I'm trying to evaluate
$$ \int_{-\infty}^\infty{\delta(t^2)} dt $$
If it were just $ \delta (t) $ the area would be $1$, but I don't think this is the case here because the area of the function should be smaller since it approaches $0$ faster.
I attempted a change of variables with $\tau = t^2$, but that ends up with
$$ \int_{-\infty}^\infty\ \frac{\delta(\tau)}{2\sqrt{\tau}} d\tau $$
However, if I do this, it should evaluate to $ \frac{1}{2\sqrt{0}}$ which would be undefined.
Is there a different change of variables I should attempt?
It is not well-defined. Consider $\delta_{\epsilon}(t) = \frac{1}{2\epsilon}\mathbf{1}_{[-\epsilon,\epsilon]}(t)$ which we know to converge to $\delta$ as distribution. Then
$$ \int_{-\infty}^{\infty} \delta_{\epsilon}(t^2) \, dt = \frac{1}{\sqrt{\epsilon}} \to \infty \quad \text{as} \quad \epsilon \to 0^+. $$
On the other hand, we can give a reasonable interpretation of $\delta(t^2)$ on a suitable space of test functions. Indeed, define $\varphi_+'(0) := \lim_{h \to 0^+} \frac{\varphi(h)-\varphi(0)}{h}$ (resp. $\varphi_-'(0) := \lim_{h \to 0^-} \frac{\varphi(h)-\varphi(0)}{h}$) if the limit exists. Then consider the space
$$\mathcal{A} = \{ \varphi \in C_c(\mathbb{R}) : \text{$\varphi(0) = 0$ and $\varphi_{\pm}'(0)$ exist.} \}$$
Then
\begin{align*} \int_{-\infty}^{\infty} \varphi(t) \delta_{\epsilon}(t^2) \, dt &= \frac{1}{2\epsilon} \int_{-\sqrt{\epsilon}}^{\sqrt{\epsilon}} \varphi(t) \, dt \\ &= \frac{1}{2} \int_{-1}^{1} \frac{\varphi(\sqrt{\epsilon}u) - \varphi(0)}{\sqrt{\epsilon}} \, du \qquad (t = \sqrt{\epsilon}u) \\ &\xrightarrow[\epsilon \to 0^+]{} \frac{1}{2} \left( \int_{0}^{1} \varphi_+'(0) u \, du + \int_{-1}^{0} \varphi_-'(0) u \, du \right) \\ &= \frac{\varphi_+'(0) - \varphi_-'(0)}{4}. \end{align*}
In view of this computation, $\delta(t^2)$ can be realized as what captures the 'jump discontinuity' of $\varphi'$ at $0$, provided $\varphi(0) = 0$.