Question 2 - Chapter 7 Class 12 Integrals (Important Question)
Last updated at Dec. 16, 2024 by Teachoo
Chapter 7 Class 12 Integrals
Ex 7.1, 10
Important
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Class 12
Important Questions for exams Class 12
- Chapter 1 Class 12 Relation and Functions
- Chapter 2 Class 12 Inverse Trigonometric Functions
- Chapter 3 Class 12 Matrices
- Chapter 4 Class 12 Determinants
- Chapter 5 Class 12 Continuity and Differentiability
- Chapter 6 Class 12 Application of Derivatives
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Chapter 7 Class 12 Integrals
- Chapter 8 Class 12 Application of Integrals
- Chapter 9 Class 12 Differential Equations
- Chapter 10 Class 12 Vector Algebra
- Chapter 11 Class 12 Three Dimensional Geometry
- Chapter 12 Class 12 Linear Programming
- Chapter 13 Class 12 Probability
Transcript
Question 2 Evaluate the definite integral β«_0^πβ(π₯ tanβ‘π₯ )/(secβ‘π₯ +γ tanγβ‘π₯ ) ππ₯ Let I=β«_0^πβ(π₯ tanβ‘π₯ )/(secβ‘π₯ +γ tanγβ‘π₯ ) ππ₯ β΄ I=β«_0^πβ((π β π₯) tanβ‘γ (π β π₯)γ)/(secβ‘(π β π₯) +γ tanγβ‘(π β π₯) ) ππ₯ I=β«_0^πβ((π β π₯)(βtanβ‘γ π₯γ) )/((βsecβ‘γ π₯γ) + γ( βtanγβ‘π₯)) ππ₯ I=β«_0^πβ(β(π β π₯) tanβ‘π₯ )/(β(secβ‘π₯ +γ tanγβ‘π₯)) ππ₯ Using The Property, P4 P4 : β«_0^πβγπ(π₯)ππ₯=γ β«_0^πβπ(πβπ₯)ππ₯ I=β«_0^πβ((π β π₯) tanβ‘π₯ )/((secβ‘π₯ +γ tanγβ‘π₯)) ππ₯ Adding (1) and (2) i.e. (1) + (2) I+I=β«_0^πβ(π₯ tanβ‘π₯ )/(secβ‘π₯ +γ tanγβ‘π₯ ) ππ₯+β«_0^πβ(π tanβ‘π₯ β π₯ tanβ‘π₯)/(secβ‘π₯ +γ tanγβ‘π₯ ) ππ₯ 2I=β«_0^πβ(π₯ tanβ‘π₯ + π tanβ‘π₯ β π₯ tanβ‘π₯)/(secβ‘π₯ +γ tanγβ‘π₯ ) ππ₯ 2I=β«_0^πβ(π tanβ‘π₯)/(secβ‘π₯ +γ tanγβ‘π₯ ) ππ₯ 2I=πβ«_0^πβtanβ‘π₯/(secβ‘π₯ +γ tanγβ‘π₯ ) ππ₯ I=π/2 β«_0^πβtanβ‘π₯/(secβ‘π₯ +γ tanγβ‘π₯ ) ππ₯ =π/2 β«_0^πβ(sinβ‘π₯/cosβ‘π₯ )/(1/cosβ‘π₯ + sinβ‘π₯/cosβ‘π₯ ) ππ₯ =π/2 β«_0^πβsinβ‘π₯/(1 + sinβ‘π₯ ) ππ₯ =π/2 β«_0^πβ(sinβ‘π₯ + 1 β 1)/(1 + sinβ‘π₯ ) ππ₯ =π/2 β«_0^πβ[(1 + sinβ‘π₯)/(1 + sinβ‘π₯ ) β1/(1 + sinβ‘π₯ )] ππ₯ =π/2 β«_0^πβ[1 β1/(1 + sinβ‘π₯ )] ππ₯ =π/2 [β«_0^πβ1 ππ₯ββ«_0^πβ1/(1 + sinβ‘π₯ ) ππ₯] =π/2 [[π₯]_0^πββ«_0^πβ1/(1 + sinβ‘π₯ ) ((1 β sinβ‘π₯)/(1 β sinβ‘π₯ )) ππ₯] =π/2 [[πβ0]ββ«_0^πβ(1 β sinβ‘π₯)/(1 β sin^2β‘π₯ ) ππ₯] =π/2 [πββ«_0^πβ(1 β sinβ‘π₯)/cos^2β‘π₯ ππ₯] =π/2 [πββ«_0^πβ[1/cos^2β‘π₯ β sinβ‘π₯/cos^2β‘π₯ ] ππ₯] =π/2 {πββ«_0^πβ[sec^2β‘π₯βtanβ‘π₯ secβ‘π₯ ] ππ₯} =π/2 {πββ«_0^πβsec^2β‘π₯ ππ₯+β«_0^πβγtanβ‘π₯ secβ‘π₯ γ ππ₯} =π/2 [πβ[tanβ‘π₯ ]_0^π+[secβ‘π₯ ]_0^π ] =π/2 {πβ[tanβ‘γ(π)βtanβ‘(0) γ ]+[sec (π)βsecβ‘(0) ]} =π/2 {πβ[0β0]+[β1β1]} =π/2 {πβ0+[β2]} =π /π (π βπ)
