Ex 7.2, 19 - Chapter 7 Class 12 Integrals (Term 2)
Last updated at Dec. 20, 2019 by Teachoo
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Chapter 7 Class 12 Integrals
Serial order wise
Transcript
Ex 7.2, 19 Integrate the function (𝑒2𝑥 − 1)/(𝑒2𝑥+ 1) Simplify the given function (𝑒^2𝑥 − 1)/(𝑒^2𝑥 + 1) Dividing numerator and denominator by ex, we obtain = (𝑒^2𝑥/𝑒^𝑥 " " −" " 𝟏/𝒆^𝒙 )/(𝑒^2𝑥/𝑒^𝑥 " " + " " 𝟏/𝒆^𝒙 ) = (𝑒^𝒙 − 𝒆^(−𝒙))/(𝑒^𝒙 + 𝒆^(−𝒙) ) Let 𝑒^𝑥 + 𝑒^(−𝑥)= 𝑡 Differentiating both sides 𝑤.𝑟.𝑡.𝑥 𝑒^𝑥+(−1) 𝑒^(−𝑥)= 𝑑𝑡/𝑑𝑥 𝑒^𝑥−𝑒^(−𝑥)= 𝑑𝑡/𝑑𝑥 𝑑𝑥=𝑑𝑡/(𝑒^𝑥 − 𝑒^(−𝑥) ) Now, Integrating the function ∫1▒〖" " (𝑒^2𝑥 − 1)/(𝑒^2𝑥 + 1) " " 〗. 𝑑𝑥 = ∫1▒〖" " (𝑒^𝑥 − 𝑒^(−𝑥))/(𝑒^𝑥 + 𝑒^(−𝑥) ) " " 〗. 𝑑𝑥 (Using (𝑒^2𝑥 − 1)/(𝑒^2𝑥 + 1)=(𝑒^𝑥 − 𝑒^(−𝑥))/(𝑒^𝑥 + 𝑒^(−𝑥) ) ) Putting 𝑒^𝑥 + 𝑒^(−𝑥)=𝑡 & 𝑑𝑥=𝑑𝑡/(𝑒^𝑥 − 𝑒^(−𝑥) ) =∫1▒〖" " (𝑒^𝑥 − 𝑒^(−𝑥))/𝑡 " " 〗. 𝑑𝑡/(𝑒^𝑥 − 𝑒^(−𝑥) ) " " =∫1▒〖" " 1/𝑡 " " 〗. 𝑑𝑡 =log|𝑡|+𝐶 =log〖 |𝑒^𝑥+𝑒^(−𝑥) |〗+𝐶 =𝒍𝒐𝒈(𝒆^𝒙 + 𝒆^(−𝒙) )+𝑪 (Using 𝑡=𝑒^𝑥 + 𝑒^(−𝑥)) (As 𝑒^𝑥+𝑒^(−𝑥)>0 )
