Find Integration of sin x sin 2x sin 3x - Ex 7.3, 6

Ex 7.3, 6 𝑠𝑖𝑛 𝑥 sin⁡2𝑥 sin⁡3𝑥 ∫1▒sin⁡〖𝑥 sin⁡〖2𝑥 sin⁡3𝑥 〗〗 𝑑𝑥 =∫1▒〖(sin⁡𝑥 sin⁡2𝑥 ) sin⁡3𝑥 〗 𝑑𝑥 We know that 2 sin⁡𝐴 sin⁡𝐵=−cos⁡(𝐴+𝐵)+cos⁡(𝐴−𝐵) sin⁡𝐴 sin⁡𝐵=1/2 [−cos⁡(𝐴+𝐵)+cos⁡(𝐴−𝐵) ] sin⁡𝐴 sin⁡𝐵=1/2 [cos⁡(𝐴−𝐵)−cos⁡(𝐴+𝐵) ] Replace A by 𝑥 & B by 2𝑥 sin⁡𝑥 sin⁡2𝑥=1/2 [cos⁡(𝑥−2𝑥)−cos⁡(𝑥+2𝑥) ] sin⁡𝑥 sin⁡2𝑥 =1/2 [cos⁡(−𝑥)−cos⁡(3𝑥) ] sin⁡𝑥 sin⁡2𝑥 =1/2 [cos⁡〖 𝑥〗−cos⁡3𝑥 ] Thus, our equation becomes ∫1▒𝐬𝐢𝐧⁡〖𝒙 𝐬𝐢𝐧⁡𝟐𝒙 sin⁡3𝑥 〗 𝑑𝑥 =∫1▒〖𝟏/𝟐 (𝒄𝒐𝒔⁡𝒙−𝒄𝒐𝒔⁡𝟑𝒙 ) 〗 . sin⁡3𝑥.𝑑𝑥 =1/2 ∫1▒(cos⁡𝑥−cos⁡3𝑥 ) sin⁡3𝑥 𝑑𝑥 =1/2 [∫1▒(cos⁡𝑥. sin⁡3𝑥−cos⁡3𝑥. sin⁡3𝑥 ) ]𝑑𝑥 =1/2 [∫1▒〖cos⁡𝑥. sin⁡3𝑥 〗 𝑑𝑥−∫1▒〖cos⁡3𝑥. sin⁡3𝑥 〗 𝑑𝑥] (∵𝑐𝑜𝑠⁡(−𝑥)=𝑐𝑜𝑠⁡𝑥) ∫1▒〖𝒄𝒐𝒔⁡𝒙. 𝒔𝒊𝒏⁡𝟑𝒙 〗 𝒅𝒙 We know that 2 𝑠𝑖𝑛⁡𝐴 𝑐𝑜𝑠⁡𝐵 =𝑠𝑖𝑛⁡(𝐴+𝐵)+𝑠𝑖𝑛⁡(𝐴−𝐵) 𝑠𝑖𝑛⁡𝐴 𝑐𝑜𝑠⁡𝐵=1/2 [𝑠𝑖𝑛⁡(𝐴+𝐵)+𝑠𝑖𝑛⁡(𝐴−𝐵) ] Replace A by 3𝑥 & B by 𝑥 sin⁡3𝑥 cos⁡𝑥 = 1/2 [𝑠𝑖𝑛⁡(𝑥+3𝑥)+sin⁡(3𝑥−𝑥) ] = 1/2 [𝑠𝑖𝑛⁡4𝑥+sin⁡2𝑥 ] ∫1▒〖𝒄𝒐𝒔⁡𝟑𝒙. 𝒔𝒊𝒏⁡𝟑𝒙 〗 𝒅𝒙 We know that 2 𝑠𝑖𝑛⁡𝐴 𝑐𝑜𝑠⁡𝐵 =𝑠𝑖𝑛⁡(𝐴+𝐵)+𝑠𝑖𝑛⁡(𝐴−𝐵) 𝑠𝑖𝑛⁡𝐴 𝑐𝑜𝑠⁡𝐵 =1/2 [𝑠𝑖𝑛⁡(𝐴+𝐵)+𝑠𝑖𝑛⁡(𝐴−𝐵) ] Replace A by 3𝑥 & B by 3𝑥 sin⁡3𝑥 cos⁡3𝑥 = 1/2 [𝑠𝑖𝑛⁡(3𝑥+3𝑥)+sin⁡(3𝑥−3𝑥) ] = 1/2 [𝑠𝑖𝑛⁡6𝑥+sin⁡0 ] =1/2 [𝑠𝑖𝑛⁡6𝑥 ] Hence ∫1▒〖sin⁡3𝑥.cos⁡𝑥 〗 𝑑𝑥 =1/2 ∫1▒[𝑠𝑖𝑛⁡4𝑥+sin⁡2𝑥 ] 𝑑𝑥 Hence ∫1▒〖cos⁡3𝑥.sin⁡3𝑥 〗 𝑑𝑥 =1/2 ∫1▒sin⁡6𝑥 𝑑𝑥 Thus, our equation becomes ∫1▒sin⁡〖𝑥 sin⁡〖2𝑥 sin⁡3𝑥 〗〗 𝑑𝑥 =1/2 [∫1▒〖sin⁡3𝑥 cos⁡3𝑥 〗 𝑑𝑥−∫1▒〖sin⁡3𝑥 cos⁡3𝑥 〗 𝑑𝑥] =1/2 [1/2 ∫1▒(sin⁡4𝑥+sin⁡2𝑥 ) 𝑑𝑥−1/2 ∫1▒(sin⁡6𝑥 ) 𝑑𝑥] =1/4 [∫1▒(sin⁡4𝑥+sin⁡2𝑥 ) 𝑑𝑥−∫1▒(sin⁡6𝑥 ) 𝑑𝑥] =1/4 [∫1▒sin⁡4𝑥 𝑑𝑥+∫1▒sin⁡2𝑥 𝑑𝑥−∫1▒sin⁡6𝑥 𝑑𝑥] ∫1▒sin⁡(𝑎𝑥+𝑏) 𝑑𝑥=−𝑐𝑜𝑠⁡(𝑎𝑥 + 𝑏)/𝑎 +𝐶 =1/4 [(−cos⁡4𝑥)/4 +(〖−cos〗⁡2𝑥/2) −((−cos⁡6𝑥)/6)]+𝐶 =1/4 [(−cos⁡4𝑥)/4 −cos⁡2𝑥/2+cos⁡6𝑥/6]+𝐶 =𝟏/𝟒 [𝒄𝒐𝒔⁡𝟔𝒙/𝟔 −𝒄𝒐𝒔⁡𝟒𝒙/𝟒 − 𝒄𝒐𝒔⁡𝟐𝒙/𝟐 ]+𝑪

Ex 7.3, 6 - Chapter 7 Class 12 Integrals (Important Question)

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