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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Transcript

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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