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  1. Chapter 7 Class 12 Integrals
  2. Serial order wise

Transcript

Example 42 Evaluate ∫1β–’(sin⁑〖2π‘₯ cos⁑2π‘₯ γ€— 𝑑π‘₯)/(√(9 βˆ’cos^4⁑(2π‘₯) ) ) 𝑑π‘₯ ∫1β–’(sin⁑〖2π‘₯ cos⁑2π‘₯ γ€— 𝑑π‘₯)/(√(9 βˆ’ cos^4⁑2π‘₯ ) ) 𝑑π‘₯ Let cos^2⁑2π‘₯=𝑑 Differentiating both sides w.r.t.π‘₯ –2.2 cos⁑2π‘₯ sin⁑2π‘₯=𝑑𝑑/𝑑π‘₯ 𝑑π‘₯=𝑑𝑑/(–4 cos⁑2π‘₯ sin⁑2π‘₯ ) Hence, our equation becomes ∫1β–’(sin⁑〖2π‘₯ cos⁑2π‘₯ γ€— )/(√(9 βˆ’ cos^4⁑2π‘₯ ) ) 𝑑π‘₯ =∫1β–’γ€–sin⁑〖2π‘₯ cos⁑2π‘₯ γ€—/√(9 βˆ’ 𝑑^2 ) 𝑑π‘₯γ€— =∫1β–’γ€–(sin⁑2π‘₯ cos⁑2π‘₯)/√(9 βˆ’ 𝑑^2 ) Γ— 𝑑𝑑/(–4 cos⁑2π‘₯ sin⁑2π‘₯ )γ€— =1/(βˆ’4) ∫1▒𝑑𝑑/√((3)^2 βˆ’ (𝑑)^2 ) =(βˆ’1)/( 4) [sin^(βˆ’1)⁑〖𝑑/3+𝐢1γ€— ] =(βˆ’1)/( 4) 𝑠𝑖𝑛^(βˆ’1) 𝑑/3βˆ’πΆ1/4 =(βˆ’πŸ)/( πŸ’) π’”π’Šπ’^(βˆ’πŸ) [𝟏/πŸ‘ 𝒄𝒐𝒔^𝟐 πŸπ’™]+π‘ͺ It is of form ∫1▒〖𝑑π‘₯/√(π‘Ž^2 βˆ’ π‘₯^2 )=𝑠𝑖𝑛^(βˆ’1) π‘₯/π‘Ž+𝐢〗 Replacing π‘₯ by 𝑑 and π‘Ž by 3, we get

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.