Question 4 (MCQ) - Miscellaneous - Chapter 7 Class 12 Integrals
Last updated at Dec. 16, 2024 by Teachoo
Miscellaneous
Misc 1
Important
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Chapter 7 Class 12 Integrals
Serial order wise
Transcript
Misc 44 The value of β«_0^1βtan^(β1)β‘((2π₯ β 1)/(1 + π₯ β π₯^2 )) ππ₯ is equal to (A) 1 (B) 0 (C) β1 (D) π/4 Let I=β«_0^1βtan^(β1)β‘((2π₯β1)/(1 + π₯ β π₯^2 )) ππ₯ I=β«_0^1βtan^(β1)β‘[(π₯ + (π₯ β 1))/(1 + π₯(1 β π₯) )] ππ₯ I=β«_0^1βtan^(β1)β‘[(π₯ + (π₯ β 1))/(1 + π₯(π₯ β 1) )] ππ₯ Using γπ‘ππγ^(β1) π₯+γπ‘ππγ^(β1) (π¦)=γπ‘ππγ^(β1)β‘[(π₯ + π¦)/(1 β π₯π¦)] β΄ I=β«_0^1β[tan^(β1) (π₯)+tan^(β1) (π₯β1)] ππ₯ I=β«_0^1βγtan^(β1) [β(π₯β1)] γ ππ₯+β«_0^1βγtan^(β1) (βπ₯) γ ππ₯ Using The Property, P4 : β«_0^πβγπ(π₯)ππ₯=γ β«_0^πβπ(πβπ₯)ππ₯ β΄ I=β«_0^1βγtan^(β1) (1βπ₯) γ ππ₯+β«_0^1βγtan^(β1) (1βπ₯β1) γ ππ₯ I=β«_0^1βγtan^(β1) [β(π₯β1)] γ ππ₯+β«_0^1βγtan^(β1) (βπ₯) γ ππ₯ I=ββ«_0^1βγtan^(β1) (π₯β1) γ ππ₯ββ«_0^1βγtan^(β1) (π₯) γ ππ₯ tan^(β1) (βπ₯)=βtan^(β1)β‘γ(π₯)γ Adding (1) and (2) I+I =β«_0^1βγtan^(β1) (π₯) γ ππ₯+β«_0^1βγtan^(β1) (π₯β1) γ ππ₯ββ«_0^1βγtan^(β1) (π₯) γ ππ₯ββ«_0^1βγtan^(β1) (π₯β1) γ ππ₯ 2I=0 β΄ I=0 Hence, correct answer is B.
