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Question 1 Integrate the function (sin^(βˆ’1)⁑√π‘₯ βˆ’ cos^(βˆ’1)⁑√π‘₯)/(sin^(βˆ’1)⁑√π‘₯ + cos^(βˆ’1)⁑√π‘₯ ) , π‘₯∈[0, 1] Let 𝐼 = ∫1β–’(sin^(βˆ’1)⁑√π‘₯ βˆ’ cos^(βˆ’1)⁑√π‘₯)/(sin^(βˆ’1)⁑√π‘₯ + cos^(βˆ’1)⁑√π‘₯ ) 𝑑π‘₯ We can write as (sin^(βˆ’1)⁑√π‘₯ βˆ’ cos^(βˆ’1)⁑√π‘₯)/(sin^(βˆ’1)⁑√π‘₯ + cos^(βˆ’1)⁑√π‘₯ ) = (sin^(βˆ’1)⁑√π‘₯ βˆ’ (πœ‹/2 " βˆ’" γ€– 𝑠𝑖𝑛〗^(βˆ’1)⁑√π‘₯ ))/(πœ‹/2) We know that 〖𝑠𝑖𝑛〗^(βˆ’1)⁑π‘₯+γ€–π‘π‘œπ‘ γ€—^(βˆ’1)⁑π‘₯=πœ‹/2 or γ€–π‘π‘œπ‘ γ€—^(βˆ’1)⁑π‘₯=πœ‹/2 βˆ’γ€– 𝑠𝑖𝑛〗^(βˆ’1)⁑π‘₯ = (2/πœ‹)(sin^(βˆ’1)⁑√π‘₯ βˆ’πœ‹/2 " +" γ€– 𝑠𝑖𝑛〗^(βˆ’1)⁑√π‘₯ ) = 2/πœ‹ (2 sin^(βˆ’1)⁑√π‘₯ βˆ’πœ‹/2) = 2/πœ‹ Γ—2 sin^(βˆ’1)⁑√π‘₯βˆ’ 2/πœ‹Γ—πœ‹/2 = 4/πœ‹ sin^(βˆ’1)⁑√π‘₯βˆ’1 Integrating 𝑀.π‘Ÿ.𝑑.π‘₯ ∫1β–’(sin^(βˆ’1)⁑√π‘₯ βˆ’ cos^(βˆ’1)⁑√π‘₯)/(sin^(βˆ’1)⁑√π‘₯ + cos^(βˆ’1)⁑√π‘₯ ) 𝑑π‘₯=∫1β–’(4/πœ‹ sin^(βˆ’1)⁑√π‘₯βˆ’1) 𝑑π‘₯ = ∫1β–’γ€–4/πœ‹ sin^(βˆ’1)⁑√π‘₯ γ€— 𝑑π‘₯βˆ’βˆ«1▒𝑑π‘₯ = 4/πœ‹ ∫1β–’sin^(βˆ’1)⁑√π‘₯ 𝑑π‘₯βˆ’π‘₯+𝐢1 Let 𝐼1=∫1β–’sin^(βˆ’1)⁑√π‘₯ 𝑑π‘₯ Hence, I = 4/πœ‹ 𝐼1βˆ’π‘₯+𝐢1 Solving 𝐈_𝟏 𝐼1 = ∫1β–’sin^(βˆ’1)⁑√π‘₯ 𝑑π‘₯ Put √π‘₯=𝑑 π‘₯=𝑑^2 Differentiating 𝑀.π‘Ÿ.𝑑.π‘₯ 𝑑π‘₯/𝑑π‘₯ = (𝑑𝑑^2)/𝑑π‘₯ 1 = 2𝑑 𝑑𝑑/𝑑π‘₯ 𝑑π‘₯ = 2𝑑 𝑑𝑑 Therefore ∫1β–’sin^(βˆ’1)⁑√π‘₯ 𝑑π‘₯=∫1β–’sin^(βˆ’1)⁑𝑑 .2𝑑 𝑑𝑑 =2∫1β–’sin^(βˆ’1)⁑𝑑 .𝑑 𝑑𝑑 =2∫1▒〖𝑑 sin^(βˆ’1)⁑〖𝑑 γ€— γ€— 𝑑𝑑 =2[sin^(βˆ’1)⁑〖𝑑 γ€— ∫1▒𝑑 π‘‘π‘‘βˆ’βˆ«1β–’((𝑑/𝑑𝑑 sin^(βˆ’1)⁑𝑑 ) ∫1▒〖𝑑 𝑑𝑑〗) 𝑑𝑑 Now we know that ∫1▒〖𝑓(π‘₯) 𝑔⁑(π‘₯) γ€— 𝑑π‘₯=𝑓(π‘₯) ∫1▒𝑔(π‘₯) 𝑑π‘₯βˆ’βˆ«1β–’(𝑓^β€² (π‘₯) ∫1▒𝑔(π‘₯) 𝑑π‘₯) 𝑑π‘₯ Putting f(x) = t and g(x) = sin–1 t =2[sin^(βˆ’1)⁑〖𝑑 γ€— 𝑑^2/2 βˆ’βˆ«1β–’1/√(1 βˆ’γ€– 𝑑〗^2 ) ×𝑑^2/2 𝑑𝑑+𝐢] =2×𝑑^2/2 γ€– sin^(βˆ’1)γ€—β‘π‘‘βˆ’2Γ—βˆ«1β–’γ€–1/2 ×𝑑^2/√(1 βˆ’γ€– 𝑑〗^2 )γ€— 𝑑𝑑+𝐢 = 𝑑^2 sin^(βˆ’1)β‘π‘‘βˆ’βˆ«1▒𝑑^2/√(1 βˆ’γ€– 𝑑〗^2 ) 𝑑𝑑+𝐢 = 𝑑^2 sin^(βˆ’1)⁑𝑑+∫1β–’(βˆ’π‘‘^2)/√(1 βˆ’γ€– 𝑑〗^2 ) 𝑑𝑑+𝐢 Solving ∫1β–’γ€–βˆ’ 𝒕〗^𝟐/√(𝟏 βˆ’γ€– 𝒕〗^𝟐 ) 𝒅𝒕 We can write (βˆ’ 𝑑^2)/√(1 βˆ’γ€– 𝑑〗^2 ) =(γ€–βˆ’ 𝑑〗^2 + 1 βˆ’ 1)/√(1 βˆ’γ€– 𝑑〗^2 ) =(γ€–1 βˆ’ 𝑑〗^2 βˆ’ 1)/√(1 βˆ’γ€– 𝑑〗^2 ) =γ€–1 βˆ’ 𝑑〗^2/√(1 βˆ’γ€– 𝑑〗^2 ) βˆ’" " 1/√(1 βˆ’γ€– 𝑑〗^2 ) =√(1 βˆ’γ€– 𝑑〗^2 ) βˆ’" " 1/√(1 βˆ’γ€– 𝑑〗^2 ) Integrating 𝑀.π‘Ÿ.𝑑.π‘₯ ∫1β–’(βˆ’ 𝑑^2)/(1 βˆ’γ€– 𝑑〗^2 ) dt = ∫1β–’γ€–(√(1 βˆ’γ€– 𝑑〗^2 ) βˆ’" " 1/√(1 βˆ’γ€– 𝑑〗^2 )) γ€— 𝑑𝑑 = ∫1β–’βˆš(1^2 βˆ’γ€– 𝑑〗^2 ) π‘‘π‘‘βˆ’βˆ«1β–’1/√(1^2 βˆ’γ€– 𝑑〗^2 ) 𝑑𝑑 = 𝑑/2 √(1^2 βˆ’γ€– 𝑑〗^2 )+1^2/2 sin^(βˆ’1)⁑〖𝑑/1γ€—βˆ’sin^(βˆ’1)⁑〖𝑑/1γ€— We know that ∫1β–’βˆš(π‘Ž^2βˆ’π‘₯^2 )=π‘₯/2 √(π‘Ž^2βˆ’π‘₯^2 )+π‘Ž^2/2 sin^(βˆ’1)⁑〖π‘₯/π‘Žγ€—+𝐢 ∫1β–’1/√(π‘Ž^2 βˆ’ π‘₯^2 )=sin^(βˆ’1)⁑〖π‘₯/π‘Žγ€—+𝐢 = 𝑑/2 √(1 βˆ’γ€– 𝑑〗^2 )+1/2 sin^(βˆ’1)β‘π‘‘βˆ’sin^(βˆ’1)⁑𝑑 = 𝑑/2 √(1 βˆ’γ€– 𝑑〗^2 ) βˆ’ 1/2 sin^(βˆ’1)⁑𝑑 Hence we can write 𝐼1 = 𝑑^2 sin^(βˆ’1)⁑𝑑+∫1β–’(βˆ’ 𝑑^2)/√(1 βˆ’γ€– 𝑑〗^2 ) 𝑑𝑑 𝐼1 = 𝑑^2 sin⁑𝑑+𝑑/2 √(1 βˆ’γ€– 𝑑〗^2 )βˆ’1/2 sin^(βˆ’1)⁑𝑑 Putting 𝑑 = √π‘₯ 𝐼1 = (√π‘₯)^2 sin⁑√π‘₯+√π‘₯/2 √(1 βˆ’γ€– (√π‘₯)γ€—^2 )βˆ’1/2 sin^(βˆ’1)⁑√π‘₯ 𝐼1 = π‘₯ sin⁑√π‘₯+√π‘₯/2 √(1βˆ’π‘₯)βˆ’1/2 sin^(βˆ’1)⁑√π‘₯ Hence 𝐼 = 4/πœ‹ γ€– 𝐼〗_(1 )βˆ’π‘₯+C_1 𝐼 = 4/πœ‹ (π‘₯ 〖𝑠𝑖𝑛〗^(βˆ’1) √π‘₯+√π‘₯/2 √(1βˆ’π‘₯)βˆ’1/2 〖𝑠𝑖𝑛〗^(βˆ’1) √π‘₯)βˆ’π‘₯+C_1 𝐼 = 4/πœ‹ (π‘₯ 〖𝑠𝑖𝑛〗^(βˆ’1) √π‘₯+√(π‘₯ βˆ’ π‘₯^2 )/2 βˆ’1/2 〖𝑠𝑖𝑛〗^(βˆ’1) √π‘₯)βˆ’π‘₯+C_1 𝐼 = 4/πœ‹ π‘₯ 〖𝑠𝑖𝑛〗^(βˆ’1) √π‘₯+2/πœ‹ √(π‘₯ βˆ’ π‘₯^2 )βˆ’2/πœ‹ 〖𝑠𝑖𝑛〗^(βˆ’1) √π‘₯βˆ’π‘₯+C_1 𝐼 = 4/πœ‹ π‘₯ 〖𝑠𝑖𝑛〗^(βˆ’1) √π‘₯βˆ’2/πœ‹ 〖𝑠𝑖𝑛〗^(βˆ’1) √π‘₯+2/πœ‹ √(π‘₯ βˆ’ π‘₯^2 )βˆ’π‘₯+C_1 𝐼 = 〖𝑠𝑖𝑛〗^(βˆ’1) √π‘₯ [4π‘₯/πœ‹βˆ’2/πœ‹]+(2 √(π‘₯ βˆ’ π‘₯^2 ))/πœ‹βˆ’π‘₯+ C_1 𝐼 = 〖𝑠𝑖𝑛〗^(βˆ’1) √π‘₯ [(4π‘₯ βˆ’ 2)/πœ‹]+(2 √(π‘₯ βˆ’ π‘₯^2 ))/πœ‹βˆ’π‘₯+ C_1 𝑰 = γ€–π’”π’Šπ’γ€—^(βˆ’πŸ) βˆšπ’™ [(𝟐(πŸπ’™ βˆ’πŸ))/𝝅]+(𝟐 √(𝒙 βˆ’ 𝒙^𝟐 ))/π…βˆ’π’™+ π‘ͺ_𝟏

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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.