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Question-137209




Question Number 137209 by JulioCesar last updated on 31/Mar/21
Answered by bemath last updated on 31/Mar/21
∫ (sec^2 x−1)^2 sec x dx  =∫(sec^5 x−2sec^3 x+sec x )dx  now it easy to solve
$$\int\:\left(\mathrm{sec}\:^{\mathrm{2}} \mathrm{x}−\mathrm{1}\right)^{\mathrm{2}} \mathrm{sec}\:\mathrm{x}\:\mathrm{dx} \\ $$$$=\int\left(\mathrm{sec}\:^{\mathrm{5}} \mathrm{x}−\mathrm{2sec}\:^{\mathrm{3}} \mathrm{x}+\mathrm{sec}\:\mathrm{x}\:\right)\mathrm{dx} \\ $$$$\mathrm{now}\:\mathrm{it}\:\mathrm{easy}\:\mathrm{to}\:\mathrm{solve} \\ $$

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