Check sibling questions


Transcript

Ex 7.4, 8 Integrate ๐‘ฅ^2/โˆš(๐‘ฅ^6 + ๐‘Ž^6 ) Let ๐‘ฅ^3=๐‘ก Differentiating both sides w.r.t. x 3๐‘ฅ^2=๐‘‘๐‘ก/๐‘‘๐‘ฅ ๐‘‘๐‘ฅ=๐‘‘๐‘ก/(3๐‘ฅ^2 ) Integrating the function โˆซ1โ–’๐‘ฅ^2/โˆš(๐‘ฅ^6 + ๐‘Ž^6 ) ๐‘‘๐‘ฅ=โˆซ1โ–’๐‘ฅ^2/โˆš((๐‘ฅ^3 )^2 + (๐‘Ž^3 )^2 ) ๐‘‘๐‘ฅ Putting values of ๐‘ฅ^3=๐‘ก and ๐‘‘๐‘ฅ=๐‘‘๐‘ก/(3๐‘ฅ^2 ) , we get =โˆซ1โ–’๐‘ฅ^2/โˆš(๐‘ก^2 + (๐‘Ž^3 )^2 ) ๐‘‘๐‘ฅ =โˆซ1โ–’๐‘ฅ^2/โˆš(๐‘ก^2 + (๐‘Ž^3 )^2 ) . ๐‘‘๐‘ก/(3๐‘ฅ^2 ) =โˆซ1โ–’1/โˆš((๐‘ก^2 + (๐‘Ž^3 )^2 ) ) . ๐‘‘๐‘ก/3 =1/3 โˆซ1โ–’๐‘‘๐‘ก/โˆš(๐‘ก^2 + (๐‘Ž^3 )^2 ) =1/3 [logโก|๐‘ก+โˆš(๐‘ก^2 + (๐‘Ž^3 )^2 )|+๐ถ1] It is of form โˆซ1โ–’๐‘‘๐‘ฅ/โˆš(๐‘ฅ^2 + ๐‘Ž^2 ) =logโก|๐‘ฅ+โˆš(๐‘ฅ^2 + ๐‘Ž^2 )|+๐ถ1 โˆด Replacing ๐‘ฅ by ๐‘ก and a by ๐‘Ž^3, we get =1/3 logโก|๐‘ก+โˆš(๐‘ก^2 + ๐‘Ž^6 ) |+๐ถ =1/3 logโก|๐‘ฅ^3+โˆš((๐‘ฅ^3 )^2 + ๐‘Ž^6 ) |+๐ถ =๐Ÿ/๐Ÿ‘ ๐’๐’๐’ˆโก|๐’™^๐Ÿ‘+โˆš(๐’™^๐Ÿ”+ ๐’‚^๐Ÿ” ) |+๐‘ช ("Using" ๐‘ก=๐‘ฅ^3 )

  1. Chapter 7 Class 12 Integrals
  2. Serial order wise

About the Author

Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo