This is an answer elaborating on my comment above, and largely addressing the
edits to the original question:
I don't think there were any "old techniques" before Hermite proved
transcendence of $e$ in the early 1870s. As far as I know, the only transcendental numbers known before then were Liouville's interesting but
somewhat artificial examples from early that century.
The subject of transcendence, and the related subject of Diophantine approximation (i.e. approximating irrational numbers, especially irrational algebraic numbers, by rationals), is relatively new. Liouville proved the first results showing that it is not so easy to approximate an irrational algebraic number by rational ones, and used this to construct his transcendental numbers, which he could recognize as being transcendental because they are too well approximated by rational numbers. ("Too well" and "not so easy" here refer to the following problem: if you try to approximate $\alpha$ by the rational number $p/q$, can you get within a distance of $O(q^{-n})$ for some given $n$ as you let the denominator $q$ get arbitrarily large. (The larger $n$ is, the smaller $q^{-n}$ is, and so the better the rate of approximation.) Liouville showed, using the pigeon hole principal more-or-less, that if $\alpha$ is algebraic of degree $d$ then you can't do better than $n = d$.)
But this left open the problem of showing that various given numbers (like $e$ and $\pi$) are transcendental.
If you like, here is one way to think of the problem: if you want to show that $\alpha$ is transcendental, then
you want to show that $f(\alpha) \neq 0$ for any non-zero polynomial $f$
with rational coefficients. The difficulty is that there will be lots of polynomials with real coefficients that have $\alpha$ as a root, and any one of them can be approximated as closely as you like by an $f$ with rational coefficients, so we can find (lots of!) $f$ with rational coefficients such that $f(\alpha)$ is as close to zero as we like.
So you have to find some way to pin down the difference between $f(\alpha)$ being zero and $f(\alpha)$ being very close to zero. This is not so easy!
(For example, computationally, you can't tell the difference between $0$ and any real number that is smaller than your computational accuracy can recognize.)
And now one sees why Diophantine approximation ideas of the type mentioned above
are relevant. They are related to quantifying how close we can make $f(\alpha)$ to zero while bounding the denominators of the rational numbers involved.
It is not coincidence that bounding the denominators is relevant: morally this is an attempt to pass from working over $\mathbb Q$ to working over $\mathbb Z$.
Why do we want to do this? Well, as I already noted, it's pretty hard to tell the difference between $\mathbb Q$ and $\mathbb R$, since the former is dense in the latter, but we can tell the difference between $\mathbb Z$ and $\mathbb R$, since the former is discrete: a non-zero integer is some definite positive distance (i.e. at least 1) away from $0$.
The preceding remarks are somewhat philosophical, and they reflect my (limited) experience of thinking about these kinds of questions. If you look at the proofs in the link above, it may not be obvious that they are relevant, but I believe that they in fact do have some relevance: e.g. you will see that the arguments reduce to considering integer rather than rational polynomials, and that growth considerations play a key role. Another thing you will see is that certain auxiliary polynomials enter the proof, and a key fact about them is that they have a high order of vanishing at their zeroes. The appearance of auxiliary polynomials, often with a high order of vanishing, is ubiquitous in this theory.
One more (somewhat cultural) remark:
Roth's theorem, for which he got the Fields medal, is the ultimate strengthening of Liouville's theorem: he shows that if $\alpha$ is irrational algebraic, then one can't do better than $O(q^{-2})$ in the problem of Diophantine approximation discussed above. The proof involves (among other things) constructions with auxiliary polynomials. So my impression is that Liouville, Hermite, and Lindemann (and there are probably other names that should be here) invented a new subject, namely Diophantine approximation and transcendence theory, whose modern methods are an outgrowth of the methods that they introduced.
P.S. Reading the first few pages of Baker's book Transcendental Number Theory
(Google Books link) might help.
As a high school student, most of the tricks I'm aware of were already stated by you, or in the comments. However, there's one more trick that I don't think anyone has mentioned: Integrating an inverse function.
$$\int\!f^{-1}(x)\ dx = x\cdot\!f^{-1}(x)\ - F(f^{-1}(x))\ + c$$where $$F(x) = \int\!f(x)\ dx$$
So for instance, if you wish to find $\int\cos^{-1}(x)\ dx,$ you will have $$f(x)= \cos x$$ and $$F(x) = \int\cos x\ dx = \sin x\ (+c)$$
So to find $\int\cos^{-1}(x)\ dx,$ use the formula as the follows:
$$\int\cos^{-1}(x)\ dx = x\cdot\cos^{-1}(x)\ - \sin(\cos^{-1}(x))\ + c$$
$$= x\cdot\cos^{-1}(x)\ - \sqrt {1-x^2}\ + C$$
I personally like this trick as it can be generalized to any inverse function. A simple way to prove it would be using the Chain Rule but it's a really nice formula that avoids working things out from scratch every time.
Hope that helped to add to your list :)
Best Answer
Here's a relevant flow chart, because as Silvanus P Thompson said:
Integration, being a inverse process, is much less systematic than differentiation.
EDIT Right after that quote the author gives an 'unsolved, but rather simple' integral. It may not be solvable though - that quality doesn't necessarily follow from simplicity when it comes to integration.