I already know that there are multiple answers of integrating $\int_{}\sec^3x$. My textbook uses an integration by parts approach and I also have seen a partial functions approach as well.
I stumbled upon this post and one of the users @N.S. posted an interesting approach that used substitution, but I wasn't able to work the problem out completely...Here is where I got to
$$\int_{} \sec^3(x) = \int \frac{1}{\cos^3x} * \frac{\cos(x)}{\cos(x)} = \int \frac{\cos(x)}{\cos^4(x)} = \int \frac{\cos(x)}{(1- \sin^2(x))^2}dx $$
Apply substitution with $u = \sin(x), du = \cos(x)$:
$$\int \frac{du}{(1-u^2)^2} = \int \frac{du}{1 - 2u^2 + u^4} = \frac{1}{\frac{2}{3}u^3 - \frac{u^5}{5}} $$
But this just seems wrong and nowhere close to the answer of $\frac{1}{2} (\sec(x) \tan(x) + ln| \sec(x) + \tan (x)|) + C$