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


Transcript

Example 35 Evaluate ∫1β–’cos⁑〖6π‘₯ √(1+sin⁑6π‘₯ )γ€— 𝑑π‘₯ ∫1β–’cos⁑〖6π‘₯ √(1+sin⁑6π‘₯ )γ€— 𝑑π‘₯ Put 𝑑 = √(1+sin⁑6π‘₯ ) 𝑑^2 = 1+sin⁑6π‘₯ Differentiate 𝑀.π‘Ÿ.𝑑.π‘₯ (𝑑𝑑^2)/𝑑π‘₯=𝑑/𝑑π‘₯ (1+sin⁑6π‘₯ ) 2𝑑. 𝑑𝑑/𝑑π‘₯=6 cos 6 π‘₯ (2𝑑 𝑑𝑑)/(6 cos⁑6π‘₯ )=𝑑π‘₯ Therefore, ∫1β–’cos⁑〖6π‘₯ √(1+sin⁑6π‘₯ )γ€— =∫1β–’cos⁑〖6π‘₯ 𝑑〗 . ( 2 𝑑 𝑑𝑑)/γ€–6 cos〗⁑6π‘₯ =∫1▒𝑑^2/3⁑𝑑𝑑 =1/3 ∫1▒𝑑^2⁑𝑑𝑑 =1/3 𝑑^(2 + 1)/(2 + 1) + 𝐢 =1/3 𝑑^3/3 + 𝐢 = 𝑑^3/9 + 𝐢 Putting back 𝑑 = √(1+𝑠𝑖𝑛⁑6π‘₯ ) = (√(1 + sin⁑6π‘₯ ))^3/9 + 𝐢 = (1 + sin⁑6π‘₯ )^(1/2 Γ— 3)/9 + 𝐢 = 𝟏/πŸ— (𝟏 + π’”π’Šπ’β‘πŸ”π’™ )^(πŸ‘/𝟐)+π‘ͺ

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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 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.