Chapter 7 Class 12 Integrals
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Ex 7.9, 3 - Chapter 7 Class 12 Integrals

Last updated at April 16, 2024 by

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Ex 7.9, 3 Evaluate the integrals using substitution ∫_0^1β–’sin^(βˆ’1)⁑(2π‘₯/(1 + π‘₯^2 )) 𝑑π‘₯ Let I = ∫_0^1β–’sin^(βˆ’1)⁑(2π‘₯/(1 + π‘₯^2 )) 𝑑π‘₯ Put x = tan Ο• Differentiating w.r.t.Ο• 𝑑π‘₯/𝑑ϕ=(𝑑 (tan⁑ϕ ))/𝑑ϕ 𝑑π‘₯/𝑑ϕ=〖𝑠𝑒𝑐〗^2 Ο• 𝑑π‘₯=〖𝑠𝑒𝑐〗^2 Ο• 𝑑ϕ Hence when x varies from 0 to 1, πœ™ varies from 0 to πœ‹/4 Therefore we can write integral as I = ∫_0^(πœ‹/4)β–’sin^(βˆ’1)⁑(2π‘‘π‘Žπ‘›πœ™/(1 +γ€–π‘‘π‘Žπ‘›γ€—^2 πœ™)) 〖𝑠𝑒𝑐〗^2 πœ™ π‘‘πœ™ I = ∫_0^(πœ‹/4)β–’sin^(βˆ’1)⁑〖(𝑠𝑖𝑛2πœ™)γ€— 〖𝑠𝑒𝑐〗^2 πœ™ π‘‘πœ™ I = ∫_0^(πœ‹/4)β–’sin^(βˆ’1)⁑〖(𝑠𝑖𝑛2πœ™)γ€— 〖𝑠𝑒𝑐〗^2 πœ™ π‘‘πœ™ I = ∫_0^(πœ‹/4)β–’γ€–2Ο• 〖𝑠𝑒𝑐〗^2 Ο• 𝑑ϕ" " γ€— I = 2∫_0^(πœ‹/4)β–’γ€–Ο• 〖𝑠𝑒𝑐〗^2 Ο• 𝑑ϕ" " γ€— I=2Γ—[Ο• βˆ«γ€–π‘ π‘’π‘γ€—^2 Ο• π‘‘Ο•βˆ’βˆ«(𝑑ϕ/π‘‘Ο•βˆ«γ€–π‘ π‘’π‘γ€—^2 Ο• 𝑑ϕ)𝑑ϕ]_0^(πœ‹/4) =2Γ— [Ο• tanβ‘Ο•βˆ’βˆ«1β–’1Γ—tan⁑ϕ 𝑑ϕ]_0^(πœ‹/4) =2Γ— [Ο• tanβ‘Ο•βˆ’π‘™π‘œπ‘”|sec⁑ϕ |]_0^(πœ‹/4) =2[πœ‹/4 π‘‘π‘Žπ‘› πœ‹/4βˆ’π‘™π‘œπ‘”|𝑠𝑒𝑐(πœ‹/4)|βˆ’(0 tan⁑(0)βˆ’π‘™π‘œπ‘”|sec⁑(0) |) ] =2 [πœ‹/4Γ—1βˆ’π‘™π‘œπ‘”|√2|βˆ’0+π‘™π‘œπ‘” |1|] = 2 (πœ‹/4βˆ’π‘™π‘œπ‘”βˆš2βˆ’0+0) = πœ‹/2βˆ’2 π‘™π‘œπ‘” √2 = πœ‹/2βˆ’log⁑〖 (√2)^2 γ€— = 𝝅/πŸβˆ’π’π’π’ˆ 𝟐

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo