Solving $\int\frac{1}{\cos x}$ without knowing anything about secant! As the title says, I am trying to solve this integral without using anything related to secant, it might sound stupid but I want to know if it can be done.   
I have tried the following approach but it didn't lead me anywhere:
$$ \int\frac{1}{\cos x} = \int\frac{\sin^2 x+\cos^2 x}{cosx}$$
$$ = \int\frac{\sin^2}{\cos^2 x} +  \int\cos x$$
$$ = -\sin x + \int\frac{\sin^2 x}{\cos x} $$

You can start out like this:
$$
\int \frac{dx}{\cos x} 
= \int \frac{\cos x \, dx}{\cos^2 x} 
= \int \frac{\cos x \, dx}{1-\sin^2 x}
.
$$
Then let $u = \sin x$.
