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Ex 7.2, 12 - Integrate (x3 - 1)1/3 .x5 - Chapter 7 CBSE - Integration by substitution - x^n

Ex 7.2, 12 - Chapter 7 Class 12 Integrals - Part 2
Ex 7.2, 12 - Chapter 7 Class 12 Integrals - Part 3 Ex 7.2, 12 - Chapter 7 Class 12 Integrals - Part 4

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Ex 7.2, 12 Integrate the function: (๐‘ฅ3 โ€“ 1)^(1/3) . ๐‘ฅ5 (๐‘ฅ3 โ€“ 1)^(1/3) . ๐‘ฅ5 Step 1: Let ๐‘ฅ3= ๐‘ก Differentiating both sides ๐‘ค.๐‘Ÿ.๐‘ก.๐‘ฅ 3๐‘ฅ^2= ๐‘‘๐‘ก/๐‘‘๐‘ฅ 3๐‘ฅ^2. d๐‘ฅ=๐‘‘๐‘ก ๐‘‘๐‘ฅ = ๐‘‘๐‘ก/(3๐‘ฅ^2 ) Step 2: Integrating the function โˆซ1โ–’ใ€–" " (๐‘ฅ3 โ€“ 1)^(1/3) . ๐‘ฅ5" " ใ€— . ๐‘‘๐‘ฅ Putting the value of ๐‘ฅ^3 & ๐‘‘๐‘ฅ=๐‘‘๐‘ก/(3๐‘ฅ^2 ) = โˆซ1โ–’ใ€–" " (๐‘ก โ€“ 1)^(1/3) . ๐‘ฅ5ใ€— . ๐‘‘๐‘ก/(3๐‘ฅ^2 ) = โˆซ1โ–’ใ€–" " (๐‘ก โ€“ 1)^(1/3) . ๐‘ฅ^2. ๐‘ฅ^3 ใ€—. ๐‘‘๐‘ก/(3๐‘ฅ^2 ) = โˆซ1โ–’ใ€–" " (๐‘ก โ€“ 1)^(1/3) ใ€— . ๐‘ฅ^3/3 . ๐‘‘๐‘ก = โˆซ1โ–’ใ€–" " (๐‘ก โ€“ 1)^(1/3) ใ€— . ๐‘ก/3 . ๐‘‘๐‘ก = 1/3 โˆซ1โ–’ใ€–" " (๐‘ก โ€“ 1)^(1/3) ใ€— . ๐‘ก . ๐‘‘๐‘ก = 1/3 โˆซ1โ–’ใ€–" " (๐‘ก โ€“ 1)^(1/3) ใ€— . (๐‘กโˆ’1+1) ๐‘‘๐‘ก = 1/3 โˆซ1โ–’ใ€–" " (๐‘ก โ€“ 1)^(1/3) ใ€— . ((๐‘กโˆ’1)+1) ๐‘‘๐‘ก = 1/3 โˆซ1โ–’((๐‘ก โ€“ 1)^(1/3) (๐‘กโˆ’1)+(๐‘กโˆ’1)^(1/3) ) ๐‘‘๐‘ก = 1/3 โˆซ1โ–’((๐‘ก โ€“ 1)^(1/3 +1)+(๐‘กโˆ’1)^(1/3) ) ๐‘‘๐‘ก = 1/3 โˆซ1โ–’((๐‘ก โ€“ 1)^(4/3 )+(๐‘กโˆ’1)^(1/3) ) ๐‘‘๐‘ก = 1/3 โˆซ1โ–’ใ€– (๐‘ก โ€“ 1)^(4/3 ). ๐‘‘๐‘กใ€— + 1/3 โˆซ1โ–’ใ€– (๐‘ก โ€“ 1)^(1/3 ). ๐‘‘๐‘กใ€— = 1/3 โˆซ1โ–’ใ€– (๐‘ก โ€“1)^(4/3 ). ๐‘‘๐‘กใ€— + 1/3 โˆซ1โ–’ใ€– (๐‘ก โ€“1)^(1/3 ). ๐‘‘๐‘กใ€— = 1/3 (๐‘ก โ€“1)^(4/3 + 1)/(4/3 + 1) + 1/3 (๐‘ก โ€“1)^(1/3 + 1)/(1/3 + 1) + ๐ถ = 1/3 (๐‘ก โˆ’ 1)^(7/3)/(7/3) + 1/3 (๐‘ก โˆ’ 1)^(4/3)/(4/3) + ๐ถ = 1/7 (๐‘กโˆ’1)^(7/3) +1/4 (๐‘กโˆ’1)^(4/3) +๐ถ = ๐Ÿ/๐Ÿ• (๐’™^๐Ÿ‘ โ€“๐Ÿ)^(๐Ÿ•/๐Ÿ‘) + ๐Ÿ/๐Ÿ’ (๐’™^๐Ÿ‘โ€“๐Ÿ)^(๐Ÿ’/๐Ÿ‘) + ๐‘ช

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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.