Example 35 - Chapter 7 Class 12 Integrals
Last updated at Dec. 16, 2024 by Teachoo
Last updated at Dec. 16, 2024 by Teachoo
Example 35 Evaluate ∫1▒cos〖6𝑥 √(1+sin6𝑥 )〗 𝑑𝑥 ∫1▒cos〖6𝑥 √(1+sin6𝑥 )〗 𝑑𝑥 Put 𝑡 = √(1+sin6𝑥 ) 𝑡^2 = 1+sin6𝑥 Differentiate 𝑤.𝑟.𝑡.𝑥 (𝑑𝑡^2)/𝑑𝑥=𝑑/𝑑𝑥 (1+sin6𝑥 ) 2𝑡. 𝑑𝑡/𝑑𝑥=6 cos 6 𝑥 (2𝑡 𝑑𝑡)/(6 cos6𝑥 )=𝑑𝑥 Therefore, ∫1▒cos〖6𝑥 √(1+sin6𝑥 )〗 =∫1▒cos〖6𝑥 𝑡〗 . ( 2 𝑡 𝑑𝑡)/〖6 cos〗6𝑥 =∫1▒𝑡^2/3𝑑𝑡 =1/3 ∫1▒𝑡^2𝑑𝑡 =1/3 𝑡^(2 + 1)/(2 + 1) + 𝐶 =1/3 𝑡^3/3 + 𝐶 = 𝑡^3/9 + 𝐶 Putting back 𝑡 = √(1+𝑠𝑖𝑛6𝑥 ) = (√(1 + sin6𝑥 ))^3/9 + 𝐶 = (1 + sin6𝑥 )^(1/2 × 3)/9 + 𝐶 = 𝟏/𝟗 (𝟏 + 𝒔𝒊𝒏𝟔𝒙 )^(𝟑/𝟐)+𝑪
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About the Author
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