Products of secants and tangents have strategies, in the same
way that the products of sines and cosines have strategies.
Based on the even and odd powers, we evaluate $$\int \sec^n(x)
\tan^m(x)\, dx.$$

If $n=2$, we can use the substitution $u=\tan(x)$,
$du=\sec^2(x)\, dx$ to get $\int u^m\, du$.

If $n$ is even, we can use the identity
$\sec^2(x)=1+\tan^2(x)$ to convert all but two powers of secant
into tangents. (If $n=0$ and $m$ is large, things get tedious,
but we can convert two powers of tangent into secants instead,
thereby reducing the power of tangent. Then repeat on the
reduce power of tangent.)

If $m=1$ (and $n>1$), we can substitute $u=\sec(x)$,
$du=\sec(x)\tan(x)\, dx$ to get $\int u^{n-1}\, du$. (If
$m$ and $n$ are both one, we know the antiderivative.)

If $m$ is odd, we can use the identity $\tan^2(x) =
\sec^2(x)-1$ to convert all but one power of tangent into
secants.

Other cases can be much harder. We look at more
challenging integrals later in this module.

Alternatively,
as we did with sines and cosines, we can observe the
combinations of secant and tangent functions in the integrand,
and look for a derivative to strip off.

Possible derivatives to strip off:
$\sec^2x\,dx$ with the remaining factors of the integrand
containing only tangent functions (which may involve
substituting $\sec^2 x=\tan^2 x+1$); and $\sec x\tan x\,dx$,
with the remaining factors of the integrand containing only
secant functions (which may involve substituting $\tan^2
x=\sec^2 x-1$). For some integrands, this cannot be done,
but we will not ask you to compute
such integrals.

Some examples:

$\int\sec^6x\tan^5x\,dx=\int\sec^4x\tan^5x\sec^2x\,dx$.
Here, we stripped off $\sec^2x\,dx$ since we know the remaining
even power of $\sec x$ can be rewritten in terms of $\tan
x$. Now we make our substitution to get $\int(\tan^2
x+1)^2\tan^5x\sec^2\,dx$ and let $u=\tan x$. This will
work for any power of tangent.

$\int\sec^5x\tan^7x\,dx$. Here, we don't
want to strip off $\sec^2x\,dx$, since the remaining power of
secant cannot be turned into tangents. So try stripping
off $\sec x\tan x\,dx$, getting $\int\sec^4x\tan^6x\sec x\tan
x\,dx$. Now we can set $\tan^6x=(\sec^2x-1)^3$ and let
$u=\sec x$.