Ex 7.2, 20 - Chapter 7 Class 12 Integrals
Last updated at April 16, 2024 by Teachoo
Integration by substitution - e^x
Integration by substitution - e^x
Last updated at April 16, 2024 by Teachoo
Ex 7.2, 20 Integrate the function (๐^2๐ฅ โ ๐^(โ2๐ฅ))/(๐^2๐ฅ + ๐^(โ2๐ฅ) ) Let ๐^2๐ฅ + ๐^(โ2๐ฅ)= ๐ก Differentiating both sides ๐ค.๐.๐ก.๐ฅ ๐^2๐ฅ. ๐(2๐ฅ)/๐๐ฅ +๐^(โ2๐ฅ) ๐(โ2๐ฅ)/๐๐ฅ= ๐๐ก/๐๐ฅ ใ2๐ใ^2๐ฅโใ2๐ใ^(โ2๐ฅ)= ๐๐ก/๐๐ฅ 2(๐^2๐ฅโ๐^(โ2๐ฅ) )=๐๐ก/๐๐ฅ " " ๐๐ฅ = ๐๐ก/2(๐^2๐ฅโ ๐^(โ2๐ฅ) ) Integrating the function โซ1โใ" " (๐^2๐ฅ โ ๐^(โ2๐ฅ))/(๐^2๐ฅ + ๐^(โ2๐ฅ) )ใ. ๐๐ฅ Putting ๐^2๐ฅ + ๐^(โ2๐ฅ)=๐ก & ๐๐ฅ=๐๐ก/2(๐^2๐ฅโ ๐^(โ2๐ฅ) ) = โซ1โใ" " (๐^2๐ฅ โ ๐^(โ2๐ฅ))/๐กใ. ๐๐ก/2(๐^2๐ฅโ ๐^(โ2๐ฅ) ) = โซ1โใ" " 1/2๐กใ. ๐๐ก = 1/2 โซ1โ1/๐ก. ๐๐ก = 1/2 logโกใ |๐ก|ใ+๐ถ = 1/2 logโกใ |๐^2๐ฅ + ๐^(โ2๐ฅ) |ใ+๐ถ = ๐/๐ ๐๐๐โกใ (๐^๐๐ + ๐^(โ๐๐) )ใ+๐ช (Using โซ1โ1/๐ฅ. ๐๐ฅ=๐๐๐โก|๐ฅ| ) (Using ๐ก=๐^2๐ฅ + ๐^(โ2๐ฅ)) (โด ๐^2๐ฅ+๐^(โ2๐ฅ)>0 )