# Ex 7.7, 2 - Chapter 7 Class 12 Integrals (Term 2)

Last updated at Dec. 20, 2019 by Teachoo

Last updated at Dec. 20, 2019 by Teachoo

Transcript

Ex 7.7, 2 (Method 1) โ(1โ4๐ฅ2) โซ1โใโ(1โ4๐ฅ^2 ).๐๐ฅใ =โซ1โใโ(4(1/4โ๐ฅ^2 ) ).๐๐ฅใ =โซ1โใโ4 โ(1/4โ๐ฅ^2 ).๐๐ฅใ =2โซ1โใโ((1/2)^2โ๐ฅ^2 ).๐๐ฅ" " ใ It is of the form โซ1โใโ(๐^2โ๐ฅ^2 ) .๐๐ฅ=1/2 ๐ฅโ(๐^2โ๐ฅ^2 )+๐^2/2 ๐ ๐๐^(โ1) ๐ฅ/๐+๐ถ1ใ Replacing a with 2 and 1/2 , we get =2[1/2 ๐ฅโ((1/2)^2โ๐ฅ^2 ) +(1/2)^2.1/2 ๐ ๐๐^(โ1) ๐ฅ/(1/2)+๐ถ1] =2[1/2 ๐ฅโ(1/4โ๐ฅ^2 ) +1/4 . 1/2 ๐ ๐๐^(โ1) 2๐ฅ+๐ถ1] =๐ฅโ(1/4โ๐ฅ^2 )+1/4 ๐ ๐๐^(โ1) 2๐ฅ+2๐ถ1 =1/4 ๐ ๐๐^(โ1) 2๐ฅ+๐ฅโ((1 โ 4๐ฅ^2)/4)+๐ถ =๐/๐ ๐๐๐^(โ๐) ๐๐+๐/๐ ๐โ(๐ โ ๐๐^๐ )++๐ช [where ๐ถ=2๐ถ1] Ex 7.7, 2 (Method 2) โ(1โ4๐ฅ2) Let 2๐ฅ=๐ก Differentiating both sides w.r.t. ๐ฅ 2=๐๐ก/๐๐ฅ ๐๐ฅ=๐๐ก/2 Integrating the function โซ1โใโ(1โ4๐ฅ^2 ) ๐๐ฅใ =โซ1โใโ(1โ(2๐ฅ)^2 ) ๐๐ฅใ Putting value of t = 2๐ฅ and ๐๐ฅ = ๐๐ก/2 =โซ1โใโ(1โ๐ก^2 ) .๐๐ก/2ใ =1/2 โซ1โใโ(1โ๐ก^2 ) ๐๐กใ =1/2 โซ1โใโ((1)^2โ(๐ก)^2 ) ๐๐กใ =1/2 [๐ก/2 โ((1)^2โ๐ก^2 )+(1)^2/2 ๐ ๐๐^(โ1) ๐ก/((1) )+๐ถ1] =๐ก/4 โ(1โ๐ก^2 )+1/4 ๐ ๐๐^(โ1) (๐ก)+๐ถ1" " /2 =๐ก/4 โ(1โ๐ก^2 )+1/4 ๐ ๐๐^(โ1) (๐ก)+๐ถ It is of the form โซ1โใโ(๐^2โ๐ฅ^2 ) ๐๐ฅ=1/2 ๐ฅโ(๐^2โ๐ฅ^2 )+๐^2/2 ๐ ๐๐^(โ1) ๐ฅ/๐ +๐ถ1ใ Replacing a by 1 and ๐ฅ by ๐ก , we get Putting back t = 2x =2๐ฅ/4 โ(1โ(2๐ฅ)^2 )+1/4 ๐ ๐๐^(โ1) 2๐ฅ+๐ถ =๐/๐ ๐๐๐^(โ๐) ๐๐+๐/๐ โ(๐โ๐๐^๐ )+๐ช

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Ex 7.7, 12 (Supplementary NCERT) Important Deleted for CBSE Board 2022 Exams

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Ex 7.7, 14 (Supplementary NCERT) Important Deleted for CBSE Board 2022 Exams

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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.