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Ex 7.3, 3 - Integrate cos 2x cos 4x cos 6x - Chapter 7 Class 12

Ex 7.3, 3 - Chapter 7 Class 12 Integrals - Part 2
Ex 7.3, 3 - Chapter 7 Class 12 Integrals - Part 3 Ex 7.3, 3 - Chapter 7 Class 12 Integrals - Part 4 Ex 7.3, 3 - Chapter 7 Class 12 Integrals - Part 5

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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 has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.