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  1. Chapter 7 Class 12 Integrals
  2. Serial order wise

Transcript

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 is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.