Slide50.JPG

Slide51.JPG
Slide52.JPG Slide53.JPG Slide54.JPG Slide55.JPG Slide56.JPG

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


Transcript

Example 42 Evaluate โˆซ_0^๐œ‹โ–’(๐‘ฅ ๐‘‘๐‘ฅ)/(๐‘Ž^2 cos^2โกใ€–๐‘ฅ + ๐‘^2 ใ€— sin^2โก๐‘ฅ )Let I= โˆซ_0^๐œ‹โ–’ใ€–๐‘ฅ/(๐‘Ž^2 ๐‘๐‘œ๐‘ ^2 ๐‘ฅ + ๐‘^2 ๐‘ ๐‘–๐‘›^2 ๐‘ฅ) ๐‘‘๐‘ฅใ€— โˆด I=โˆซ_0^๐œ‹โ–’ใ€–((๐œ‹ โˆ’ ๐‘ฅ))/(๐‘Ž^2 ๐‘๐‘œ๐‘ ^2 (๐œ‹ โˆ’ ๐‘ฅ) + ๐‘^2 ๐‘ ๐‘–๐‘›^2 (๐œ‹ โˆ’ ๐‘ฅ) ) ๐‘‘๐‘ฅใ€— I=โˆซ_0^๐œ‹โ–’ใ€–(๐œ‹ โˆ’ ๐‘ฅ)/(๐‘Ž^2 [๐‘๐‘œ๐‘ (๐œ‹ โˆ’ ๐‘ฅ)]^2 + ๐‘^2 [๐‘ ๐‘–๐‘›(๐œ‹ โˆ’ ๐‘ฅ)]^2 ) ๐‘‘๐‘ฅใ€— I=โˆซ_0^๐œ‹โ–’ใ€–(๐œ‹ โˆ’ ๐‘ฅ)/(๐‘Ž^2 [โˆ’ ๐‘๐‘œ๐‘  ๐‘ฅ]^2 + ๐‘^2 [๐‘ ๐‘–๐‘› ๐‘ฅ]^2 ) ๐‘‘๐‘ฅใ€— I=โˆซ_0^๐œ‹โ–’ใ€–(๐œ‹ โˆ’ ๐‘ฅ)/(๐‘Ž^2 cos^2โก๐‘ฅ + ๐‘^2 sin^2โก๐‘ฅ ) ๐‘‘๐‘ฅใ€— Adding (1) and (2) i.e. (1) + (2) I+I=โˆซ_0^๐œ‹โ–’ใ€–๐‘ฅ/(๐‘Ž^2 cos^2โก๐‘ฅ + ๐‘^2 sin^2โก๐‘ฅ ) ๐‘‘๐‘ฅใ€—+โˆซ1โ–’(๐œ‹ โˆ’ ๐‘ฅ)/(๐‘Ž^2 cos^2โก๐‘ฅ + ๐‘^2 sin^2โก๐‘ฅ ) ๐‘‘๐‘ฅ 2I=โˆซ_0^๐œ‹โ–’(๐‘ฅ + ๐œ‹ โˆ’ ๐‘ฅ)/(๐‘Ž^2 cos^2โก๐‘ฅ + ๐‘^2 sin^2โก๐‘ฅ ) ๐‘‘๐‘ฅ 2I=โˆซ_0^๐œ‹โ–’(๐œ‹ )/(๐‘Ž^2 cos^2โก๐‘ฅ + ๐‘^2 sin^2โก๐‘ฅ ) ๐‘‘๐‘ฅ I=๐œ‹/2 โˆซ_0^๐œ‹โ–’ใ€–1/(๐‘Ž^2 cos^2โก๐‘ฅ + ๐‘^2 sin^2โก๐‘ฅ ) ๐‘‘๐‘ฅใ€— Dividing numerator and denominator by ๐‘๐‘œ๐‘ ^2 ๐‘ฅ, we get I=๐œ‹/2 โˆซ_0^๐œ‹โ–’ใ€–(1/cos^2โก๐‘ฅ )/((๐‘Ž^2 cos^2โกใ€–๐‘ฅ + ๐‘^2 sin^2โก๐‘ฅ ใ€—)/cos^2โก๐‘ฅ ) ๐‘‘๐‘ฅใ€— I=๐œ‹/2 โˆซ_0^๐œ‹โ–’ใ€–(๐‘ ๐‘’๐‘^2 ๐‘ฅ)/((๐‘Ž^2 cos^2โก๐‘ฅ)/cos^2โก๐‘ฅ + (๐‘^2 sin^2โก๐‘ฅ)/cos^2โก๐‘ฅ ) ๐‘‘๐‘ฅใ€— I=๐œ‹/2 โˆซ_0^๐œ‹โ–’ใ€–(๐‘ ๐‘’๐‘^2 ๐‘ฅ)/(๐‘Ž^2 + ๐‘^2 tan^2โก๐‘ฅ ) ๐‘‘๐‘ฅใ€— Let ๐‘“(๐‘ฅ)=sec^2โก๐‘ฅ/(๐‘Ž^2 + ๐‘^2 tan^2โก๐‘ฅ ) and a = ฯ€ Now, ๐‘“(2๐‘Žโˆ’๐‘ฅ)=sec^2โก(๐œ‹ โˆ’ ๐‘ฅ)/(๐‘Ž^2 + ๐‘^2 tan^2โก(๐œ‹ โˆ’ ๐‘ฅ) ) ๐‘“(2๐‘Žโˆ’๐‘ฅ)=[โˆ’๐‘ ๐‘’๐‘ ๐‘ฅ]^2/(๐‘Ž^2 + ๐‘^2 [โˆ’tanโก๐‘ฅ ]^2 ) ๐‘“(2๐‘Žโˆ’๐‘ฅ)=(๐‘ ๐‘’๐‘^2 ๐‘ฅ)/(๐‘Ž^2 + ๐‘^2 tan^2โก๐‘ฅ ) Therefore, ๐‘“(๐‘ฅ)=๐‘“(2๐‘Žโˆ’๐‘ฅ) Therefore, I=๐œ‹/2 โˆซ_0^๐œ‹โ–’ใ€–(๐‘ ๐‘’๐‘^2 ๐‘ฅ)/(๐‘Ž^2 + ๐‘^2 tan^2โก๐‘ฅ ) ๐‘‘๐‘ฅใ€— =๐œ‹/2 ร— 2 โˆซ_0^(๐œ‹/2)โ–’ใ€–(๐‘ ๐‘’๐‘^2 ๐‘ฅ)/(๐‘Ž^2 + ๐‘^2 tan^2โก๐‘ฅ ) ๐‘‘๐‘ฅใ€— =๐œ‹โˆซ_0^(๐œ‹/2)โ–’ใ€–(๐‘ ๐‘’๐‘^2 ๐‘ฅ)/(๐‘Ž^2 + ๐‘^2 tan^2โก๐‘ฅ ) ๐‘‘๐‘ฅใ€— Let ๐‘ tanโกใ€–๐‘ฅ=๐‘กใ€— Differentiating both sides w.r.t. ๐‘ฅ ๐‘ ๐‘ ๐‘’๐‘^2 ๐‘ฅ ๐‘‘๐‘ฅ=๐‘‘๐‘ก ๐‘‘๐‘ก=๐‘‘๐‘ก/(๐‘^2 ๐‘ ๐‘’๐‘^2 ๐‘ฅ) Putting the values of tan ๐‘ฅ and ๐‘‘๐‘ฅ , we get ๐ผ=๐œ‹โˆซ1_0^(๐œ‹/2)โ–’ใ€–(๐‘ ๐‘’๐‘^2 ๐‘ฅ)/(๐‘Ž^2 + ๐‘ก^2 ) . ๐‘‘๐‘ฅใ€— ๐ผ=๐œ‹ โˆซ1_0^โˆžโ–’ใ€–(๐‘ ๐‘’๐‘^2 ๐‘ฅ)/(๐‘Ž^2 + ๐‘ก^2 ) .๐‘‘๐‘ก/(๐‘ ๐‘ ๐‘’๐‘^2 ๐‘ฅ)ใ€— ๐ผ=๐œ‹/๐‘ โˆซ1_0^โˆžโ–’๐‘‘๐‘ก/(๐‘Ž^2 + ๐‘ก^2 ) ๐ผ= ใ€–๐œ‹/๐‘ [1/๐‘Ž tan^(โˆ’1)โก(๐‘ก/๐‘Ž) ]ใ€—_0^โˆž Putting limits, I=๐œ‹/๐‘ [1/๐‘Ž ใ€–๐‘ก๐‘Ž๐‘›ใ€—^(โˆ’1) (โˆž/๐‘Ž)โˆ’1/๐‘Ž ใ€–๐‘ก๐‘Ž๐‘›ใ€—^(โˆ’1) (0/๐‘Ž)] I =๐œ‹/๐‘ [ใ€–1/๐‘Ž tan^(โˆ’1)ใ€—โกใ€–(โˆž)โˆ’1/๐‘Ž tan^(โˆ’1)โก(0) ใ€— ] I =๐œ‹/๐‘ (1/๐‘Ž (๐œ‹/2)โˆ’0) I =๐…^๐Ÿ/๐Ÿ๐’‚๐’ƒ

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.