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Ex 7.3
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Ex 7.3, 3 Important You are here
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Ex 7.3, 23 (MCQ)
Ex 7.3, 24 (MCQ) Important
Last updated at May 29, 2023 by Teachoo
Ex 7.3, 3 Integrate the function - cos 2x cos 4x cos 6x We know that 2 cos A cos B=[cos(𝐴+𝐵)+cos(𝐴−𝐵) ] Replace A by 2𝑥 & B by 4𝑥 2 cos 2𝑥 cos 4𝑥=cos(2𝑥+4𝑥)+cos(2𝑥−4𝑥) 2(cos 2𝑥 cos 4𝑥)=cos〖 (6𝑥)〗+cos(−2𝑥) 2 cos 2𝑥 cos 4𝑥=cos6𝑥+cos2𝑥 cos 2𝑥 cos 4𝑥=1/2 (cos6𝑥+cos2𝑥 ) (∵𝑐𝑜𝑠(−𝑥)=𝑐𝑜𝑠𝑥) Now, ∫1▒(cos2𝑥 cos4𝑥 cos6𝑥 ) 𝑑𝑥 =∫1▒(1/2 (cos6𝑥+cos2𝑥 ) cos6𝑥 ) 𝑑𝑥 =1/2 ∫1▒〖(cos6𝑥+cos2𝑥 ) cos6𝑥 〗 𝑑𝑥 =1/2 ∫1▒〖(cos6𝑥 )^2+(cos2𝑥.cos6𝑥 ) 〗 𝑑𝑥 =1/2 [∫1▒〖(cos6𝑥 )^2 𝑑𝑥+〗 ∫1▒〖cos2𝑥.cos6𝑥 〗 𝑑𝑥] Solving both these integrals separately ∫1▒(〖𝐜𝐨𝐬〗^𝟐𝟔𝒙 ) We know that cos 2𝜃=2 cos^2𝜃−1 cos^2 𝜃=(cos2𝜃 + 1)/2 Replace 𝜃 by 6𝑥 cos^2 6𝑥=cos〖2(6𝑥) + 1〗/2 cos^2 6𝑥=cos〖12𝑥 + 1〗/2 ∫1▒cos^2〖6𝑥 𝑑𝑥〗 =(∫1▒cos〖12𝑥 + 1〗 )/2 𝑑𝑥 =1/2 ∫1▒(cos〖12𝑥+1〗 ) 𝑑𝑥 ∫1▒𝐜𝐨𝐬𝟐𝒙 𝐜𝐨𝐬 𝟔𝒙 𝒅𝒙 We know that 2 cos A cos B=cos (𝐴+𝐵)+cos(𝐴−𝐵) cos A cos B=1/2 [cos (𝐴+𝐵)+cos(𝐴−𝐵) ] Replace A by 2𝑥 & B by 6𝑥 cos 2𝑥 cos 6𝑥 =1/2 [cos(2𝑥+6𝑥)+cos(2𝑥−6𝑥) ] =1/2 [cos8𝑥+cos(−4𝑥) ] =1/2 [cos8𝑥+cos4𝑥 ] 𝑑𝑥 ∫1▒〖𝑐𝑜𝑠2𝑥 cos6𝑥 〗 𝑑𝑥 =1/2 ∫1▒(cos8𝑥+cos4𝑥 ) 𝑑𝑥 Thus, our equation becomes ∫1▒(cos2𝑥 cos4𝑥 cos6𝑥 ) 𝑑𝑥 =1/2 [∫1▒cos^26𝑥 𝑑𝑥+∫1▒𝑐𝑜𝑠2𝑥 cos6𝑥 𝑑𝑥] =1/2 [1/2 ∫1▒(𝑐𝑜𝑠〖12𝑥+1〗 ) 𝑑𝑥+1/2 ∫1▒(𝑐𝑜𝑠8𝑥+cos4𝑥 ) 𝑑𝑥] =1/4 [∫1▒(𝑐𝑜𝑠〖12𝑥+1〗 ) 𝑑𝑥+∫1▒〖(𝑐𝑜𝑠8𝑥 〗+cos4𝑥)𝑑𝑥] =1/4 [∫1▒𝑐𝑜𝑠12𝑥 𝑑𝑥+∫1▒1 𝑑𝑥+∫1▒cos8𝑥 𝑑𝑥+∫1▒cos4𝑥 𝑑𝑥] =𝟏/𝟒 [𝐬𝐢𝐧𝟏𝟐𝒙/𝟏𝟐 +𝒙+𝒔𝒊𝒏𝟖𝒙/𝟖 + 𝒔𝒊𝒏𝟒𝒙/𝟒]+𝑪 ∫1▒cos(𝑎𝑥+𝑏) 𝑑𝑥=𝑠𝑖𝑛(𝑎𝑥 + 𝑏)/𝑎 =1/4 [∫1▒(𝑐𝑜𝑠〖12𝑥+1〗 ) 𝑑𝑥+∫1▒〖(𝑐𝑜𝑠8𝑥 〗+cos4𝑥)𝑑𝑥] =1/4 [∫1▒𝑐𝑜𝑠12𝑥 𝑑𝑥+∫1▒1 𝑑𝑥+∫1▒cos8𝑥 𝑑𝑥+∫1▒cos4𝑥 𝑑𝑥] =𝟏/𝟒 [𝐬𝐢𝐧𝟏𝟐𝒙/𝟏𝟐 +𝒙+𝒔𝒊𝒏𝟖𝒙/𝟖 + 𝒔𝒊𝒏𝟒𝒙/𝟒]+𝑪