Check sibling questions

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


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.

Davneet Singh's photo - Co-founder, Teachoo

Made by

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.