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Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class


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 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.