



Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class
Ex 7.3
Ex 7.3, 2
Ex 7.3, 3 Important
Ex 7.3, 4 Important
Ex 7.3, 5
Ex 7.3, 6 Important You are here
Ex 7.3, 7
Ex 7.3, 8
Ex 7.3, 9 Important
Ex 7.3, 10 Important
Ex 7.3, 11
Ex 7.3, 12
Ex 7.3, 13 Important
Ex 7.3, 14
Ex 7.3, 15
Ex 7.3, 16 Important
Ex 7.3, 17
Ex 7.3, 18 Important
Ex 7.3, 19 Important
Ex 7.3, 20 Important
Ex 7.3, 21
Ex 7.3, 22 Important
Ex 7.3, 23 (MCQ)
Ex 7.3, 24 (MCQ) Important
Last updated at May 29, 2023 by Teachoo
Ex 7.3, 6 𝑠𝑖𝑛 𝑥 sin2𝑥 sin3𝑥 ∫1▒sin〖𝑥 sin〖2𝑥 sin3𝑥 〗 〗 𝑑𝑥 =∫1▒〖(sin𝑥 sin2𝑥 ) sin3𝑥 〗 𝑑𝑥 We know that 2 sin𝐴 sin𝐵=−cos(𝐴+𝐵)+cos(𝐴−𝐵) sin𝐴 sin𝐵=1/2 [−cos(𝐴+𝐵)+cos(𝐴−𝐵) ] sin𝐴 sin𝐵=1/2 [cos(𝐴−𝐵)−cos(𝐴+𝐵) ] Replace A by 𝑥 & B by 2𝑥 sin𝑥 sin2𝑥=1/2 [cos(𝑥−2𝑥)−cos(𝑥+2𝑥) ] sin𝑥 sin2𝑥 =1/2 [cos(−𝑥)−cos(3𝑥) ] sin𝑥 sin2𝑥 =1/2 [cos〖 𝑥〗−cos3𝑥 ] Thus, our equation becomes ∫1▒𝐬𝐢𝐧〖𝒙 𝐬𝐢𝐧𝟐𝒙 sin3𝑥 〗 𝑑𝑥 =∫1▒〖𝟏/𝟐 (𝒄𝒐𝒔𝒙−𝒄𝒐𝒔𝟑𝒙 ) 〗 . sin3𝑥.𝑑𝑥 =1/2 ∫1▒(cos𝑥−cos3𝑥 ) sin3𝑥 𝑑𝑥 =1/2 [∫1▒(cos𝑥. sin3𝑥−cos3𝑥. sin3𝑥 ) ]𝑑𝑥 =1/2 [∫1▒〖cos𝑥. sin3𝑥 〗 𝑑𝑥−∫1▒〖cos3𝑥. sin3𝑥 〗 𝑑𝑥] (∵𝑐𝑜𝑠(−𝑥)=𝑐𝑜𝑠𝑥) ∫1▒〖𝒄𝒐𝒔𝒙. 𝒔𝒊𝒏𝟑𝒙 〗 𝒅𝒙 We know that 2 𝑠𝑖𝑛𝐴 𝑐𝑜𝑠𝐵 =𝑠𝑖𝑛(𝐴+𝐵)+𝑠𝑖𝑛(𝐴−𝐵) 𝑠𝑖𝑛𝐴 𝑐𝑜𝑠𝐵=1/2 [𝑠𝑖𝑛(𝐴+𝐵)+𝑠𝑖𝑛(𝐴−𝐵) ] Replace A by 3𝑥 & B by 𝑥 sin3𝑥 cos𝑥 = 1/2 [𝑠𝑖𝑛(𝑥+3𝑥)+sin(3𝑥−𝑥) ] = 1/2 [𝑠𝑖𝑛4𝑥+sin2𝑥 ] ∫1▒〖𝒄𝒐𝒔𝟑𝒙. 𝒔𝒊𝒏𝟑𝒙 〗 𝒅𝒙 We know that 2 𝑠𝑖𝑛𝐴 𝑐𝑜𝑠𝐵 =𝑠𝑖𝑛(𝐴+𝐵)+𝑠𝑖𝑛(𝐴−𝐵) 𝑠𝑖𝑛𝐴 𝑐𝑜𝑠𝐵 =1/2 [𝑠𝑖𝑛(𝐴+𝐵)+𝑠𝑖𝑛(𝐴−𝐵) ] Replace A by 3𝑥 & B by 3𝑥 sin3𝑥 cos3𝑥 = 1/2 [𝑠𝑖𝑛(3𝑥+3𝑥)+sin(3𝑥−3𝑥) ] = 1/2 [𝑠𝑖𝑛6𝑥+sin0 ] =1/2 [𝑠𝑖𝑛6𝑥 ] Hence ∫1▒〖sin3𝑥.cos𝑥 〗 𝑑𝑥 =1/2 ∫1▒[𝑠𝑖𝑛4𝑥+sin2𝑥 ] 𝑑𝑥 Hence ∫1▒〖cos3𝑥.sin3𝑥 〗 𝑑𝑥 =1/2 ∫1▒sin6𝑥 𝑑𝑥 Thus, our equation becomes ∫1▒sin〖𝑥 sin〖2𝑥 sin3𝑥 〗 〗 𝑑𝑥 =1/2 [∫1▒〖sin3𝑥 cos3𝑥 〗 𝑑𝑥−∫1▒〖sin3𝑥 cos3𝑥 〗 𝑑𝑥] =1/2 [1/2 ∫1▒(sin4𝑥+sin2𝑥 ) 𝑑𝑥−1/2 ∫1▒(sin6𝑥 ) 𝑑𝑥] =1/4 [∫1▒(sin4𝑥+sin2𝑥 ) 𝑑𝑥−∫1▒(sin6𝑥 ) 𝑑𝑥] =1/4 [∫1▒sin4𝑥 𝑑𝑥+∫1▒sin2𝑥 𝑑𝑥−∫1▒sin6𝑥 𝑑𝑥] ∫1▒sin(𝑎𝑥+𝑏) 𝑑𝑥=−𝑐𝑜𝑠(𝑎𝑥 + 𝑏)/𝑎 +𝐶 =1/4 [(−cos4𝑥)/4 +(〖−cos〗2𝑥/2) −((−cos6𝑥)/6)]+𝐶 =1/4 [(−cos4𝑥)/4 −cos2𝑥/2+cos6𝑥/6]+𝐶 =𝟏/𝟒 [𝒄𝒐𝒔𝟔𝒙/𝟔 −𝒄𝒐𝒔𝟒𝒙/𝟒 − 𝒄𝒐𝒔𝟐𝒙/𝟐 ]+𝑪