Integrating an inner product by parts I am trying to integrate the following:
$$ \int \langle F, -(\partial_u \kappa) T\rangle du $$
where $F: S^1 \times [0,T) \to \mathbb{R}^2$ is a simple closed plane curve given
by spatial parameter $u$ (mod $2\pi$) and time parameter $t$. The paper I
am reading states that integrating the above by parts yields:
$$ \int \langle \partial_u F, \kappa T\rangle + \langle F, \kappa \partial_u T\rangle du $$
I am not even sure how you would integrate an inner product, let alone integrate one by parts. Any help is appreciated.
Edit: Thought I would share some of the work I have done here. The closest I have come to solving this is getting the equation
$$ \partial_u \langle F, \kappa T\rangle = \langle \partial_u F, \kappa T\rangle + \langle F, (\partial_u \kappa)T\rangle + \langle F,\kappa\partial_u T\rangle $$
which gives us:
$$  \langle F, -(\partial_u \kappa)T\rangle + \partial_u \langle F, \kappa T\rangle = \langle \partial_u F, \kappa T\rangle +  + \langle F,\kappa\partial_u T\rangle $$
But I have an extra term and haven't done any integration.
2nd edit: Thought I would link the paper just in case anyone wanted to get some more context. https://projecteuclid.org/download/pdf_1/euclid.jdg/1214439902. Page 7 Lemma 3.1.7
 A: We have a simple closed curve in the plane $F \colon \mathbb{S}^1 \times I \to \mathbb{R}^2$ that is evolving in time $t \in I$ according to the heat equation.  We are letting $u$ denote the spatial parameter.  We let $T$ and $N$ be the unit tangent and inward-pointing normal vectors, respectively.
The OP has correctly calculated that
$$\frac{\partial}{\partial u} \langle F, \kappa T \rangle = \left\langle \frac{\partial F}{\partial u}, \kappa T\right\rangle + \left\langle F, \frac{\partial \kappa}{\partial u} T \right\rangle + \left\langle F, \kappa \frac{\partial T}{\partial u} \right \rangle\!,$$
whence
$$\left\langle F, -\frac{\partial \kappa}{\partial u} T \right\rangle  = - \frac{\partial}{\partial u} \langle F, \kappa T\rangle + \left\langle \frac{\partial F}{\partial u}, \kappa T\right\rangle + \left\langle F, \kappa \frac{\partial T}{\partial u} \right \rangle\!,$$
whence
$$\int_0^{2\pi}\left\langle F, -\frac{\partial \kappa}{\partial u} T \right\rangle du  = - \int_0^{2\pi}\frac{\partial}{\partial u} \langle F, \kappa T\rangle\, du + \int_0^{2\pi}\left\langle \frac{\partial F}{\partial u}, \kappa T\right\rangle du + \int_0^{2\pi}\left\langle F, \kappa \frac{\partial T}{\partial u} \right \rangle du.$$
By the Fundamental Theorem of Calculus, the first integral on the right side is zero (because we're working with a simple closed curve):
$$\int_0^{2\pi}\frac{\partial}{\partial u} \langle F, \kappa T\rangle\, du = \left. \frac{}{} \langle F, \kappa T\rangle \right|^{2\pi}_0 = \langle F(2\pi), \kappa(2\pi)T(2\pi) \rangle - \langle F(0), \kappa(0)T(0) \rangle = 0.$$
