Slide6.JPG

Slide7.JPG
Slide8.JPG Slide9.JPG

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


Transcript

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

Ask a doubt
Davneet Singh's photo - Co-founder, Teachoo

Made by

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.