Ex 7.3, 6 - Integrate sin x sin 2x sin 3x - CBSE - Integration using trigo identities - CD and CD inv formulae

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  1. Class 12
  2. Important Question for exams Class 12

Transcript

Ex 7.3, 6 ๐‘ ๐‘–๐‘› ๐‘ฅ sinโก2๐‘ฅ sinโก3๐‘ฅ โˆซ1โ–’sinโกใ€–๐‘ฅ sinโกใ€–2๐‘ฅ sinโก3๐‘ฅ ใ€— ใ€— ๐‘‘๐‘ฅ =โˆซ1โ–’ใ€–(sinโก๐‘ฅ sinโก2๐‘ฅ ) sinโก3๐‘ฅ ใ€— ๐‘‘๐‘ฅ We know that 2 sinโก๐ด sinโก๐ต=โˆ’cosโก(๐ด+๐ต)+cosโก(๐ดโˆ’๐ต) sinโก๐ด sinโก๐ต=1/2 [โˆ’cosโก(๐ด+๐ต)+cosโก(๐ดโˆ’๐ต) ] sinโก๐ด sinโก๐ต=1/2 [cosโก(๐ดโˆ’๐ต)โˆ’cosโก(๐ด+๐ต) ] Replace A by ๐‘ฅ & B by 2๐‘ฅ sinโก๐‘ฅ sinโก3๐‘ฅ=1/2 [cosโก(๐‘ฅโˆ’2๐‘ฅ)โˆ’cosโก(๐‘ฅ+2๐‘ฅ) ] sinโก๐‘ฅ sinโก3๐‘ฅ =1/2 [cosโก(โˆ’๐‘ฅ)โˆ’cosโก(3๐‘ฅ) ] sinโก๐‘ฅ sinโก3๐‘ฅ =1/2 [cosโก(๐‘ฅ)โˆ’cosโก3๐‘ฅ ] Thus, โˆซ1โ–’sinโกใ€–๐‘ฅ sinโกใ€–2๐‘ฅ sinโก3๐‘ฅ ใ€— ใ€— ๐‘‘๐‘ฅ =โˆซ1โ–’ใ€–1/2 [(cosโก๐‘ฅโˆ’cosโก3๐‘ฅ )] ใ€— . sinโก3๐‘ฅ.๐‘‘๐‘ฅ =1/2 โˆซ1โ–’(cosโก๐‘ฅโˆ’cosโก3๐‘ฅ ) sinโก3๐‘ฅ ๐‘‘๐‘ฅ =1/2 [โˆซ1โ–’(cosโก๐‘ฅ. sinโก3๐‘ฅโˆ’cosโก3๐‘ฅ. sinโก3๐‘ฅ ) ]๐‘‘๐‘ฅ =1/2 [โˆซ1โ–’ใ€–cosโก๐‘ฅ. sinโก3๐‘ฅ ใ€— ๐‘‘๐‘ฅโˆ’โˆซ1โ–’ใ€–cosโก3๐‘ฅ. sinโก3๐‘ฅ ใ€— ๐‘‘๐‘ฅ] โˆซ1โ–’ใ€–๐’„๐’๐’”โก๐’™. ๐’”๐’Š๐’โก๐Ÿ‘๐’™ ใ€— ๐’…๐’™ We know that 2 ๐‘ ๐‘–๐‘›โก๐ด ๐‘๐‘œ๐‘ โก๐ต =๐‘ ๐‘–๐‘›โก(๐ด+๐ต)+๐‘ ๐‘–๐‘›โก(๐ดโˆ’๐ต) ๐‘ ๐‘–๐‘›โก๐ด ๐‘๐‘œ๐‘ โก๐ต =1/2 [๐‘ ๐‘–๐‘›โก(๐ด+๐ต)+๐‘ ๐‘–๐‘›โก(๐ดโˆ’๐ต) ] Replace A by 3๐‘ฅ & B by ๐‘ฅ sinโก3๐‘ฅ cosโก๐‘ฅ = 1/2 [๐‘ ๐‘–๐‘›โก(๐‘ฅ+3๐‘ฅ)+sinโก(3๐‘ฅโˆ’๐‘ฅ) ] = 1/2 [๐‘ ๐‘–๐‘›โก4๐‘ฅ+sinโก2๐‘ฅ ] โˆซ1โ–’ใ€–๐’„๐’๐’”โก๐Ÿ‘๐’™. ๐’”๐’Š๐’โก๐Ÿ‘๐’™ ใ€— ๐’…๐’™ We know that 2 ๐‘ ๐‘–๐‘›โก๐ด ๐‘๐‘œ๐‘ โก๐ต =๐‘ ๐‘–๐‘›โก(๐ด+๐ต)+๐‘ ๐‘–๐‘›โก(๐ดโˆ’๐ต) ๐‘ ๐‘–๐‘›โก๐ด ๐‘๐‘œ๐‘ โก๐ต =1/2 [๐‘ ๐‘–๐‘›โก(๐ด+๐ต)+๐‘ ๐‘–๐‘›โก(๐ดโˆ’๐ต) ] Replace A by 3๐‘ฅ & B by 3๐‘ฅ sinโก3๐‘ฅ cosโก3๐‘ฅ = 1/2 [๐‘ ๐‘–๐‘›โก(3๐‘ฅ+3๐‘ฅ)+sinโก(3๐‘ฅโˆ’3๐‘ฅ) ] = 1/2 [๐‘ ๐‘–๐‘›โก6๐‘ฅ+sinโก0 ]=1/2 [๐‘ ๐‘–๐‘›โก6๐‘ฅ ] Hence โˆซ1โ–’ใ€–sinโก3๐‘ฅ.cosโก๐‘ฅ ใ€— ๐‘‘๐‘ฅ =1/2 โˆซ1โ–’[๐‘ ๐‘–๐‘›โก4๐‘ฅ+sinโก2๐‘ฅ ] ๐‘‘๐‘ฅ Hence โˆซ1โ–’ใ€–cosโก3๐‘ฅ.sinโก3๐‘ฅ ใ€— ๐‘‘๐‘ฅ =1/2 โˆซ1โ–’sinโก6๐‘ฅ ๐‘‘๐‘ฅ Thus, our equation becomes โˆซ1โ–’sinโกใ€–๐‘ฅ sinโกใ€–2๐‘ฅ sinโก3๐‘ฅ ใ€— ใ€— ๐‘‘๐‘ฅ =1/2 [โˆซ1โ–’ใ€–sinโก3๐‘ฅ cosโก3๐‘ฅ ใ€— ๐‘‘๐‘ฅโˆ’โˆซ1โ–’ใ€–sinโก3๐‘ฅ cosโก3๐‘ฅ ใ€— ๐‘‘๐‘ฅ] =1/2 [1/2 โˆซ1โ–’(sinโก4๐‘ฅ+sinโก2๐‘ฅ ) ๐‘‘๐‘ฅโˆ’1/2 โˆซ1โ–’(sinโก6๐‘ฅ ) ๐‘‘๐‘ฅ] =1/4 [โˆซ1โ–’(sinโก4๐‘ฅ+sinโก2๐‘ฅ ) ๐‘‘๐‘ฅโˆ’โˆซ1โ–’(sinโก6๐‘ฅ ) ๐‘‘๐‘ฅ] =1/4 [โˆซ1โ–’sinโก4๐‘ฅ ๐‘‘๐‘ฅ+โˆซ1โ–’sinโก2๐‘ฅ ๐‘‘๐‘ฅโˆ’โˆซ1โ–’sinโก6๐‘ฅ ๐‘‘๐‘ฅ] โˆซ1โ–’sinโก(๐‘Ž๐‘ฅ+๐‘) ๐‘‘๐‘ฅ=โˆ’๐‘๐‘œ๐‘ โก(๐‘Ž๐‘ฅ + ๐‘)/๐‘Ž +๐ถ =1/4 [(โˆ’cosโก4๐‘ฅ)/4 +(ใ€–โˆ’cosใ€—โก2๐‘ฅ/2) โˆ’((โˆ’cosโก6๐‘ฅ)/6)]+๐ถ =1/4 [(โˆ’cosโก4๐‘ฅ)/4 โˆ’cosโก2๐‘ฅ/2+cosโก6๐‘ฅ/6]+๐ถ =๐Ÿ/๐Ÿ’ [๐’„๐’๐’”โก๐Ÿ”๐’™/๐Ÿ” โˆ’๐’„๐’๐’”โก๐Ÿ’๐’™/๐Ÿ’ โˆ’ ๐’„๐’๐’”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.