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Ex 7.2
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Ex 7.2, 37 You are here
Ex 7.2, 38 (MCQ) Important
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Last updated at May 29, 2023 by Teachoo
Ex 7.2, 37 Integrate (๐ฅ3 ๐ ๐๐ (tan^(โ1)โกใ๐ฅ^4 ใ))/(1 + ๐ฅ8) โซ1โใ" " (๐ฅ3 ๐ ๐๐ (tan^(โ1)โกใ๐ฅ^4 ใ ))/(1 + ๐ฅ8)ใ . ๐๐ฅ Let tan^(โ1)โกใ๐ฅ^4 ใ= ๐ก Differentiating both sides ๐ค.๐.๐ก.๐ฅ 1/(1 +(๐ฅ^4 )^2 ). ๐(๐ฅ^4 )/๐๐ฅ= ๐๐ก/๐๐ฅ 1/(1 +๐ฅ^8 ). 4๐ฅ^3=๐๐ก/๐๐ฅ (4๐ฅ^3)/(1 + ๐ฅ^8 )=๐๐ก/๐๐ฅ (Using (๐(ใ๐ก๐๐ใ^(โ1)โก๐ฅ))/๐๐ฅ=1/(1 + ๐ฅ^2 ) and chain rule ) ๐๐ฅ=(1 + ๐ฅ^8)/(4๐ฅ^3 ) . ๐๐ก Now, our function becomes โซ1โใ" " (๐ฅ3 ๐ ๐๐ (tan^(โ1)โกใ๐ฅ^4 ใ ))/(1 + ๐ฅ8)ใ . ๐๐ฅ Putting ใ๐ก๐๐ใ^(โ1)โกใ๐ฅ^4 ใ=๐ก & ๐๐ฅ=(1 + ๐ฅ^8)/(4๐ฅ^3 ) . ๐๐ก = โซ1โใ" " (๐ฅ^3 ๐ ๐๐ (๐ก))/(1 + ๐ฅ^8 )ใ. (1 + ๐ฅ^8)/(4๐ฅ^3 ) ๐๐ก" " = โซ1โใ" " sinโก๐ก/4ใ ๐๐ก" " = 1/4 โซ1โsinโก๐ก . ๐๐ก" " = (โ1)/4 cosโก๐ก+ ๐ถ = (โ๐)/๐ ใ๐๐๐ ใโก(ใ๐ญ๐๐งใ^(โ๐)โกใ๐^๐ ใ )+๐ช (Using ๐ก=ใ๐ก๐๐ใ^(โ1)โกใ๐ฅ^4 ใ)