Example 7 - Chapter 7 Class 12 Integrals - Part 4

Example 7 - Chapter 7 Class 12 Integrals - Part 5
Example 7 - Chapter 7 Class 12 Integrals - Part 6

Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class


Transcript

Example 7 Find (ii) ∫1▒〖sin⁡〖2𝑥 〗 cos⁡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 2𝑥 & B by 3𝑥 sin⁡2𝑥 cos⁡3𝑥=1/2 [sin⁡(2𝑥+3𝑥)+sin⁡(2𝑥−3𝑥) ] sin⁡2𝑥 cos⁡3𝑥=1/2 [sin⁡(5𝑥)+sin⁡(−𝑥) ] sin⁡2𝑥 cos⁡3𝑥=1/2 [sin⁡5𝑥−sin⁡𝑥 ] ∫1▒(sin⁡〖2𝑥 〗 cos⁡3𝑥 ) 𝑑𝑥=1/2 ∫1▒(sin⁡5𝑥−sin⁡𝑥 ) 𝑑𝑥 =1/2 [∫1▒sin⁡5𝑥 𝑑𝑥−∫1▒sin⁡𝑥 𝑑𝑥] ∫1▒𝒔𝒊𝒏⁡𝟓𝒙 𝒅𝒙 Let 5𝑥=𝑡 5 = 𝑑𝑡/𝑑𝑥 𝑑𝑥=1/5 𝑑𝑡 =∫1▒sin⁡𝑡 . 1/5 𝑑𝑡 =1/5 ∫1▒sin⁡𝑡 . 𝑑𝑡 =1/5 (〖−cos〗⁡𝑡+𝐶1) =−1/5 cos⁡𝑡+1/5 𝐶1 Putting value of 𝑡 =−1/5 cos⁡5𝑥+1/5 𝐶1 ∫1▒sin⁡𝑥 𝑑𝑥 =−cos⁡𝑥+𝐶2 Thus, ∫1▒〖sin⁡〖2𝑥 〗 cos⁡3𝑥 〗=1/2 [∫1▒〖sin⁡5𝑥 𝑑𝑥〗−∫1▒sin⁡𝑥 𝑑𝑥] =1/2 [−1/5 cos⁡5𝑥+1/5 𝐶1−(−cos⁡𝑥 )+𝐶2] =1/2 [−1/5 cos⁡5𝑥+cos⁡𝑥+1/5 𝐶1−𝐶2] =1/2 [−1/5 cos⁡5𝑥+cos⁡𝑥+𝐶] =(−𝟏)/𝟏𝟎 𝐜𝐨𝐬⁡𝟓𝒙+𝟏/𝟐 𝐜𝐨𝐬⁡𝒙+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 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.