Misc 22 - Chapter 7 Class 12 Integrals

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Misc 22 Integrate the function tan^(βˆ’1)⁑√((1 βˆ’ π‘₯)/(1 + π‘₯)) Let x = cos 2πœƒ 𝑑π‘₯/π‘‘πœƒ=βˆ’2 sin⁑〖2πœƒ γ€— dx = βˆ’2 sin 2πœƒ dπœƒ Substituting, ∫1β–’γ€–tan^(βˆ’1)⁑√((1 βˆ’ π‘₯)/(1 + π‘₯)) 𝑑π‘₯γ€— = ∫1β–’γ€–π‘‘π‘Žπ‘›γ€—^(βˆ’1) √((1 βˆ’ cos⁑2πœƒ)/(1 + cos⁑2πœƒ ))Γ—(βˆ’2 sin⁑〖2 πœƒ)γ€— 𝑑 πœƒ = βˆ’2∫1β–’γ€–π‘‘π‘Žπ‘›γ€—^(βˆ’1) √((1 βˆ’ (1 βˆ’ 2〖𝑠𝑖𝑛〗^2 πœƒ))/(1 + (2γ€–π‘π‘œπ‘ γ€—^2 πœƒ βˆ’ 1) ))Γ—("sin 2πœƒ dπœƒ " ) = βˆ’2∫1β–’γ€–π‘‘π‘Žπ‘›γ€—^(βˆ’1) √((sin^2β‘πœƒ/cos^2β‘πœƒ ) )Γ—sin⁑〖2πœƒ π‘‘πœƒγ€— = βˆ’2∫1β–’γ€–π‘‘π‘Žπ‘›γ€—^(βˆ’1) (sinβ‘πœƒ/cosβ‘πœƒ )Γ—sin⁑〖2πœƒ π‘‘πœƒγ€— = βˆ’2∫1β–’γ€–π‘‘π‘Žπ‘›γ€—^(βˆ’1) (π‘‘π‘Žπ‘›β‘πœƒ )Γ—sin⁑〖2πœƒ π‘‘πœƒγ€— = βˆ’ 2 ∫1β–’πœƒ sin⁑〖2πœƒ π‘‘πœƒγ€— =βˆ’2(πœƒβˆ«1β–’γ€–sin⁑2πœƒ π‘‘πœƒβˆ’(∫1▒𝑑(πœƒ)/π‘‘πœƒ ∫1β–’sin⁑2πœƒ π‘‘πœƒ) π‘‘πœƒγ€—) =βˆ’2(πœƒ((βˆ’cos⁑2πœƒ)/2)βˆ’βˆ«1β–’1((βˆ’cos⁑2πœƒ)/2) π‘‘πœƒ) =βˆ’2(βˆ’πœƒ(cos⁑2πœƒ/2)•+∫1β–’cos⁑2πœƒ/2 π‘‘πœƒ) =βˆ’2(βˆ’(πœƒ cos⁑2πœƒ)/2•+sin⁑2πœƒ/4) 1/2 γ€–π‘π‘œπ‘ γ€—^(βˆ’1) (π‘₯)=πœƒ πœƒ = 1/2 γ€–π‘π‘œπ‘ γ€—^(βˆ’1) π‘₯ π‘₯^2=γ€–π‘π‘œπ‘ γ€—^2 2πœƒ π‘₯^2=1βˆ’γ€–π‘ π‘–π‘›γ€—^2 2πœƒ 〖𝑠𝑖𝑛〗^2 2πœƒ="1 βˆ’ " π‘₯^2 sin 2πœƒ = √(1βˆ’π‘₯^2 ) Now, x = cos 2πœƒ Putting the values = βˆ’2 (βˆ’1/2 (1/2 γ€–π‘π‘œπ‘ γ€—^(βˆ’1) π‘₯)π‘₯+√(1 βˆ’ π‘₯^2 )/4) = βˆ’2 (√(1 βˆ’ π‘₯^2 )/4βˆ’(π‘₯ γ€–π‘π‘œπ‘ γ€—^(βˆ’1) π‘₯)/4)+ C = 𝟏/𝟐 (𝒙〖 𝒄𝒐𝒔〗^(βˆ’πŸ) π’™βˆ’βˆš(πŸβˆ’π’™^𝟐 ) )+ C

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo