$\int_{-1}^{1}x^{2}\delta(x^3)dx$ How to solve 
$$\int_{-1}^{1}x^{2}\delta(x^3)dx$$
where $\delta(x)$ is dirac delta.
I tried a substitution $$y=x^3$$$$\frac{1}{3x^2}dy=dx$$
$$\int_{-1}^{1} \frac{1}{3}\delta(y)dy=\frac{1}{3}$$
But I know the result is wrong. How am I wrong?
Thank you so much.
 A: Here is a way of arriving at the result.  We use the simple regularization of the Dirac Delta given by
$$\delta_\epsilon(x)=\begin{cases}\frac1{2\epsilon}&,x\in[-\epsilon,\epsilon]\\\\0&,\text{elsewhere} \end{cases}$$
where $\lim_{\epsilon\to0}\delta_\epsilon(x) \sim \delta(x)$ in the sense of distributions.  That is, for every smooth function with compact support (i.e., test function), we have 
$$\lim_{\epsilon\to0}\int_{-\infty}^\infty f(x)\delta_\epsilon(x)\,dx=f(0)$$
Then, we have 
$$\begin{align}
\int_{-1}^1 x^2\delta(x^3)\,dx&=\lim_{\epsilon\to0}\int_{-1}^1 x^2 \delta_\epsilon(x^3)\,dx\\\\
&=\lim_{\epsilon\to0}\int_{-\sqrt[3]{\epsilon}}^\sqrt[3]{{\epsilon}}x^2\left(\frac1{2\epsilon}\right)\,dx\\\\
&=\frac13
\end{align}$$
as expected!

Just to remark, the regularization we used herein in not unique. 
A: Your solution is fine. Indeed you should perform the substitution $y=x^3$. With that you obtain
$$\int_{-1}^1\!dx\,x^2\delta(x^3) = \frac{1}{3}\int_{-1}^1\,dy\,\delta(y). \tag{1}$$
Now it is the defining property of the delta-distribution that 
$$\int\!dy\,f(y) \delta(y) =f(0).$$
Using for $f(y)$ the indicator function on $[-1,1]$, you obtain the result
$$\int_{-1}^1\!dx\,x^2\delta(x^3) = \frac{1}{3}\int_{-1}^1\,dy\,\delta(y) = \frac13.$$
