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Misc 19 - Integrate sin-1 root x - cos-1 root x - CBSE - Miscellaneous

Misc 19 - Chapter 7 Class 12 Integrals - Part 2
Misc 19 - Chapter 7 Class 12 Integrals - Part 3
Misc 19 - Chapter 7 Class 12 Integrals - Part 4
Misc 19 - Chapter 7 Class 12 Integrals - Part 5
Misc 19 - Chapter 7 Class 12 Integrals - Part 6
Misc 19 - Chapter 7 Class 12 Integrals - Part 7
Misc 19 - Chapter 7 Class 12 Integrals - Part 8
Misc 19 - Chapter 7 Class 12 Integrals - Part 9

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Misc 19 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 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.