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:
$$ \int\frac{1}{\cos x} = \int\frac{\sin^2 x+\cos^2 x}{cosx}$$
$$ = \int\frac{\sin^2}{\cos^2 x} +  \int\cos x$$
$$ = -\sin x + \int\frac{\sin^2 x}{\cos x} $$

A general rule for finding $\int\cos^n\sin^m$ that works for 3/4 of the $(n,m)\in\Bbb Z^2$: If $n$ is odd let $u=\sin$; if $m$ is odd let $u=\cos$.
That works if $n=-1$, $m=0$.
