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Example 42 - Integral sin 2x cos 2x / root (9 - cos4 2x)

Example 42 - Chapter 7 Class 12 Integrals - Part 2


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 has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.