(๐œ‹/4)|โˆ’ (โˆ’logโก|๐‘๐‘œ๐‘ ๐‘’๐‘(๐œ‹/6)+ ๐‘๐‘œ๐‘ก(๐œ‹/6)| ) =โˆ’๐‘™๐‘œ๐‘”|โˆš2+1|+๐‘™๐‘œ๐‘”|2+โˆš3| =โˆ’๐‘™๐‘œ๐‘”|2+โˆš3|โˆ’ ๐‘™๐‘œ๐‘”|โˆš2+1| =๐‘™๐‘œ๐‘”|(2 + โˆš3)/(โˆš2 + 1)| =๐‘™๐‘œ๐‘”|(2 + โˆš3)/(โˆš2 + 1)ร—(2 โˆ’ โˆš3)/(2 โˆ’ โˆš3)| =๐‘™๐‘œ๐‘”[((2)^2 โˆ’ (โˆš3)^2)/((โˆš2 + 1) ร— (2 โˆ’ โˆš3) )] =๐‘™๐‘œ๐‘”[(4 โˆ’ 3)/(โˆš2 + 1)(2 โˆ’ โˆš3) ] =๐‘™๐‘œ๐‘”[(1 ร— (โˆš2 โˆ’ 1))/((2 โˆ’ โˆš3)(โˆš2 + 1) ร—(โˆš2 โˆ’ 1) )] =๐‘™๐‘œ๐‘”[(โˆš2 โˆ’ 1)/(2 โˆ’ โˆš3)[(โˆš2)^2โˆ’ 1^2 ] ] =๐’๐’๐’ˆ[(โˆš๐Ÿ โˆ’ ๐Ÿ)/(๐Ÿ โˆ’ โˆš๐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 and Computer Science at Teachoo