I'm trying to figure out if my result is missing something. The problem is the following. I'm trying to prove the solution of the integral equation: $$P.V.\int_a^b \frac{g(t)}{t-s}dt =1$$ where $a<s<b$ ,is the function $$g(s)=\frac{1}{\pi^2\sqrt{(s-a)(b-s)}}\left[\pi c+\int_a^b \frac{\sqrt{(t-a)(b-t)}}{s-t}dt \right]$$ so that $$g(s)=\frac{1}{\pi^2\sqrt{(s-a)(b-s)}}\left[\pi c+\pi\left(s-\frac{a+b}{2}\right)\right]$$
But this is my result on solving the integral inside square brackets:
$$\int_a^b \frac{\sqrt{(t-a)(b-t)}}{s-t}dt = -\int_a^b \frac{\sqrt{(t-a)(b-t)}}{t-s}dt$$ making $t-s=(b-a)x$
$$-\int_{\frac{a-s}{b-a}}^{\frac{b-s}{b-a}} \frac{\sqrt{(s+(b-a)x-a)(b-s-(b-a)x)}}{(b-a)x}(b-a)dx$$
$$-\int_{\frac{a-s}{b-a}}^{\frac{b-s}{b-a}} \frac{\sqrt{(b-a)(\frac{s-a}{b-a}+x)(b-a)(\frac{b-s}{b-a}-x)}}{x}dx$$
and making $\frac{a-s}{b-a}=\alpha$ and $=\frac{b-s}{b-a}=\beta$, then
$$-(b-a) \int_{\alpha}^{\beta} \frac{\sqrt{(-\alpha+x)(\beta-x)}}{x} dx$$ $$-(b-a) \int_{\alpha}^{\beta} \frac{\sqrt{(x-\alpha)(\beta-x)}}{x} dx$$
and using this it yields:
$$=-(b-a) \pi \left[ \frac{\alpha+\beta}{2} - \sqrt{\alpha \beta} \right]$$
Substituting the original values for $\alpha$ and $\beta$
$$=-(b-a)\pi \left[ \frac{a-s+b-s}{2(b-a)} - \sqrt{\frac{a-s}{b-a} \frac{b-s}{b-a} } \right]$$
$$=-\pi \left[ \frac{a+b}{2} -s - \color{blue}{\sqrt{(a-s) (b-s) }} \right]$$ $$=\pi \left[s- \frac{a+b}{2} + \color{blue}{\sqrt{(a-s) (b-s) }} \right]$$
which is pretty close to the term inside brackets for the solution, but that $\color{blue}{\sqrt{(a-s) (b-s) }}$ term is bothering me.
So the question here is, am I doing something wrong in the steps for solving the integral or my result seems right and maybe the book solution is missing that square root term? My bet here is it has something to do with the fact that $a<s<b$ and that square root term then would be a pure imaginary value, or anything of that sort.
Book ref: Singular integral equations - Estrada, Kanwal; p.75, eqs. 3.19 and 3.20
