Are the smooth functions dense in either $\mathcal L_2$ or $\mathcal L_1$?

Is the subset consisting of all integrable (or square integrable) smooth functions of the set of all integrable (or square integrable) functions, dense under the usual Euclidean or integral of absolute difference metric?

By smooth I mean derivatives of all orders exist.

Yes. In fact, by the Stone-Weierstrass theorem and the existence of smooth bump functions, smooth functions with compact support are uniformly dense in the space of continuous functions with compact support. Uniform density implies $L^2$ and $L^1$ density for functions with compact (and therefore finite measure) support, and since continuous functions with compact support are dense in $L^2$ and $L^1$, the result follows.

If you wanted to see this more directly, you can go through the iterations of approximating an arbitrary ($L^1$ or $L^2$) function with a bounded function with bounded support, then with a simple function, then with a step function, and finally approximate the step function with a smooth function using bump functions.

Jonas's argument is good. Another proof is: given $f \in L^p$ (here $p=1,2$), take the convolution of $f$ with a sequence of mollifiers $\eta_\epsilon$. Using properties of convolutions, it's easy to check that $f * \eta_\epsilon$ is a smooth function, and that $f * \eta_\epsilon \to f$ in $L^p$ as $\epsilon \to 0$. This has the advantage of being a little more direct.

Edit: For a reference, see Folland's Real Analysis, section 8.2.

The smoothness of $f * \eta_\epsilon$ is Proposition 8.10 and comes from differentiating under the integral sign in the convolution (with justification!), and choosing to put the derivative on $\eta_\epsilon$. Intuitively, it comes from the idea that convolution is an "averaging" operation and tends to smooth, smear, or blur rough areas of $f$ together, and so should be a smoothing operation. (The wikipedia article has a nice animation illustrating this.)

The fact that $f * \eta_\epsilon \to f$ in $L^p$ is Folland's Theorem 8.14 (a), and it's pretty elementary. He also has Proposition 8.17 which proves that $C^\infty_c$ is dense in $L^p$, but it sort of inexplicably starts by using the fact that $C_c$ is dense in $L^p$. I suppose this is used to get a compactly supported function, so that you can approximate $f \in L^p$ by functions which are not only smooth (which $f * \eta_\epsilon$ is) but also compactly supported (which $f * \eta_\epsilon$ need not be, although $\eta_\epsilon$ is). But an easier argument would be to first approximate $f$ in $L^p$ norm by a function $g$ which is compactly supported but not necessarily continuous; for example, $g = f 1_{[-N,N]}$ for large $N$ (this works by dominated convergence), and then apply mollifiers to $g$. Unless, of course, there is some subtlety that I've missed.

Edit 2: Indeed there is. Folland's 8.14 (a) relies upon the fact that translation is strongly continuous in $L^p$, which uses the density of $C_c$. So apparently it is not so easy to bypass this step, and that destroys a lot of the "directness" of my argument.