If the integral as written in my comment is accurate, then we have
[tex]I=\displaystyle\int_{3/2}^2\sqrt{(x-1)(3-x)}\,\mathrm dx[/tex]
Expand the polynomial, then complete the square within the square root:
[tex](x-1)(3-x)=-x^2+4x-3=1-(x-2)^2[/tex]
[tex]I=\displaystyle\int_{3/2}^2\sqrt{1-(x-2)^2}\,\mathrm dx[/tex]
Let [tex]x=2-\cos\theta[/tex] and [tex]\mathrm dx=\sin\theta\,\mathrm d\theta[/tex]:
[tex]I=\displaystyle\int_{\pi/3}^{\pi/2}\sqrt{1-(2-\cos\theta-2)^2}\sin\theta\,\mathrm d\theta[/tex]
[tex]I=\displaystyle\int_{\pi/3}^{\pi/2}\sqrt{1-\cos^2\theta}\sin\theta\,\mathrm d\theta[/tex]
[tex]I=\displaystyle\int_{\pi/3}^{\pi/2}\sqrt{\sin^2\theta}\sin\theta\,\mathrm d\theta[/tex]
Recall that [tex]\sqrt{x^2}=|x|[/tex] for all [tex]x[/tex], but for all [tex]\theta[/tex] in the integration interval we have [tex]\sin\theta>0[/tex]. So [tex]\sqrt{\sin^2\theta}=\sin\theta[/tex]:
[tex]I=\displaystyle\int_{\pi/3}^{\pi/2}\sin^2\theta\,\mathrm d\theta[/tex]
Recall the double angle identity,
[tex]\sin^2\theta=\dfrac{1-\cos(2\theta)}2[/tex]
[tex]I=\displaystyle\frac12\int_{\pi/3}^{\pi/2}(1-\cos(2\theta))\,\mathrm d\theta[/tex]
[tex]I=\dfrac\theta2-\dfrac{\sin(2\theta)}4\bigg|_{\pi/3}^{\pi/2}[/tex]
[tex]I=\dfrac\pi4-\left(\dfrac\pi6-\dfrac{\sqrt3}8\right)=\boxed{\dfrac\pi{12}+\dfrac{\sqrt3}8}[/tex]
You can determine the more general result in the same way.
[tex]I=\displaystyle\int_p^q\sqrt{(x-a)(b-x)}\,\mathrm dx[/tex]
Complete the square to get
[tex](x-a)(b-x)=-(x-a)(x-b)=-x^2+(a+b)x-ab=\dfrac{(a+b)^2}4-ab-\left(x-\dfrac{a+b}2\right)^2[/tex]
and let [tex]c=\frac{(a+b)^2}4-ab[/tex] for brevity. Note that
[tex]c=\dfrac{(a+b)^2}4-ab=\dfrac{a^2-2ab+b^2}4=\dfrac{(a-b)^2}4[/tex]
[tex]I=\displaystyle\int_p^q\sqrt{c-\left(x-\dfrac{a+b}2\right)^2}\,\mathrm dx[/tex]
Make the following substitution,
[tex]x=\dfrac{a+b}2-\sqrt c\,\cos\theta[/tex]
[tex]\mathrm dx=\sqrt c\,\sin\theta\,\mathrm d\theta[/tex]
and the integral reduces like before to
[tex]I=\displaystyle\int_P^Q\sqrt{c-c\cos^2\theta}\,\sin\theta\,\mathrm d\theta[/tex]
where
[tex]p=\dfrac{a+b}2-\sqrt c\,\cos P\implies P=\cos^{-1}\dfrac{\frac{a+b}2-p}{\sqrt c}[/tex]
[tex]q=\dfrac{a+b}2-\sqrt c\,\cos Q\implies Q=\cos^{-1}\dfrac{\frac{a+b}2-q}{\sqrt c}[/tex]
[tex]I=\displaystyle\frac{\sqrt c}2\int_P^Q(1-\cos(2\theta))\,\mathrm d\theta[/tex]
(Depending on the interval [p, q] and thus [P, Q], the square root of cosine squared may not always reduce to sine.)
Resolving the integral and replacing c, with
[tex]c=\dfrac{(a-b)^2}4\implies\sqrt c=\dfrac{|a-b|}2=\dfrac{b-a}2[/tex]
because [tex]a<b[/tex], gives
[tex]I=\dfrac{b-a}2(\cos(2P)-\cos(2Q)-(P-Q))[/tex]
Without knowing p and q explicitly, there's not much more to say.
