Often while solving integrals, it becomes easier to solve it, by transforming into polar, spherical and other coordinate systems, from our standard cartesian coordinates.
My question is, is a simple $u$-substitution essentially the same thing, just in one dimension ? Is it nothing more than a simple change of coordinates, where we move from $x$ coordinate to $u$ coordinates ?
Moreover, is coordinate transformation or $u$ substitution, just a trivial change of basis, while integral transforms represent a standard change of basis. Can someone explain why one is considered trivial while the other is 'standard' ?
Yes, integration by substitution is a special case of the following theorem: $\newcommand{\d}{\mathop{}\!\mathrm{d}}$ Let $U,V\subset\mathbb{R}^n$ be open and $\Phi\colon U\to V$ a $C^1$-diffeomorphism. Then a measurable function $F\colon V\to \mathbb{R}$ is integrable iff $$G:=|\det\circ\d\Phi|\cdot(F\circ\Phi)$$ is integrable$^1$ and$$\int_V F=\int_U G$$ (Both integrals are defined with respect to the Lebesgue measure.) More generally, we can consider the Bochner integral and then this theorem also holds if $f$ is a function to a Banach space.
$^1$ Since $\phi\in C^1(U,V)$, $\d\Phi\in C(U,\mathbb R^{n\times n})$ and hence $|\det\circ\d\Phi|\in C(U,\mathbb R)$ is the composition of continuous functions.