There's no geometric interpretation of $\int f(x) \mbox{d}(g(x))$ as far as I'm concern. However, it does have a meaning. Say, you want to take the differentiation of $\sin(x)$ w.r.t x, and $\sin (x)$, they give 2 different results:
- $\frac{\mbox{d} \sin x}{\mbox{d}x} = \cos x$
- $\frac{\mbox{d} \sin x}{\mbox{d} \sin x} = 1$
Another example is to take the derivatives of $x^4$ w.r.t x, and $x^2$ respectively.
- $\frac{\mbox{d} x^4}{\mbox{d}x} = 4x^3$
- $\frac{\mbox{d} x^4}{\mbox{d} x^2} = \frac{\mbox{d} (x^2)^2}{\mbox{d} x^2} = 2x^2$
An antiderivative of $f$ w.r.t $x$ is some function $F$, such that $\dfrac{\mbox{d}F}{\mbox{d}x} = f$
So:
- $\int \mbox{d}(x^2)$ is some function, such that its derivative w.r.t $x^2$ is 1, this family of function is, of course, $x^2 + C$.
- $\int x^5\mbox{d}(x^5)$ is some function, such that its derivative w.r.t $x^5$ is $x^5$, this family of function is, of course, $\dfrac{x^{10}}{2} + C$, since: $\frac{1}{2}\dfrac{\mbox{d}(x^{10} + C)}{\mbox{d}(x^5)} = \frac{1}{2}\dfrac{\mbox{d}((x^5)^2)}{\mbox{d}(x^5)} + 0 = x^5$
When solving problems, you just need to remember that $\int f(x)\mbox{d}(g(x)) = \int f(x).g'(x) \mbox{d}x$, e.g, when you take some function out of d, you have to differentiate it, and vice versa, when you put some function into d, you'll have to integrate it, like this:
- $\int x^5 \mbox{d}(x^2) = \int 2x^6 \mbox{d}(x) = \dfrac{2x^7}{7} + C$. (Take $x^2$ out of d, we differentiate it, and have $2x$)
- $\int \sin^2 x \cos x \mbox{d}x = \int \sin^2 x \mbox{d}(\sin x) = \frac{\sin ^ 3 x}{3} + C$ (Put $\cos x$ into d, we have to integrate it to get $\sin x$).