Why is the reduction-formula for $\int\sec^n(x)dx$ only valid for $n\ge3$? I have a question regarding the following reduction formula:
$$I_n=\int\sec^n(x)dx=\frac{1}{n-1}(\sec^{n-2}x\tan x)+\frac{n-2}{n-1}I_{n-2}+C$$
My calculus book states that it is only valid for $n\ge3$. Why is this the case? How does one intuit such a result?
Sure enough, $I_1$ breaks down because of division by zero. But what about $n=2$? And why can't $n=0$ be a base-case? Or even $n=-1$ etc.?
Finally: How do I know for sure it will actually work with all $n\ge3$? 
 A: Consider the beginning of the derivation:
$$\int \sec^n(x) dx = \int \sec(x)^{n-2} \sec(x)^2 dx = \sec(x)^{n-2} \tan(x) - \int (n-2) \sec(x)^{n-2} \tan(x)^2 dx.$$
In the case $n=1$, this is true but it is not useful, because you get the same multiple of $I_1$ on the right side as you already had, so you can't isolate $I_1$. You just get the trivial equation $I_1=I_1$.
In the case $n=2$, this technically works (the second term is just zero). But it only works because you've already used the answer in taking the first step (you had to integrate $\sec(x)^2$ to do the integration by parts in the first place). So it doesn't make sense to think of it as part of the recursion for even $n$, instead it is the base case of the recursion for even $n$.
A: As the comments are pointing out, it actually does work for $n=2$.
To see why certain values of $n$ are excluded, you have to look at how the formula is proven.
Probably the formula is proven by integration by parts. We have 
$$ \int \sec^n(x)dx = \int\sec^2(x)\sec^{n-2}(x)dx = \tan(x)\sec^{n-2}(x) - \int\tan (x)\frac{d\sec^{n-2}(x)}{dx}dx $$
. The derivative of $\sec^{n-2}(x) = (\sec(x))^{n-2}$ is $(n-2)\sec^{n-3}(x)\sec(x)\tan(x)$ .
Plugging this in gives 
$$ \int \sec^n(x)dx = \tan(x)\sec^{n-2}(x) - (n-2)\int\tan^2 (x)\sec^{n-2}(x)dx $$
Now we use the identity $\tan^2(x) = \sec^2(x) - 1$ to make this 
$$ \int \sec^n(x)dx = \tan(x)\sec^{n-2}(x) - (n-2)\int\sec^{n}(x)dx + (n-2)\int\sec^{n-2}(x)dx $$
or 
$$I_n = \tan(x)\sec^{n-2}(x) - (n-2)I_n + (n-2)I_{n-2}$$
Adding $(n-2)I_{n}$ to both sides gives
$$(n-1)I_n = \tan(x)\sec^{n-2}(x) + (n-2)I_{n-2}$$
Now we get the formula by dividing by $n-1$ -- which we can do for any value of $n$, except $n=1$. (And, of course, I've dropped the $+C$ throughout, but it belongs too.)
So, to answer your question: $n=1$ is special because the division-by-$n-1$ step in the proof won't be possible.
