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Misc 27 - Definite integral cos2 x dx / cos2 x + 4 sin2 x

Misc 27 - Chapter 7 Class 12 Integrals - Part 2
Misc 27 - Chapter 7 Class 12 Integrals - Part 3 Misc 27 - Chapter 7 Class 12 Integrals - Part 4 Misc 27 - Chapter 7 Class 12 Integrals - Part 5

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Misc 27 Evaluate the definite integral โˆซ_0^(๐œ‹/2)โ–’ใ€–(cos^2โก๐‘ฅ ๐‘‘๐‘ฅ)/(cos^2โก๐‘ฅ + 4 sin^2โก๐‘ฅ ) ใ€— Let I = โˆซ1_0^(๐œ‹/2)โ–’(ใ€–๐‘๐‘œ๐‘ ใ€—^2 ๐‘ฅ)/(ใ€–๐‘๐‘œ๐‘ ใ€—^2 ๐‘ฅ + 4ใ€–๐‘ ๐‘–๐‘›ใ€—^2 ๐‘ฅ) ๐‘‘๐‘ฅ = โˆซ1_0^(๐œ‹/2)โ–’(ใ€–๐‘๐‘œ๐‘ ใ€—^2 ๐‘ฅ)/(ใ€–๐‘๐‘œ๐‘ ใ€—^2 ๐‘ฅ + 4(ใ€–1 โˆ’ ๐‘๐‘œ๐‘ ใ€—^2 ๐‘ฅ) ๐‘‘๐‘ฅ) = โˆซ1_0^(๐œ‹/2)โ–’(ใ€–๐‘๐‘œ๐‘ ใ€—^2 ๐‘ฅ)/(4 โˆ’ 3 ใ€–๐‘๐‘œ๐‘ ใ€—^2 ๐‘ฅ) ๐‘‘๐‘ฅ = (โˆ’1)/3 โˆซ1_0^(๐œ‹/4)โ–’ใ€– (โˆ’3 ใ€–๐‘๐‘œ๐‘ ใ€—^2 ๐‘ฅ )/(4 โˆ’ 3 ใ€–๐‘๐‘œ๐‘ ใ€—^2 ๐‘ฅ) ใ€— ๐‘‘๐‘ฅ = (โˆ’1)/3 โˆซ1_0^(๐œ‹/2)โ–’(ใ€–โˆ’ 3 ๐‘๐‘œ๐‘ ใ€—^2 ๐‘ฅ + 4 โˆ’ 4)/(4 โˆ’ 3 ใ€–๐‘๐‘œ๐‘ ใ€—^2 ๐‘ฅ) ๐‘‘๐‘ฅ = (โˆ’1)/3 โˆซ1_0^(๐œ‹/2)โ–’(ใ€–4 โˆ’ 3 ๐‘๐‘œ๐‘ ใ€—^2 ๐‘ฅ โˆ’ 4)/(4 โˆ’ 3 ใ€–๐‘๐‘œ๐‘ ใ€—^2 ๐‘ฅ) ๐‘‘๐‘ฅ = (โˆ’1)/3 โˆซ1_0^(๐œ‹/2)โ–’ใ€–1โˆ’4/(4 โˆ’ 3 ใ€–๐‘๐‘œ๐‘ ใ€—^2 ๐‘ฅ)ใ€— ๐‘‘๐‘ฅ Dividing numerator and denominator by ใ€–๐‘๐‘œ๐‘ ใ€—^2 ๐‘ฅ = (โˆ’1)/3 (๐œ‹/2)+4/3 โˆซ1_0^(๐œ‹/2)โ–’(๐‘‘๐‘ฅ/(ใ€–๐‘๐‘œ๐‘ ใ€—^2 ๐‘ฅ))/((4 โˆ’ 3 ใ€–๐‘๐‘œ๐‘ ใ€—^2 ๐‘ฅ )/(ใ€–๐‘๐‘œ๐‘ ใ€—^2 ๐‘ฅ)) ๐‘‘๐‘ฅ = (โˆ’1)/3 (๐œ‹/2)+4/3 โˆซ1_0^(๐œ‹/2)โ–’(ใ€–๐‘ ๐‘’๐‘ใ€—^2 ๐‘ฅ)/(4 ใ€–๐‘ ๐‘’๐‘ใ€—^2 ๐‘ฅ โˆ’ 3) ๐‘‘๐‘ฅ = (โˆ’๐œ‹)/6+4/3 โˆซ1_0^(๐œ‹/2)โ–’(ใ€–๐‘ ๐‘’๐‘ใ€—^2 ๐‘ฅ)/(4 (1 + ใ€–๐‘ก๐‘Ž๐‘›ใ€—^2 ๐‘ฅ) โˆ’ 3) ๐‘‘๐‘ฅ = (โˆ’๐œ‹)/6+4/3 โˆซ1_0^(๐œ‹/2)โ–’(ใ€–๐‘ ๐‘’๐‘ใ€—^2 ๐‘ฅ)/(4 + 4 ใ€–๐‘ก๐‘Ž๐‘›ใ€—^2 ๐‘ฅ โˆ’ 3) ๐‘‘๐‘ฅ = (โˆ’๐œ‹)/6+4/3 โˆซ1_0^(๐œ‹/2)โ–’(ใ€–๐‘ ๐‘’๐‘ใ€—^2 ๐‘ฅ)/( 4 ใ€–๐‘ก๐‘Ž๐‘›ใ€—^2 ๐‘ฅ + 1) ๐‘‘๐‘ฅ Let tan x = t Differentiating w.r.t x ใ€–๐‘ ๐‘’๐‘ใ€—^2 x dx = dt Thus, When x = 0, t = 0, & when x = ๐œ‹/2, ๐‘ก= โˆž Substituting values and limit I = (โˆ’๐œ‹)/6+4/3 โˆซ1_0^(๐œ‹/2)โ–’(ใ€–๐‘ ๐‘’๐‘ใ€—^2 ๐‘ฅ)/( 4 ใ€–๐‘ก๐‘Ž๐‘›ใ€—^2 ๐‘ฅ + 1) ๐‘‘๐‘ฅ โˆด I =(โˆ’๐œ‹)/6+4/3 โˆซ1_0^โˆžโ–’๐‘‘๐‘ก/(ใ€–4๐‘กใ€—^2+1) = (โˆ’๐œ‹)/6+4/3 โˆ™1/4 โˆซ1_0^โˆžโ–’๐‘‘๐‘ก/(ใ€– ๐‘กใ€—^2+1/4) =(โˆ’๐œ‹)/6+4/3โˆ™1/4 ร— 1/((1/2) ) [ใ€–๐‘ก๐‘Ž๐‘›ใ€—^(โˆ’1) ๐‘ก/(1/2)]_0^โˆž = (โˆ’๐œ‹)/6+2/3โˆ™ [ใ€–๐‘ก๐‘Ž๐‘›ใ€—^(โˆ’1) 2๐‘ก]_0^โˆž =(โˆ’๐œ‹)/6+2/3โˆ™ใ€–[๐‘ก๐‘Ž๐‘›ใ€—^(โˆ’1) โˆžโˆ’ใ€–๐‘ก๐‘Ž๐‘›ใ€—^(โˆ’1) 0] = โˆ’๐œ‹/6+2/3โˆ™[๐œ‹/2โˆ’0] =โˆ’๐œ‹/6+๐œ‹/3 =๐…/๐Ÿ” We know that โˆซ1โ–’1/(๐‘ฅ^2 + ๐‘Ž^2 )=ใ€–1/๐‘Ž tan^(โˆ’1)ใ€—โกใ€–๐‘ฅ/๐‘Žใ€—+๐ถ = (โˆ’๐œ‹)/6+2/3โˆ™[๐œ‹/2โˆ’0] =(โˆ’๐œ‹)/6+๐œ‹/3 =๐œ‹/3โˆ’๐œ‹/6 =๐…/๐Ÿ”

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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.