I tried to prove the volume of a cone. If you let the radius be $r$ and let the height be equal to the radius, then all you need to do is integrate the area of a circle with radius $r$ by $dr$.
$$\int_{}^{} \pi r^2 \:dr= \frac{1}{3} \pi r^3$$
Now you have proven the volume of a cone, where the height equals the radius.
Here is what I have tried to do next:
Let the height of the cone equal $h$. Let $h=xr$. (Where $r$ still equals the radius of the circle).
$$\int_{}^{} \pi r^2 \:d(xr)= \frac{1}{3} \pi xr^3 = \frac{1}{3} \pi\: xr\:r^2 = \frac{1}{3} \pi hr^2$$
This is actually the volume of a cone, so I assumed that what I did was correct, but when I tried to do the same on Mathematica, the following error appeared:
So now I'm wondering, what was my mistake? And also, if you can't integrate by $d(rx)$, why not?
There are several problems with integration by $d(rx)$. The first is the meaning, as usually taught in first-year calculus. The $d(xr)$ in an integral means that the variable of integration is $xr$. However, you have no $xr$ in your integrand, even though your variable $r$ does change when $xr$ changes. In an integral, you need your $d$ term to point out the only thing that changes in your integral during the integration, which we call the variable of integration. You violated this.
Another problem is that Mathematica, despite its power, has no creativity. It must follow and enforce certain standards, and it does not know what non-standard terminology means. That is no slam agains Mathematica, that is the case for software in general. You violated the standard that the $d$ in an integral is followed immediately by a single letter, denoting the variable of integration. Mathematica does not know what to do with your double-letter $rx$ surrounded by parentheses.
The standard way of handling this is to replace your $rx$ after the $d$ with a new variable, and convert all the changing quantities in the integrand to use this variable. Of course, we call this integration by substitution, taught to all first-year calculus students. In your case we can substitute $u=rx$ to get the single variable after the $d$ and we then replace all occurrences of the variable $r$ with $\dfrac ux$. Now the only changing quantity in the integrand is $u$ (since $x$ and $\pi$ are constants here), and we can properly get
$$\begin{align} \int \pi r^2\,d(xr) &= \int \frac{\pi}{x^2}(rx)^2\,d(rx) \\ &= \int \frac{\pi}{x^2}u^2\,du \\ &= \frac{\pi}{x^2}\frac 13u^3 \quad\text{(leaving out the arbitrary constant)} \\ &= \frac{\pi}{x^2}\frac 13(rx)^3 \\ &= \frac 13\pi(xr)r^2 \\ &= \frac 13\pi hr^2 \end{align}$$