Misc 44 - Value fo tan- 1 (2x - 1 / 1 + x - x2) dx is - Miscellaneous

Misc 44 - Chapter 7 Class 12 Integrals - Part 2
Misc 44 - Chapter 7 Class 12 Integrals - Part 3

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Misc 44 The value of ∫_0^1β–’tan^(βˆ’1)⁑((2π‘₯ βˆ’ 1)/(1 + π‘₯ βˆ’ π‘₯^2 )) 𝑑π‘₯ is equal to (A) 1 (B) 0 (C) βˆ’1 (D) πœ‹/4 Let I=∫_0^1β–’tan^(βˆ’1)⁑((2π‘₯βˆ’1)/(1 + π‘₯ βˆ’ π‘₯^2 )) 𝑑π‘₯ I=∫_0^1β–’tan^(βˆ’1)⁑[(π‘₯ + (π‘₯ βˆ’ 1))/(1 + π‘₯(1 βˆ’ π‘₯) )] 𝑑π‘₯ I=∫_0^1β–’tan^(βˆ’1)⁑[(π‘₯ + (π‘₯ βˆ’ 1))/(1 + π‘₯(π‘₯ βˆ’ 1) )] 𝑑π‘₯ Using γ€–π‘‘π‘Žπ‘›γ€—^(βˆ’1) π‘₯+γ€–π‘‘π‘Žπ‘›γ€—^(βˆ’1) (𝑦)=γ€–π‘‘π‘Žπ‘›γ€—^(βˆ’1)⁑[(π‘₯ + 𝑦)/(1 βˆ’ π‘₯𝑦)] ∴ I=∫_0^1β–’[tan^(βˆ’1) (π‘₯)+tan^(βˆ’1) (π‘₯βˆ’1)] 𝑑π‘₯ I=∫_0^1β–’γ€–tan^(βˆ’1) [βˆ’(π‘₯βˆ’1)] γ€— 𝑑π‘₯+∫_0^1β–’γ€–tan^(βˆ’1) (βˆ’π‘₯) γ€— 𝑑π‘₯ Using The Property, P4 : ∫_0^π‘Žβ–’γ€–π‘“(π‘₯)𝑑π‘₯=γ€— ∫_0^π‘Žβ–’π‘“(π‘Žβˆ’π‘₯)𝑑π‘₯ ∴ I=∫_0^1β–’γ€–tan^(βˆ’1) (1βˆ’π‘₯) γ€— 𝑑π‘₯+∫_0^1β–’γ€–tan^(βˆ’1) (1βˆ’π‘₯βˆ’1) γ€— 𝑑π‘₯ I=∫_0^1β–’γ€–tan^(βˆ’1) [βˆ’(π‘₯βˆ’1)] γ€— 𝑑π‘₯+∫_0^1β–’γ€–tan^(βˆ’1) (βˆ’π‘₯) γ€— 𝑑π‘₯ I=βˆ’βˆ«_0^1β–’γ€–tan^(βˆ’1) (π‘₯βˆ’1) γ€— 𝑑π‘₯βˆ’βˆ«_0^1β–’γ€–tan^(βˆ’1) (π‘₯) γ€— 𝑑π‘₯ tan^(βˆ’1) (βˆ’π‘₯)=βˆ’tan^(βˆ’1)⁑〖(π‘₯)γ€— Adding (1) and (2) I+I =∫_0^1β–’γ€–tan^(βˆ’1) (π‘₯) γ€— 𝑑π‘₯+∫_0^1β–’γ€–tan^(βˆ’1) (π‘₯βˆ’1) γ€— 𝑑π‘₯βˆ’βˆ«_0^1β–’γ€–tan^(βˆ’1) (π‘₯) γ€— 𝑑π‘₯βˆ’βˆ«_0^1β–’γ€–tan^(βˆ’1) (π‘₯βˆ’1) γ€— 𝑑π‘₯ 2I=0 ∴ I=0 Hence, correct answer is B.

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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 14 years. He provides courses for Maths, Science and Computer Science at Teachoo