Question 4 (MCQ) - Chapter 7 Class 12 Integrals (Important Question)
Last updated at Dec. 16, 2024 by Teachoo
Chapter 7 Class 12 Integrals
Ex 7.1, 18 Important
Ex 7.1, 20
Ex 7.2, 20 Important
Ex 7.2, 26 Important
Ex 7.2, 35
Ex 7.2, 36 Important
Ex 7.3, 6 Important
Ex 7.3, 13 Important
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Ex 7.3, 24 (MCQ) Important
Example 9 (i)
Example 10 (i)
Ex 7.4, 8 Important
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Ex 7.4, 22
Ex 7.4, 25 (MCQ) Important
Example 15 Important
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Ex 7.5, 18 Important
Ex 7.5, 21 Important
Example 20 Important
Example 22 Important
Ex 7.6, 13 Important
Ex 7.6, 14 Important
Ex 7.6, 18 Important
Ex 7.6, 19
Ex 7.6, 24 (MCQ) Important
Ex 7.7, 5 Important
Ex 7.7, 10
Ex 7.7, 11 Important
Question 1 Important
Question 4 Important
Question 6 Important
Example 25 (i)
Ex 7.8, 15
Ex 7.8, 16 Important
Ex 7.8, 20 Important
Ex 7.8, 22 (MCQ)
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Ex 7.9, 8
Ex 7.9, 9 (MCQ) Important
Example 28 Important
Example 32 Important
Example 34 Important
Ex 7.10,8 Important
Ex 7.10, 18 Important
Example 38 Important
Example 39 Important
Example 42 Important
Misc 18 Important
Misc 8 Important
Question 1 Important
Misc 23 Important
Misc 29 Important
Question 2 Important
Misc 38 (MCQ) Important
Question 4 (MCQ) Important You are here
Integration Formula Sheet Important
Chapter 7 Class 12 Integrals
Last updated at Dec. 16, 2024 by Teachoo
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.