Example 35 - Chapter 7 Class 12 Integrals

Last updated at Dec. 16, 2024 by Teachoo

Example 35 - Evaluate integral cos 6x root 1 + sin 6x dx - Examples

part 2 - Example 35 - Examples - Serial order wise - Chapter 7 Class 12 Integrals

part 3 - Example 35 - Examples - Serial order wise - Chapter 7 Class 12 Integrals

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Example 35 Evaluate ∫1▒cos⁡〖6𝑥 √(1+sin⁡6𝑥 )〗 𝑑𝑥 ∫1▒cos⁡〖6𝑥 √(1+sin⁡6𝑥 )〗 𝑑𝑥 Put 𝑡 = √(1+sin⁡6𝑥 ) 𝑡^2 = 1+sin⁡6𝑥 Differentiate 𝑤.𝑟.𝑡.𝑥 (𝑑𝑡^2)/𝑑𝑥=𝑑/𝑑𝑥 (1+sin⁡6𝑥 ) 2𝑡. 𝑑𝑡/𝑑𝑥=6 cos 6 𝑥 (2𝑡 𝑑𝑡)/(6 cos⁡6𝑥 )=𝑑𝑥 Therefore, ∫1▒cos⁡〖6𝑥 √(1+sin⁡6𝑥 )〗 =∫1▒cos⁡〖6𝑥 𝑡〗 . ( 2 𝑡 𝑑𝑡)/〖6 cos〗⁡6𝑥 =∫1▒𝑡^2/3⁡𝑑𝑡 =1/3 ∫1▒𝑡^2⁡𝑑𝑡 =1/3 𝑡^(2 + 1)/(2 + 1) + 𝐶 =1/3 𝑡^3/3 + 𝐶 = 𝑡^3/9 + 𝐶 Putting back 𝑡 = √(1+𝑠𝑖𝑛⁡6𝑥 ) = (√(1 + sin⁡6𝑥 ))^3/9 + 𝐶 = (1 + sin⁡6𝑥 )^(1/2 × 3)/9 + 𝐶 = 𝟏/𝟗 (𝟏 + 𝒔𝒊𝒏⁡𝟔𝒙 )^(𝟑/𝟐)+𝑪

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Davneet Singh

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

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