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

Transcript

Ex 7.3, 3 Integrate the function - cos 2x cos 4x cos 6x We know that 2 cos A cos B=[cosโก(๐ด+๐ต)+cosโก(๐ดโˆ’๐ต) ] Replace A by 2๐‘ฅ & B by 4๐‘ฅ 2 cos 2๐‘ฅ cos 4๐‘ฅ=cosโก(2๐‘ฅ+4๐‘ฅ)+cosโก(2๐‘ฅโˆ’4๐‘ฅ) 2(cos 2๐‘ฅ cos 4๐‘ฅ)=cosโกใ€– (6๐‘ฅ)ใ€—+cosโก(โˆ’2๐‘ฅ) 2 cos 2๐‘ฅ cos 4๐‘ฅ=cosโก6๐‘ฅ+cosโก2๐‘ฅ cos 2๐‘ฅ cos 4๐‘ฅ=1/2 (cosโก6๐‘ฅ+cosโก2๐‘ฅ ) (โˆต๐‘๐‘œ๐‘ โก(โˆ’๐‘ฅ)=๐‘๐‘œ๐‘ โก๐‘ฅ) Now, โˆซ1โ–’(cosโก2๐‘ฅ cosโก4๐‘ฅ cosโก6๐‘ฅ ) ๐‘‘๐‘ฅ =โˆซ1โ–’(1/2 (cosโก6๐‘ฅ+cosโก2๐‘ฅ ) cosโก6๐‘ฅ ) ๐‘‘๐‘ฅ =1/2 โˆซ1โ–’ใ€–(cosโก6๐‘ฅ+cosโก2๐‘ฅ ) cosโก6๐‘ฅ ใ€— ๐‘‘๐‘ฅ =1/2 โˆซ1โ–’ใ€–(cosโก6๐‘ฅ )^2+(cosโก2๐‘ฅ.cosโก6๐‘ฅ ) ใ€— ๐‘‘๐‘ฅ =1/2 [โˆซ1โ–’ใ€–(cosโก6๐‘ฅ )^2 ๐‘‘๐‘ฅ+ใ€— โˆซ1โ–’ใ€–cosโก2๐‘ฅ.cosโก6๐‘ฅ ใ€— ๐‘‘๐‘ฅ] Solving both these integrals separately โˆซ1โ–’(ใ€–๐œ๐จ๐ฌใ€—^๐Ÿโก๐Ÿ”๐’™ ) We know that cos 2๐œƒ=2 cos^2โก๐œƒโˆ’1 cos^2 ๐œƒ=(cosโก2๐œƒ + 1)/2 Replace ๐œƒ by 6๐‘ฅ cos^2 6๐‘ฅ=cosโกใ€–2(6๐‘ฅ) + 1ใ€—/2 cos^2 6๐‘ฅ=cosโกใ€–12๐‘ฅ + 1ใ€—/2 โˆซ1โ–’cos^2โกใ€–6๐‘ฅ ๐‘‘๐‘ฅใ€— =(โˆซ1โ–’cosโกใ€–12๐‘ฅ + 1ใ€— )/2 ๐‘‘๐‘ฅ =1/2 โˆซ1โ–’(cosโกใ€–12๐‘ฅ+1ใ€— ) ๐‘‘๐‘ฅ โˆซ1โ–’๐œ๐จ๐ฌโก๐Ÿ๐’™ ๐œ๐จ๐ฌ ๐Ÿ”๐’™ ๐’…๐’™ We know that 2 cos A cos B=cos (๐ด+๐ต)+cosโก(๐ดโˆ’๐ต) cos A cos B=1/2 [cos (๐ด+๐ต)+cosโก(๐ดโˆ’๐ต) ] Replace A by 2๐‘ฅ & B by 6๐‘ฅ cos 2๐‘ฅ cos 6๐‘ฅ =1/2 [cosโก(2๐‘ฅ+6๐‘ฅ)+cosโก(2๐‘ฅโˆ’6๐‘ฅ) ] =1/2 [cosโก8๐‘ฅ+cosโก(โˆ’4๐‘ฅ) ] =1/2 [cosโก8๐‘ฅ+cosโก4๐‘ฅ ] ๐‘‘๐‘ฅ โˆซ1โ–’ใ€–๐‘๐‘œ๐‘ โก2๐‘ฅ cosโก6๐‘ฅ ใ€— ๐‘‘๐‘ฅ =1/2 โˆซ1โ–’(cosโก8๐‘ฅ+cosโก4๐‘ฅ ) ๐‘‘๐‘ฅ Thus, our equation becomes โˆซ1โ–’(cosโก2๐‘ฅ cosโก4๐‘ฅ cosโก6๐‘ฅ ) ๐‘‘๐‘ฅ =1/2 [โˆซ1โ–’cos^2โก6๐‘ฅ ๐‘‘๐‘ฅ+โˆซ1โ–’๐‘๐‘œ๐‘ โก2๐‘ฅ cosโก6๐‘ฅ ๐‘‘๐‘ฅ] =1/2 [1/2 โˆซ1โ–’(๐‘๐‘œ๐‘ โกใ€–12๐‘ฅ+1ใ€— ) ๐‘‘๐‘ฅ+1/2 โˆซ1โ–’(๐‘๐‘œ๐‘ โก8๐‘ฅ+cosโก4๐‘ฅ ) ๐‘‘๐‘ฅ] =1/4 [โˆซ1โ–’(๐‘๐‘œ๐‘ โกใ€–12๐‘ฅ+1ใ€— ) ๐‘‘๐‘ฅ+โˆซ1โ–’ใ€–(๐‘๐‘œ๐‘ โก8๐‘ฅ ใ€—+cosโก4๐‘ฅ)๐‘‘๐‘ฅ] =1/4 [โˆซ1โ–’๐‘๐‘œ๐‘ โก12๐‘ฅ ๐‘‘๐‘ฅ+โˆซ1โ–’1 ๐‘‘๐‘ฅ+โˆซ1โ–’cosโก8๐‘ฅ ๐‘‘๐‘ฅ+โˆซ1โ–’cosโก4๐‘ฅ ๐‘‘๐‘ฅ] =๐Ÿ/๐Ÿ’ [๐ฌ๐ข๐งโก๐Ÿ๐Ÿ๐’™/๐Ÿ๐Ÿ +๐’™+๐’”๐’Š๐’โก๐Ÿ–๐’™/๐Ÿ– + ๐’”๐’Š๐’โก๐Ÿ’๐’™/๐Ÿ’]+๐‘ช โˆซ1โ–’cosโก(๐‘Ž๐‘ฅ+๐‘) ๐‘‘๐‘ฅ=๐‘ ๐‘–๐‘›โก(๐‘Ž๐‘ฅ + ๐‘)/๐‘Ž =1/4 [โˆซ1โ–’(๐‘๐‘œ๐‘ โกใ€–12๐‘ฅ+1ใ€— ) ๐‘‘๐‘ฅ+โˆซ1โ–’ใ€–(๐‘๐‘œ๐‘ โก8๐‘ฅ ใ€—+cosโก4๐‘ฅ)๐‘‘๐‘ฅ] =1/4 [โˆซ1โ–’๐‘๐‘œ๐‘ โก12๐‘ฅ ๐‘‘๐‘ฅ+โˆซ1โ–’1 ๐‘‘๐‘ฅ+โˆซ1โ–’cosโก8๐‘ฅ ๐‘‘๐‘ฅ+โˆซ1โ–’cosโก4๐‘ฅ ๐‘‘๐‘ฅ] =๐Ÿ/๐Ÿ’ [๐ฌ๐ข๐งโก๐Ÿ๐Ÿ๐’™/๐Ÿ๐Ÿ +๐’™+๐’”๐’Š๐’โก๐Ÿ–๐’™/๐Ÿ– + ๐’”๐’Š๐’โก๐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.