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To find the integral of the sec(x), you have to substitute a term that is not immediately obvious.

[tex]\int sec(x) dx = \int sec(x) \frac{sec(x)+tan(x)}{sec(x)+tan(x)} dx[/tex]

[tex]u= sec(x)+tan(x)[/tex]

[tex]du= (sec(x)tan(x)+sec^{2}(x))dx[/tex]

[tex]\int sec(x) \frac{sec(x)+tan(x)}{sec(x)+tan(x)} dx = \int \frac{sec^{2}(x)+sec(x)tan(x)dx}{sec(x)+tan(x)} dx = \int \frac{du}{u} = ln |u| + C = ln |sec(x)+tan(x)| + C[/tex]

My question is: What method is used to find this clever substitution term? All the online resources tell me what the term is, but not how they arrived at it. I understand that the goal is to eliminate the x variable and be left with u and du, but for some reason when I try to "reverse engineer" the process, I cannot arrive at u. I ask for a method that would work with other situations where a substitution term is not immediately obvious.

Thanks!

[tex]\int sec(x) dx = \int sec(x) \frac{sec(x)+tan(x)}{sec(x)+tan(x)} dx[/tex]

[tex]u= sec(x)+tan(x)[/tex]

[tex]du= (sec(x)tan(x)+sec^{2}(x))dx[/tex]

[tex]\int sec(x) \frac{sec(x)+tan(x)}{sec(x)+tan(x)} dx = \int \frac{sec^{2}(x)+sec(x)tan(x)dx}{sec(x)+tan(x)} dx = \int \frac{du}{u} = ln |u| + C = ln |sec(x)+tan(x)| + C[/tex]

My question is: What method is used to find this clever substitution term? All the online resources tell me what the term is, but not how they arrived at it. I understand that the goal is to eliminate the x variable and be left with u and du, but for some reason when I try to "reverse engineer" the process, I cannot arrive at u. I ask for a method that would work with other situations where a substitution term is not immediately obvious.

Thanks!

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