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Ex 7.2, 33 - Integrate 1 / 1 - tan x - Chapter 7 Class 12

Ex 7.2, 33 - Chapter 7 Class 12 Integrals - Part 2
Ex 7.2, 33 - Chapter 7 Class 12 Integrals - Part 3 Ex 7.2, 33 - Chapter 7 Class 12 Integrals - Part 4 Ex 7.2, 33 - Chapter 7 Class 12 Integrals - Part 5

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Ex 7.2, 33 Integrate 1/(1 βˆ’ π‘‘π‘Žπ‘›β‘π‘₯ ) ∫1β–’1/(1 βˆ’ tan⁑π‘₯ ) 𝑑π‘₯ = ∫1β–’1/(1 βˆ’ sin⁑π‘₯/cos⁑π‘₯ ) 𝑑π‘₯ = ∫1β–’1/(γ€–cos⁑π‘₯ βˆ’ sin〗⁑π‘₯/cos⁑π‘₯ ) 𝑑π‘₯ = ∫1β–’cos⁑π‘₯/γ€–cos⁑π‘₯ βˆ’ sin〗⁑π‘₯ 𝑑π‘₯ = ∫1β–’(2 cos⁑π‘₯)/2(γ€–cos⁑π‘₯ βˆ’ sin〗⁑π‘₯ ) 𝑑π‘₯ = ∫1β–’(cos⁑π‘₯ + cos⁑π‘₯)/2(γ€–cos⁑π‘₯ βˆ’ sin〗⁑π‘₯ ) 𝑑π‘₯ = ∫1β–’(cos⁑π‘₯ + cos⁑π‘₯ + sin⁑π‘₯ βˆ’ sin⁑π‘₯)/2(γ€–cos⁑π‘₯ βˆ’ sin〗⁑π‘₯ ) 𝑑π‘₯ = ∫1β–’(cos⁑π‘₯ βˆ’ sin⁑π‘₯ + cos⁑π‘₯ + sin⁑π‘₯)/2(γ€–cos⁑π‘₯ βˆ’ sin〗⁑π‘₯ ) 𝑑π‘₯ = 1/2 ∫1β–’((cos⁑π‘₯ βˆ’ sin⁑π‘₯ + cos⁑π‘₯ + sin⁑π‘₯)/γ€–cos⁑π‘₯ βˆ’ sin〗⁑π‘₯ ) 𝑑π‘₯ = 1/2 ∫1β–’((cos⁑π‘₯ + sin⁑π‘₯)/γ€–cos⁑π‘₯ βˆ’ sin〗⁑π‘₯ +(cos⁑π‘₯ βˆ’ sin⁑π‘₯)/γ€–cos⁑π‘₯ βˆ’ sin〗⁑π‘₯ ) 𝑑π‘₯ = 1/2 ∫1β–’(1+ (cos⁑π‘₯ + sin⁑π‘₯)/γ€–cos⁑π‘₯ βˆ’ sin〗⁑π‘₯ ) 𝑑π‘₯ = 1/2 ∫1▒𝑑π‘₯ + 1/2 ∫1β–’(cos⁑π‘₯ + sin⁑π‘₯)/γ€–cos⁑π‘₯ βˆ’ sin〗⁑π‘₯ 𝑑π‘₯ (Adding & subtracting 𝑠𝑖𝑛⁑π‘₯ in numerator) = π‘₯/2+1/2 ∫1β–’(cos⁑π‘₯ + sin⁑π‘₯)/γ€–cos⁑π‘₯ βˆ’ sin〗⁑π‘₯ 𝑑π‘₯+𝐢_1 Solving 𝐈1 I1 = ∫1β–’(cos⁑π‘₯ + sin⁑π‘₯)/γ€–cos⁑π‘₯ βˆ’ sin〗⁑π‘₯ .𝑑π‘₯ Let cos⁑π‘₯ βˆ’ sin⁑π‘₯=𝑑 Differentiating both sides 𝑀.π‘Ÿ.𝑑.π‘₯ γ€–γ€– βˆ’π‘ π‘–π‘›γ€—β‘π‘₯βˆ’π‘π‘œπ‘ γ€—β‘π‘₯=𝑑𝑑/𝑑π‘₯ βˆ’(〖𝑠𝑖𝑛⁑π‘₯+π‘π‘œπ‘ γ€—β‘π‘₯ )=𝑑𝑑/𝑑π‘₯ 𝑑π‘₯=𝑑𝑑/(βˆ’(γ€–sin⁑π‘₯ βˆ’ cos〗⁑π‘₯ ) ) Thus, our equation becomes I1 = ∫1β–’(cos⁑π‘₯ + sin⁑π‘₯)/γ€–cos⁑π‘₯ βˆ’ sin〗⁑π‘₯ . 𝑑π‘₯ I1 = ∫1β–’(cos⁑π‘₯ βˆ’ sin⁑π‘₯)/𝑑 . 𝑑𝑑/(βˆ’(γ€–sin⁑π‘₯ βˆ’ cos〗⁑π‘₯ ) ) I1 = βˆ’1∫1▒𝑑𝑑/𝑑 I1 = βˆ’log⁑|𝑑|+𝐢2 I1 = βˆ’log⁑|cos⁑π‘₯βˆ’sin⁑π‘₯ |+𝐢2 (Using ∫1β–’1/π‘₯. 𝑑π‘₯=π‘™π‘œπ‘”β‘|π‘₯|) (Using 〖〖𝑑=π‘π‘œπ‘ γ€—β‘π‘₯βˆ’π‘ π‘–π‘›γ€—β‘π‘₯ ) Putting the value of I1 in (1) ∴ ∫1β–’γ€–1/(1 + tan⁑π‘₯ ) " " γ€—= π‘₯/2+1/2 ∫1β–’(cos⁑π‘₯ + sin⁑π‘₯)/γ€–cos⁑π‘₯ βˆ’ sin〗⁑π‘₯ 𝑑π‘₯+𝐢_1 = π‘₯/2+1/2(βˆ’π₯𝐨𝐠⁑〖 |π’„π’π’”β‘π’™βˆ’π’”π’Šπ’β‘π’™ |γ€—)+𝐢2/2+𝐢1 = 𝒙/𝟐 βˆ’πŸ/𝟐 π’π’π’ˆβ‘γ€– |π’„π’π’”β‘π’™βˆ’π’”π’Šπ’β‘π’™ |γ€—+π‘ͺ

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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.