Ex 7.2, 37 - Integrate x^3 sin (tan-1 x^4) / 1 + x^8 - Teachoo

Ex 7.2, 37 - Chapter 7 Class 12 Integrals - Part 2

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Transcript

Ex 7.2, 37 Integrate (๐‘ฅ3 ๐‘ ๐‘–๐‘› (tan^(โˆ’1)โกใ€–๐‘ฅ^4 ใ€—))/(1 + ๐‘ฅ8) โˆซ1โ–’ใ€–" " (๐‘ฅ3 ๐‘ ๐‘–๐‘› (tan^(โˆ’1)โกใ€–๐‘ฅ^4 ใ€— ))/(1 + ๐‘ฅ8)ใ€— . ๐‘‘๐‘ฅ Let tan^(โˆ’1)โกใ€–๐‘ฅ^4 ใ€—= ๐‘ก Differentiating both sides ๐‘ค.๐‘Ÿ.๐‘ก.๐‘ฅ 1/(1 +(๐‘ฅ^4 )^2 ). ๐‘‘(๐‘ฅ^4 )/๐‘‘๐‘ฅ= ๐‘‘๐‘ก/๐‘‘๐‘ฅ 1/(1 +๐‘ฅ^8 ). 4๐‘ฅ^3=๐‘‘๐‘ก/๐‘‘๐‘ฅ (4๐‘ฅ^3)/(1 + ๐‘ฅ^8 )=๐‘‘๐‘ก/๐‘‘๐‘ฅ (Using (๐‘‘(ใ€–๐‘ก๐‘Ž๐‘›ใ€—^(โˆ’1)โก๐‘ฅ))/๐‘‘๐‘ฅ=1/(1 + ๐‘ฅ^2 ) and chain rule ) ๐‘‘๐‘ฅ=(1 + ๐‘ฅ^8)/(4๐‘ฅ^3 ) . ๐‘‘๐‘ก Now, our function becomes โˆซ1โ–’ใ€–" " (๐‘ฅ3 ๐‘ ๐‘–๐‘› (tan^(โˆ’1)โกใ€–๐‘ฅ^4 ใ€— ))/(1 + ๐‘ฅ8)ใ€— . ๐‘‘๐‘ฅ Putting ใ€–๐‘ก๐‘Ž๐‘›ใ€—^(โˆ’1)โกใ€–๐‘ฅ^4 ใ€—=๐‘ก & ๐‘‘๐‘ฅ=(1 + ๐‘ฅ^8)/(4๐‘ฅ^3 ) . ๐‘‘๐‘ก = โˆซ1โ–’ใ€–" " (๐‘ฅ^3 ๐‘ ๐‘–๐‘› (๐‘ก))/(1 + ๐‘ฅ^8 )ใ€—. (1 + ๐‘ฅ^8)/(4๐‘ฅ^3 ) ๐‘‘๐‘ก" " = โˆซ1โ–’ใ€–" " sinโก๐‘ก/4ใ€— ๐‘‘๐‘ก" " = 1/4 โˆซ1โ–’sinโก๐‘ก . ๐‘‘๐‘ก" " = (โˆ’1)/4 cosโก๐‘ก+ ๐ถ = (โˆ’๐Ÿ)/๐Ÿ’ ใ€–๐’„๐’๐’” ใ€—โก(ใ€–๐ญ๐š๐งใ€—^(โˆ’๐Ÿ)โกใ€–๐’™^๐Ÿ’ ใ€— )+๐‘ช (Using ๐‘ก=ใ€–๐‘ก๐‘Ž๐‘›ใ€—^(โˆ’1)โกใ€–๐‘ฅ^4 ใ€—)

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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.