Check sibling questions

    Slide8.JPG

Slide9.JPG
Slide10.JPG Slide11.JPG Slide12.JPG

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


Transcript

Misc 26 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”

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.