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Ex 7.10, 3 - Chapter 7 Class 12 Integrals (Term 2)
Last updated at March 16, 2023 by Teachoo
Chapter 7 Class 12 Integrals
Serial order wise
Transcript
Ex 7.10, 3 Evaluate the integrals using substitution β«_0^1βsin^(β1)β‘(2π₯/(1 + π₯^2 )) ππ₯ Let I = β«_0^1βsin^(β1)β‘(2π₯/(1 + π₯^2 )) ππ₯ Put x = tan Ο Differentiating w.r.t.Ο ππ₯/πΟ=(π (tanβ‘Ο ))/πΟ ππ₯/πΟ=γπ ππγ^2 Ο ππ₯=γπ ππγ^2 Ο πΟ Hence when x varies from 0 to 1, π varies from 0 to π/4 Therefore we can write integral as I = β«_0^(π/4)βsin^(β1)β‘(2π‘πππ/(1 +γπ‘ππγ^2 π)) γπ ππγ^2 π ππ I = β«_0^(π/4)βsin^(β1)β‘γ(π ππ2π)γ γπ ππγ^2 π ππ I = β«_0^(π/4)βsin^(β1)β‘γ(π ππ2π)γ γπ ππγ^2 π ππ I = β«_0^(π/4)βγ2Ο γπ ππγ^2 Ο πΟ" " γ (Using 2π‘πππ/(1 +γπ‘ππγ^2 π)=sinβ‘2π ) I = 2β«_0^(π/4)βγΟ γπ ππγ^2 Ο πΟ" " γ I=2Γ[Ο β«γπ ππγ^2 Ο πΟββ«(πΟ/πΟβ«γπ ππγ^2 Ο πΟ)πΟ]_0^(π/4) =2Γ [Ο tanβ‘Οββ«1β1Γtanβ‘Ο πΟ]_0^(π/4) =2Γ [Ο tanβ‘Οβπππ|secβ‘Ο |]_0^(π/4) =2[π/4 π‘ππ π/4βπππ|π ππ(π/4)|β(0 tanβ‘(0)βπππ|secβ‘(0) |) ] Using by parts β«_π^πβγπ(π₯)π(π₯)ππ₯=π(π₯) β«_π^πβγ9(π₯)ββ«_π^πβγ(π(π₯)γ γ β«_π^πβγπ(π₯)γ ππ₯) ππ₯γ Putting π(π₯) = Ο , g(π)=γπ ππγ^2 Ο =2 [π/4Γ1βπππ|β2|β0+πππ |1|] = 2 (π/4βπππβ2β0+0) = π/2β2 πππ β2 = π/2βlogβ‘γ (β2)^2 γ = π /πβπππ π (β΅logβ‘1=0) (β΅π ππππ₯=ππππ₯^π)
