Example 7 (ii) - Chapter 7 Class 12 Integrals

Last updated at August 2, 2026 by

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

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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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