Check sibling questions

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

Example 42 - Chapter 7 Class 12 Integrals - Part 2

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


Transcript

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 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

Ask a doubt
Davneet Singh's photo - Co-founder, Teachoo

Made by

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.