I tried to prove the following statement:
$\displaystyle\int_{h(\alpha)}^{h(\cdot)}g=\int_{\alpha}^{(\cdot)}(g\circ h )\cdot h'$
I am still new to functions involving integrals, so I was hoping I could receive a thumbs-up / thumbs-down regarding the proof. It is not clear to me what assumptions I need to prove this...for my proof, I just assumed that $g$ was continuous and that $(g\circ h)\cdot h'$ was continuous (maybe I can get away with fewer assumptions?)
First, suppose $f$ is a continuous function. Then by the First Fundamental Theorem of Calculus (FFTC), the following function $F_{\alpha}(\cdot)=\int_{\alpha}^{(\cdot)}f$ is an antiderivative of $f$ because $F_{\alpha}'(\cdot)=f(\cdot)$
Now, suppose $f(\cdot)=\left[g\circ h (\cdot)\right]\cdot h'(\cdot)$, where $g$ is a continuous function. By applying the Chain Rule and the FFTC, we see that the following function $G_{\alpha}(\cdot)$ must also be an antiderivative of $f$: $G_{\alpha}(\cdot)=\int_{\alpha}^{h(\cdot)}g$.
The above two paragraphs show that $F_{\alpha}'(\cdot)=G_{\alpha}'(\cdot)$, which, by the Mean Value Theorem, implies that $G_{\alpha}(\cdot)=F_{\alpha}(\cdot)+c$, where $c$ is some constant. Letting $x=\alpha$, and noting that $\int_{\alpha}^{\alpha}f=0$ we can solve for $c$ the following way:
$$G_{\alpha}(\alpha)=F_{\alpha}(\alpha)+c \implies G_{\alpha}(\alpha)=c$$
We can therefore rewrite the above expression as:
$$G_{\alpha}(\cdot)=F_{\alpha}(\cdot)+G_{\alpha}(\alpha)\iff \int_{\alpha}^{h(\cdot)}g=\int_{\alpha}^{(\cdot)}f+\int_{\alpha}^{h(\alpha)}g \quad (\dagger_1)$$
$(\dagger_1)$ can be rearranged as:
$$\int_{\alpha}^{h(\cdot)}g-\int_{\alpha}^{h(\alpha)}g=\int_{\alpha}^{(\cdot)}f$$
We can now apply some useful rules of integration, which let us claim:
$-\int_{\alpha}^{h(\alpha)}g=\int_{h(\alpha)}^{\alpha}g$
$\int^{h(\cdot)}_{\alpha}g+\int_{h(\alpha)}^{\alpha}g=\int_{h(\alpha)}^{h(\cdot)}g$
As such, we conclude that:
$$\int_{h(\alpha)}^{h(\cdot)}g=\int_{\alpha}^{(\cdot)}f = \int_{\alpha}^{(\cdot)} \left[g \circ h\right]\cdot h'$$