Integration Full Chapter Explained -


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  1. Chapter 7 Class 12 Integrals
  2. Serial order wise


Ex 7.10, 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€— (Using 2π‘‘π‘Žπ‘›πœ™/(1 +γ€–π‘‘π‘Žπ‘›γ€—^2 πœ™)=sin⁑2πœ™ ) 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) |) ] Using by parts ∫_π‘Ž^𝑏▒〖𝑓(π‘₯)𝑔(π‘₯)𝑑π‘₯=𝑓(π‘₯) ∫_π‘Ž^𝑏▒〖9(π‘₯)βˆ’βˆ«_π‘Ž^𝑏▒〖(𝑓(π‘₯)γ€— γ€— ∫_π‘Ž^𝑏▒〖𝑔(π‘₯)γ€— 𝑑π‘₯) 𝑑π‘₯γ€— Putting 𝑓(π‘₯) = Ο• , g(πœ™)=〖𝑠𝑒𝑐〗^2 Ο• =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 (∡log⁑1=0) (βˆ΅π‘Ž π‘™π‘œπ‘”π‘₯=π‘™π‘œπ‘”π‘₯^π‘Ž)

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.