$\int_0^\infty \phi(x)\delta(x^2-1)\, dx$ I need to evaluate $\int_0^\infty \phi(x)\delta(x^2-1)\, dx$ but have very little context for knowing which steps are valid when working with $\delta$.  A $u$-substitution seems appropriate, so instead finding 
$$\int_0^\infty \phi(x)\delta(u)\frac{du}{2x}$$
however, I'm not sure of how to deal with the $\phi(x)/x$.  Should I next infer $x = \pm\sqrt{u+1}$ and then infer that this integral is 
$$\frac{\phi(-1)}{-2} + \frac{\phi(1)}{2}$$
But that seems to rely on $\phi$ being defined for negative values, which seems like it shouldn't be necessary.

For context it may help to know that I only have by definition that 
$$\int_{-\infty}^\infty \phi(x)\delta(x-x_0)\ dx = \phi(x_0)$$

Consider the integral $\int_0^\infty \phi(x) \delta(x^2-1)dx$. Proceed with a $u$ substitution setting $u = x^2 -1$, which yields $du = 2xdx$.
We select $x = \sqrt{u+1}$.  We only need the positive square root, since $x \in [0,\infty)$ according to the integration bounds. This gives $dx = \frac{du}{2\sqrt{u+1}}$
Making the substitutions: $$\int_0^\infty \phi(x) \delta(x^2-1)dx = \int_{-1}^\infty \frac{\phi(\sqrt{u+1})}{2\sqrt{u+1}} \delta(u) du = \frac{\phi(1)}{2}$$

I should mention that $\int_{a}^b \phi(x) \delta(x-x_0) dx = \phi(x_0)$ for any interval $[a,b]$ that contains $x_0$. The integral is zero when $x_0 \not \in [a,b]$. We don't require the knowledge of $\phi$ over all of $(-\infty, \infty)$, since the delta function is zero everywhere but at the origin (in the undergrad ODE sense, that is. the delta function isn't technically a "function").
