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Misc 20 - Integrate root 1 - root x / 1 + root x - Class 12

Misc 20 - Chapter 7 Class 12 Integrals - Part 2
Misc 20 - Chapter 7 Class 12 Integrals - Part 3 Misc 20 - Chapter 7 Class 12 Integrals - Part 4 Misc 20 - Chapter 7 Class 12 Integrals - Part 5


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Misc 20 Integrate the function √((1 βˆ’ √π‘₯)/(1 + √π‘₯)) ∫1β–’γ€–βˆš((1 βˆ’ √π‘₯)/(1 + √π‘₯)) 𝑑π‘₯γ€— Let x = 〖𝒄𝒐𝒔〗^𝟐 𝟐𝜽 dx = βˆ’4 cos 2πœƒ sin 2πœƒ dπœƒ Substituting, = ∫1β–’βˆš((1 βˆ’ √((γ€–π‘π‘œπ‘ γ€—^2 2πœƒ) ))/(1 + √((γ€–π‘π‘œπ‘ γ€—^2 2πœƒ) )))Γ—βˆ’4 cos⁑2ΞΈ sin⁑2ΞΈ π‘‘πœƒ = ∫1β–’βˆš((1 βˆ’ cos⁑2πœƒ)/(1 + π‘π‘œπ‘  2πœƒ))Γ—(βˆ’4) cos⁑2ΞΈ sin⁑2ΞΈ π‘‘πœƒ = βˆ’4∫1β–’βˆš((1 βˆ’ (1 βˆ’ 2〖𝑠𝑖𝑛〗^(2 ) πœƒ))/(1 + (2γ€–π‘π‘œπ‘ γ€—^(2 ) πœƒ βˆ’ 1) )) cos⁑2ΞΈ (2 sin⁑θ cosβ‘γ€–πœƒ)γ€— π‘‘πœƒ = βˆ’8∫1β–’βˆš((2〖𝑠𝑖𝑛〗^(2 ) πœƒ)/(2γ€–π‘π‘œπ‘ γ€—^(2 ) πœƒ)) cos⁑2ΞΈ cosβ‘πœƒ sinβ‘πœƒ π‘‘πœƒ = βˆ’8∫1β–’sinβ‘πœƒ/cos⁑θ cos⁑2ΞΈ cosβ‘πœƒ sinβ‘πœƒ π‘‘πœƒ = βˆ’8∫1▒〖〖𝑠𝑖𝑛〗^2 πœƒγ€— cos⁑2ΞΈ π‘‘πœƒ Using cos⁑2π‘₯=1βˆ’2〖𝑠𝑖𝑛〗^2 π‘₯=2γ€–π‘π‘œπ‘ γ€—^2 π‘₯βˆ’1 sin 2π‘₯=2 sin⁑〖π‘₯ cos⁑π‘₯ γ€— = βˆ’8∫1β–’((1 βˆ’ cos⁑2ΞΈ)/2) cos⁑2ΞΈ π‘‘πœƒ = –4 ∫1β–’(π‘π‘œπ‘  2ΞΈβˆ’cos^2⁑2ΞΈ ) π‘‘πœƒ = 4 ∫1β–’(γ€–π‘π‘œπ‘ γ€—^2 2ΞΈβˆ’cos⁑2ΞΈ ) π‘‘πœƒ = 4 ∫1β–’γ€–γ€–π‘π‘œπ‘ γ€—^2 2ΞΈγ€— π‘‘πœƒβˆ’4∫1β–’cos⁑2ΞΈ π‘‘πœƒ = 4 ∫1β–’(cos⁑4πœƒ + 1)/2 π‘‘πœƒ βˆ’ 4∫1β–’γ€–π‘π‘œπ‘  2πœƒγ€— π‘‘πœƒ = 2 ∫1β–’γ€–(cos⁑4πœƒ + 1)γ€— π‘‘πœƒ βˆ’ 4∫1β–’γ€–π‘π‘œπ‘  2πœƒγ€— π‘‘πœƒ = 2 [(sin⁑4 πœƒ)/4+πœƒ] βˆ’4 [(sin⁑2 πœƒ)/2]+C Using cos 2x = 1 βˆ’ 2 sin^2 x 〖𝑠𝑖𝑛〗^2 π‘₯=(1 βˆ’ cos⁑2π‘₯)/2 = sin⁑4πœƒ/2+2πœƒ βˆ’2 𝑠𝑖𝑛 2πœƒ+ C Now x = γ€–π‘π‘œπ‘ γ€—^2 2πœƒ √π‘₯ " = " cos⁑2πœƒ γ€–π‘π‘œπ‘ γ€—^(βˆ’1) √π‘₯=2πœƒ 1/2 γ€–π‘π‘œπ‘ γ€—^(βˆ’1) √π‘₯=πœƒ And, sin 4πœƒ = 2 sin 2πœƒ cos 2πœƒ 1 βˆ’ π‘₯=1βˆ’γ€–π‘π‘œπ‘ γ€—^2 2πœƒ 1 βˆ’ π‘₯=〖𝑠𝑖𝑛〗^2 2πœƒ sin 2πœƒ = √(1βˆ’π‘₯) = 2√(1βˆ’π‘₯)Γ—βˆšπ‘₯ = 2 √π‘₯ √(1βˆ’π‘₯) Putting the values. = sin⁑4πœƒ/2+2ΞΈβˆ’2 sin⁑2ΞΈ+ C = (2√π‘₯ √(1 βˆ’ π‘₯))/2+2 (γ€–π‘π‘œπ‘ γ€—^(βˆ’1) √π‘₯)/2βˆ’2√(1βˆ’π‘₯)+C = √π‘₯ √(1βˆ’π‘₯)+γ€–π‘π‘œπ‘ γ€—^(βˆ’1) √π‘₯βˆ’2√(1βˆ’π‘₯)+ C = √(π‘₯βˆ’π‘₯^2 )+γ€–π‘π‘œπ‘ γ€—^(βˆ’1) √π‘₯βˆ’2√(1βˆ’π‘₯)+C = –2√(πŸβˆ’π’™)+〖𝒄𝒐𝒔〗^(βˆ’πŸ) βˆšπ’™+√(π’™βˆ’π’™^𝟐 )+𝐂

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