

Examples
Last updated at Dec. 16, 2024 by Teachoo
Example 7 Find (ii) ∫1▒〖sin〖2𝑥 〗 cos3𝑥 〗 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 2𝑥 & B by 3𝑥 sin2𝑥 cos3𝑥=1/2 [sin(2𝑥+3𝑥)+sin(2𝑥−3𝑥) ] sin2𝑥 cos3𝑥=1/2 [sin(5𝑥)+sin(−𝑥) ] sin2𝑥 cos3𝑥=1/2 [sin5𝑥−sin𝑥 ] ∫1▒(sin〖2𝑥 〗 cos3𝑥 ) 𝑑𝑥=1/2 ∫1▒(sin5𝑥−sin𝑥 ) 𝑑𝑥 =1/2 [∫1▒sin5𝑥 𝑑𝑥−∫1▒sin𝑥 𝑑𝑥] ∫1▒𝒔𝒊𝒏𝟓𝒙 𝒅𝒙 Let 5𝑥=𝑡 5 = 𝑑𝑡/𝑑𝑥 𝑑𝑥=1/5 𝑑𝑡 =∫1▒sin𝑡 . 1/5 𝑑𝑡 =1/5 ∫1▒sin𝑡 . 𝑑𝑡 =1/5 (〖−cos〗𝑡+𝐶1) =−1/5 cos𝑡+1/5 𝐶1 Putting value of 𝑡 =−1/5 cos5𝑥+1/5 𝐶1 ∫1▒sin𝑥 𝑑𝑥 =−cos𝑥+𝐶2 Thus, ∫1▒〖sin〖2𝑥 〗 cos3𝑥 〗=1/2 [∫1▒〖sin5𝑥 𝑑𝑥〗−∫1▒sin𝑥 𝑑𝑥] =1/2 [−1/5 cos5𝑥+1/5 𝐶1−(−cos𝑥 )+𝐶2] =1/2 [−1/5 cos5𝑥+cos𝑥+1/5 𝐶1−𝐶2] =1/2 [−1/5 cos5𝑥+cos𝑥+𝐶] =(−𝟏)/𝟏𝟎 𝐜𝐨𝐬𝟓𝒙+𝟏/𝟐 𝐜𝐨𝐬𝒙+𝑪