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  1. Chapter 7 Class 12 Integrals (Term 2)
  2. Serial order wise

Transcript

Example 10 Find the following integrals: (ii) โˆซ1โ–’(๐‘ฅ + 3)/โˆš(5 โˆ’ 4๐‘ฅ โˆ’ ๐‘ฅ^2 ) ๐‘‘๐‘ฅ We can write numerator as ๐‘ฅ+3= A๐‘‘/๐‘‘๐‘ฅ [โˆ’๐‘ฅ^2โˆ’4๐‘ฅ+5]+ B ๐‘ฅ+3= A[โˆ’2๐‘ฅโˆ’4]+ B ๐‘ฅ+3=โˆ’2๐ด๐‘ฅโˆ’4A+B Finding A & B Comparing coefficient of ๐‘ฅ ๐‘ฅ=โˆ’2๐ด๐‘ฅ 1 =โˆ’2A A=(โˆ’1)/( 2) Comparing constant term 3=โˆ’4A+B 3=โˆ’4((โˆ’1)/( 2))+B 3=2+B B=3โˆ’2=1 Now, we know that ๐‘ฅ+3=A[โˆ’2๐‘ฅโˆ’4]+B ๐‘ฅ+3=(โˆ’1)/( 2) [โˆ’2๐‘ฅโˆ’4]+1 Now, our equation is โˆซ1โ–’(๐‘ฅ+3)/โˆš(5 โˆ’ 4๐‘ฅ + ๐‘ฅ^2 ).๐‘‘๐‘ฅ=โˆซ1โ–’(1/2 [โˆ’2๐‘ฅ โˆ’ 4]+1)/โˆš(5 โˆ’ 4๐‘ฅ โˆ’ ๐‘ฅ^2 ).๐‘‘๐‘ฅ =โˆซ1โ–’((โˆ’1)/2 [โˆ’2๐‘ฅ โˆ’ 4]+1)/โˆš(5 โˆ’ 4๐‘ฅ + ๐‘ฅ^2 ) ๐‘‘๐‘ฅ+โˆซ1โ–’1/โˆš(5 โˆ’ 4๐‘ฅ โˆ’ ๐‘ฅ^2 ) ๐‘‘๐‘ฅ =(โˆ’1)/2 โˆซ1โ–’(โˆ’2๐‘ฅ โˆ’ 4)/โˆš(5 โˆ’ 4๐‘ฅ + ๐‘ฅ^2 ) ๐‘‘๐‘ฅ+โˆซ1โ–’1/โˆš(5 โˆ’ 4๐‘ฅ โˆ’ ๐‘ฅ^2 ) ๐‘‘๐‘ฅ Solving I1 I1 = (โˆ’1)/2 โˆซ1โ–’ใ€–(โˆ’2๐‘ฅ โˆ’ 4)/โˆš(โˆ’๐‘ฅ^2 โˆ’ 4๐‘ฅ + 5).๐‘‘๐‘ฅใ€— Let t = โˆ’๐‘ฅ^2 โˆ’ 4๐‘ฅ + 5 Differentiating both sides w.r.t. ๐‘ฅ โˆ’2๐‘ฅ โˆ’ 4 =๐‘‘๐‘ก/๐‘‘๐‘ฅ ๐‘‘๐‘ฅ=๐‘‘๐‘ก/(โˆ’ 2๐‘ฅ โˆ’ 4) Now, I1 = (โˆ’1)/( 2) โˆซ1โ–’ใ€–(โˆ’2๐‘ฅ โˆ’ 4)/โˆš(โˆ’๐‘ฅ^2 โˆ’ 4๐‘ฅ + 5).๐‘‘๐‘ฅใ€— Putting the values of (โˆ’๐‘ฅ^2โˆ’4๐‘ฅ+5) and ๐‘‘๐‘ฅ I1 = (โˆ’1)/( 2) โˆซ1โ–’ใ€–(โˆ’2๐‘ฅ โˆ’ 4)/โˆš๐‘ก.๐‘‘๐‘ก/((โˆ’2๐‘ฅ โˆ’4) )ใ€— I1 = (โˆ’1)/( 2) โˆซ1โ–’ใ€–1/โˆš๐‘ก.๐‘‘๐‘กใ€— I1 = (โˆ’1)/( 2) โˆซ1โ–’ใ€–1/(๐‘ก)^(1/2) .๐‘‘๐‘กใ€— I1 = (โˆ’1)/( 2) โˆซ1โ–’ใ€–(๐‘ก)^((โˆ’1)/2) ๐‘‘๐‘กใ€— I1 = (โˆ’1)/( 2) ๐‘ก^((โˆ’1)/( 2) + 1)/(((โˆ’1)/( 2) + 1) )+๐ถ1 I1 = (โˆ’1)/( 2) ๐‘ก^(1/( 2) )/((1/2) )+๐ถ1 I1 = โˆ’ ๐‘ก^(1/( 2) )+๐ถ1 I1 = โˆ’โˆš๐‘ก+๐ถ1 Putting back ๐‘ก=๐‘ฅ^2โˆ’4๐‘ฅ+5) I1 = โˆ’โˆš(โˆ’๐‘ฅ^2โˆ’4๐‘ฅ+5)+๐ถ1 Solving ๐‘ฐ๐Ÿ I2 = โˆซ1โ–’ใ€–1/โˆš(5 โˆ’ 4๐‘ฅ โˆ’ ๐‘ฅ^2 ).๐‘‘๐‘ฅใ€— I2 = โˆซ1โ–’ใ€–1/โˆš(โˆ’(๐‘ฅ^2 + 4๐‘ฅ โˆ’ 5) ) ๐‘‘๐‘ฅใ€— I2 = โˆซ1โ–’ใ€–1/โˆš(โˆ’[๐‘ฅ^2 + 2(2)(๐‘ฅ) โˆ’ 5] ) ๐‘‘๐‘ฅใ€— I2 = โˆซ1โ–’ใ€–1/โˆš(โˆ’[๐‘ฅ^2 + 2(2)(๐‘ฅ) + (2)^2 โˆ’ (2)^2 โˆ’ 5] ) ๐‘‘๐‘ฅใ€— I2 = โˆซ1โ–’ใ€–1/โˆš(โˆ’[(๐‘ฅ + 2)^2 โˆ’ 4 โˆ’ 5] ) ๐‘‘๐‘ฅใ€— I2 = โˆซ1โ–’ใ€–1/โˆš(โˆ’[(๐‘ฅ + 2)^2 โˆ’ 9] ) ๐‘‘๐‘ฅใ€— I2 = โˆซ1โ–’ใ€–1/โˆš(9 โˆ’ (๐‘ฅ + 2)^2 ) ๐‘‘๐‘ฅใ€— I2 = โˆซ1โ–’ใ€–1/โˆš((3)^2 โˆ’ (๐‘ฅ + 2)^2 ) ๐‘‘๐‘ฅใ€— It is of form โˆซ1โ–’ใ€–1/โˆš(๐‘Ž^2 โˆ’ ๐‘ฅ^2 ).๐‘‘๐‘ฅ=ใ€–๐‘ ๐‘–๐‘›ใ€—^(โˆ’1)โกใ€–๐‘ฅ/๐‘Ž+๐ถ2ใ€— ใ€— Replacing ๐‘ฅ by (๐‘ฅ+2) and a by 3, we get I2 = sin^(โˆ’1)โกใ€–((๐‘ฅ + 2)/3)+๐ถ2ใ€— Now, Putting values of I1 and I2 in (1) โˆซ1โ–’ใ€–(๐‘ฅ + 3)/โˆš(5 โˆ’ 4๐‘ฅ โˆ’ ๐‘ฅ^2 ).๐‘‘๐‘ฅ=(โˆ’1)/( 2) โˆซ1โ–’(โˆ’ 2๐‘ฅ โˆ’ 4)/โˆš(5 โˆ’ 4๐‘ฅ โˆ’ ๐‘ฅ^2 )ใ€—+โˆซ1โ–’1/โˆš(5 โˆ’ 4๐‘ฅ โˆ’ ๐‘ฅ^2 ) ๐‘‘๐‘ฅ =โˆ’โˆš(5 โˆ’ 4๐‘ฅ โˆ’ ๐‘ฅ^2 )+๐ถ1+sin^(โˆ’1)โก((๐‘ฅ + 2)/3)+๐ถ2 =โˆ’โˆš(๐Ÿ“ โˆ’ ๐Ÿ’๐’™ โˆ’ ๐’™^๐Ÿ )+ใ€–๐’”๐’Š๐’ใ€—^(โˆ’๐Ÿ)โกใ€–((๐’™ + ๐Ÿ)/๐Ÿ‘)+๐‘ชใ€—

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.