Example 7 (ii) - Chapter 7 Class 12 Integrals
Last updated at Dec. 16, 2024 by Teachoo
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
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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo