Example 35 - Chapter 7 Class 12 Integrals

Transcript

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