Can we guarantee that $$\int_{t=0}^{\tau}F(r)\cos(\omega t)\mathrm{d}t=0$$ where $\tau$ is the period of the cosine function with angular frequency defined as $$\omega=\frac{2\pi}{\tau}$$ and $F(r)$ is some unknown function of position $r$ only?
In other words, $r$ does not depend on time $t$.
I know that in general $$\fbox{$\color{red}{\int_{t=0}^{\tau}F(r)\,\mathrm{d}t\ne 0}$}\tag{$\color{red}{1}$}$$ since we will get a function of $\tau$; and $$\int_{t=0}^{\tau}\cos(\omega t)\mathrm{d}t= 0$$ since we have integrated over 1 full period.
But I'm not sure about whether $$\int_{t=0}^{\tau}F(r)\cos(\omega t)\mathrm{d}t$$ is zero or not.
It was my understanding that I may take out factors of the integrand independent of the integration variable $t$: $$\int_{t=0}^{\tau}F(r)\cos(\omega t)\mathrm{d}t=F(r)\int_{t=0}^{\tau}\cos(\omega t)\mathrm{d}t=0$$ as $F(r)$ is being treated as a constant.
$\fbox{$\color{red}{\text{But it is equation (1) that is confusing me here}}$}$.
So can I conclude that since $r$ is not a function of $t$ $$\int_{t=0}^{\tau}F(r)\cos(\omega t)\mathrm{d}t=0?$$
EDIT:
One user has agreed that I may pull the factor $F(r)$ out as a constant since it does not depend on $t$.
Now, I just need confirmation:
Does $$\int_{t=0}^{\tau}F(r)\cos(\omega t)\mathrm{d}t=0$$ or not?
A simple yes or no answer/comment will suffice; thanks very much.