So there is a certain proof for $\delta(x^2-a^2)$ property in my book which has a part that states 
So i would like a proof to this general statemement which i could not find anwhere $$\int_{0}^{\infty}f(g(x))\delta{(x-a)}dx=f(g(a))$$ I know that the definition of delta function states $$\int_{0}^{\infty}f(x)\delta{(x-a)}dx=f(a)$$ but this is completely different
The "sifting property" of the Dirac delta function tells you that for any function $h(x)$, we have $$ \int_{-\infty}^\infty h(x) \delta(x - a)\,dx = h(a). $$ With that, if $a > 0$ and we set $$ h(x) = \begin{cases} 0 & x < 0\\ f(g(x)) & x \geq 0, \end{cases} $$ then we find that $$ \int_0^\infty f(g(x)) \delta(x-a)\,dx = \int_{-\infty}^\infty h(x) \delta(x - a)\,dx = h(a) = f(g(a)). $$