Evaluating integral delta function I am trying to evaluate the integral below (delta function) and not sure if I evaluated correctly? The integral is the following:
$$\int^{100}_{-100}x^3\sin(2x)\delta(x^2-9)dx$$

I have the following $x=\pm 3$, $$\therefore \int^{100}_{-100}x^3\sin(2x)\delta(x^2-9)dx = \int^{0}_{-100}x^3\sin(2x)\delta(x^2-9)dx + \int^{100}_{0}x^3\sin(2x)\delta(x^2-9)dx= \ (-3)^3\sin(2\cdot(-3)) + (3)^3\sin(2\cdot(3)) = \ -27\sin(-6) + 27\sin(6)$$
 A: Not quite. Delta function has the property that 
$$\delta(f(x)) = \sum_i \frac{1}{|f'(x_i)|}\delta(x-x_i)$$
where $x_i$ are the zeros of $f$.
A: Start from from property of delta function: 
$\delta(x^2-3^2)=\frac{1}{2*3}[\delta(x+3)+\delta(x-3)]$
Your function is even, $f(x)=f(-x)$ due to symmetry you can write
$\int_{-100}^{100} x^3 \sin(2x) \delta(x^2-9)dx = 2\int_{0}^{100} x^3 \sin(2x) \delta(x^2-9)dx = 2\frac{1}{2*3}\left(\int_{0}^{100} x^3 \sin(2x)\delta(x+3)dx+\int_{0}^{100} x^3 \sin(2x)\delta(x-3)dx\right)=2\frac{1}{2*3}\left(0+27\sin(6)\right)=9\sin(6)$
A: You need to do a change of variables to standardise the argument of the $\delta$ near points where it is $0$: if $a>0$ and $0<\varepsilon \ll a$, we have
$$ \int_{a-\varepsilon}^{a+\varepsilon} g(x) \delta(x^2-a^2)\, dx = \int_{-(2a-\varepsilon)\varepsilon}^{(2a+\varepsilon)\varepsilon} g(\sqrt{u+a^2})\frac{1}{2\sqrt{u+a^2}} \delta(u) \, du = \frac{g(a)}{2a} $$
putting $x = \sqrt{u+a^2}$, and similarly for the integral over the negative axis.
