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Ex 7.9, 3 - Chapter 7 Class 12 Integrals
Last updated at April 16, 2024 by Teachoo
Transcript
Ex 7.9, 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 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) |) ] =2 [π/4Γ1βπππ|β2|β0+πππ |1|] = 2 (π/4βπππβ2β0+0) = π/2β2 πππ β2 = π/2βlogβ‘γ (β2)^2 γ = π /πβπππ π
