Solving $\int\frac{1}{\cos x}$ without knowing anything about secant!

Question

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

Answer 1

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 $u = \sin x$.

Answer 2

ΗΙΝΤ

Make the substitution $t=\tan{\frac{x}{2}}$

From this you have that $$dx=\frac{2}{1+t^2}dt$$ $$\cos{x}=\frac{1-t^2}{1+t^2}$$

Answer 3

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 $n=-1$, $m=0$.