I'm trying to understand the proof of the Holditch's Theorem written by Arne Broman here. The proof of the generalization is really short and there is a single, important step that I don't get. In the words of the article, they use an "unusual" but legitimate operation of line integrals.
For context, they define $\alpha(t), \beta(t), \delta(t)$ the curves traced by the points $A,B$ and $D$ respectively. From their definitions, because the three points are in the same line with direction $\theta(t)$, we have two connections: $$\delta(t) = \alpha(t)+a(\cos(\theta(t)), \sin(\theta(t)))$$ $$\beta(t) = \delta(t)+b(\cos(\theta(t)), \sin(\theta(t)))$$ So the first step of the proof is just writing $\beta$ with this expression, which changes a line integral from $\beta$ to $\delta$, I understand that. Now, the second step claims that: $$\int_\delta (x+b\cos\theta)d(y+b\sin\theta) = \int_\delta x dy + b\int_\delta xd(\sin\theta)+b\int_\delta\cos \theta dy+b^2\int_\delta \cos\theta d(\sin\theta)$$
And I'm honestly really lost in this equality. It's like they're applying a distributive property with the integrand and the variable of integration, but I don't know what lets you do that. My intuition is that is just a consequence of the definition of the line integral, but right now I don't see it. Any help is appreciated.
There's nothing unusual about the distributive property here. $d$ is linear, so $d(y+b\sin\theta) = dy + b\,d(\sin\theta)$. Now you just expand it out. You can plug in parametrizations and it will be the same: If $\delta$ is parametrized by $(x(t),y(t))$, $a\le t\le b$, then you end up with \begin{multline*} \int_a^b \big(x(t)+b\cos\theta(t)\big)\big(y'(t)+b\cos\theta(t)\theta'(t)\big)\,dt = \int_a^b \big(x(t)y'(t) +bx(t)\cos\theta(t)\theta'(t)+b\cos(\theta(t))y'(t) + b^2\cos^2\theta(t)\theta'(t)\big)\,dt, \end{multline*} and now you break it up into four integrals.