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

Ex 7.4, 17 Integrate the function (๐‘ฅ + 2)/โˆš(๐‘ฅ^2 โˆ’ 1) โˆซ1โ–’(๐‘ฅ + 2)/โˆš(๐‘ฅ^2 โˆ’ 1) . ๐‘‘๐‘ฅ=โˆซ1โ–’(1/2 (2๐‘ฅ) + 2" " )/โˆš(๐‘ฅ^2 โˆ’ 1) . ๐‘‘๐‘ฅ =โˆซ1โ–’(1/2 (2๐‘ฅ))/โˆš(๐‘ฅ^2 โˆ’ 1) . ๐‘‘๐‘ฅ+โˆซ1โ–’2/โˆš(๐‘ฅ^2 โˆ’ 1) . ๐‘‘๐‘ฅ =1/2 โˆซ1โ–’( 2๐‘ฅ)/โˆš(๐‘ฅ^2 โˆ’ 1) . ๐‘‘๐‘ฅ+โˆซ1โ–’2/โˆš(๐‘ฅ^2 โˆ’ 1) . ๐‘‘๐‘ฅ Solving ๐‘ฐ๐Ÿ I1=1/2 โˆซ1โ–’( 2๐‘ฅ)/โˆš(๐‘ฅ^2 โˆ’ 1) ๐‘‘๐‘ฅ โ€ฆ(1) Let ๐‘ฅ^2โˆ’1=๐‘ก Differentiating w.r.t. x 2๐‘ฅโˆ’0=๐‘‘๐‘ก/๐‘‘๐‘ฅ ๐‘‘๐‘ฅ=๐‘‘๐‘ก/2๐‘ฅ Thus, our equation becomes I1=1/2 โˆซ1โ–’( 2๐‘ฅ)/โˆš(๐‘ฅ^2 โˆ’ 1) ๐‘‘๐‘ฅ Put the values of (๐‘ฅ^2โˆ’1)=๐‘ก and ๐‘‘๐‘ฅ, we get I1=1/2 โˆซ1โ–’( 2๐‘ฅ)/โˆš๐‘ก ๐‘‘๐‘ฅ I1=1/2 โˆซ1โ–’( 2๐‘ฅ)/โˆš๐‘ก ร— ๐‘‘๐‘ก/2๐‘ฅ I1=1/2 โˆซ1โ–’( 1)/โˆš๐‘ก ๐‘‘๐‘ก I1=1/2 โˆซ1โ–’1/๐‘ก^(1/2) ๐‘‘๐‘ก I1=1/2 โˆซ1โ–’๐‘ก^((โˆ’1)/2) ๐‘‘๐‘ก I1=1/2 ๐‘ก^((โˆ’1)/2 + 1)/((โˆ’1)/2 + 1) +๐ถ1 I1=1/2 ๐‘ก^(1/2)/(1/2) +๐ถ1 I1=๐‘ก^(1/2)+๐ถ1 I1=โˆš๐‘ก+๐ถ1 I1=โˆš(๐‘ฅ^2 โˆ’ 1) + ๐ถ1 Solving ๐‘ฐ๐Ÿ I2=โˆซ1โ–’2/โˆš(๐‘ฅ^2 โˆ’ 1) . ๐‘‘๐‘ฅ I2=2โˆซ1โ–’1/โˆš(๐‘ฅ^2 โˆ’ (1)^2 ) . ๐‘‘๐‘ฅ I2=2 ๐‘™๐‘œ๐‘”โก|๐‘ฅ+โˆš(๐‘ฅ^2 โˆ’1)|+๐ถ2 It is of form โˆซ1โ–’๐‘‘๐‘ฅ/โˆš(๐‘ฅ^2 โˆ’ ๐‘Ž^2 ) =๐‘™๐‘œ๐‘”โก|๐‘ฅ+โˆš(๐‘ฅ^2 โˆ’ ๐‘Ž^2 )|+๐ถ โˆด Replacing a by 1 , we get ("Using " ๐‘ก=๐‘ฅ^2โˆ’1) Now, Putting the values of I1 and I2 in (1) โˆซ1โ–’(๐‘ฅ + 2)/โˆš(๐‘ฅ^2 โˆ’ 1) . ๐‘‘๐‘ฅ=1/2 โˆซ1โ–’( 2๐‘ฅ)/โˆš(๐‘ฅ^2 โˆ’ 1) . ๐‘‘๐‘ฅ+2โˆซ1โ–’1/โˆš(๐‘ฅ^2 โˆ’ 1) . ๐‘‘๐‘ฅ =โˆš(๐‘ฅ^2 โˆ’ 1) + ๐ถ1+2 ๐‘™๐‘œ๐‘”โก|๐‘ฅ+โˆš(๐‘ฅ^2 โˆ’1) |+๐ถ2 =โˆš(๐’™^๐Ÿ โˆ’ ๐Ÿ)+๐Ÿ ๐’๐’๐’ˆโก|๐’™+โˆš(๐’™^๐Ÿ โˆ’๐Ÿ)|+ ๐‘ช

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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.