Integration Full Chapter Explained - https://you.tube/Integration-Class-12

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

Transcript

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
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.