How to find area under sines without calculus? In the section establishing that integrals and derivatives are inverse to each other, James Stewart's Calculus textbook says (pp325--pp326, Sec.4.3, 8Ed):

When the French mathematician Gilles de Roberval first found the area under the sine and cosine curves in 1635, this was a very challenging problem that required a great deal of ingenuity. If we didn’t have the benefit of the Fundamental Theorem of Calculus, we would have to compute a difficult limit of sums using obscure trigonometric identities. It was even more difficult for Roberval because the apparatus of limits had not been invented in 1635.

I wonder how Gilles de Roberval did it. Wikipedia and MacTutor do not contain much info on that. How to apply the method of quadrature is exactly the real challenge I suppose.
This is mainly a history question, but I'm also curious as to how one would approach this in modern days. Thank you.

Let us show that
$$ I(a,b)=\int_{a}^{b}\cos(x)\,dx = \sin(b)-\sin(a) \tag{1}$$
through Riemann sums. We have to compute:
$$ \lim_{n\to +\infty}\frac{b-a}{n}\sum_{k=1}^{n}\cos\left(a+\frac{(b-a)k}{n}\right) \tag{2}$$
but the RHS of $(2)$ is a telescopic sum in disguise, hence $(1)$ boils down to proving
$$ \lim_{n\to +\infty}\frac{(b-a)}{n}\,\cos\left(\frac{(b-a)+(a+b) n}{2 n}\right)\frac{\sin\left(\frac{b-a}{2}\right)}{\sin\left(\frac{b-a}{2 n}\right)}=\sin(b)-\sin(a)\tag{3}$$
which is not that difficult. This is my guess on de Roberval's method.
