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


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Example 40 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

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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.