Question about integral of an odd function

I am studying something and encountered this:
" Let $$R(\theta,T) = \int_{-T}^{T} \frac{(\sin \theta t)}{t}dt, S(T) = \int_0^T\frac{(\sin x)}{x}dx$$, then for $$\theta > 0$$ and changing variables $$t=x/\theta$$ shows that

$$R(\theta,T)=2\int_0^{T\theta}\frac{\sin x}{x}dx = 2S(T\theta)$$ while for $$\theta<0$$, $$R(\theta,T) = -R(|\theta|,T)$$" which I don't understand.

If $$\theta<0$$, then $$R(\theta,T)=2\int_{-T|\theta|}^{0}\frac{\sin x}{x}dx =2\int_0^{T|\theta|}\frac{\sin x}{x}dx = R(|\theta|,T)$$ as $$\frac{\sin x}{x}$$ is an even function, right? I am missing something simple here, thanks and appreciate an explanation.

When $$t=-T$$ we get $$x=t\theta =-T\theta =T|\theta|$$ and not $$-T|\theta|$$.
\begin{align}\theta <0 & \Rightarrow R(\theta ,T)=\int_{-T}^T\frac{\sin (\theta t)}{t}dt\\ &\Rightarrow R(\theta ,T)=\int_{-T}^T\frac{\sin (-|\theta| t)}{t}dt\\ &\Rightarrow R(\theta ,T)=-\int_{-T}^T\frac{\sin (|\theta| t)}{t}dt \ \ \ [\because \sin(-x)=-\sin (x)\big]\\ &\Rightarrow R(\theta ,T)=-R(|\theta|,T)\\ \end{align}