$\int_a^b \frac{d x}{\sqrt{(x-a)(b-x)}}$
$$
\begin{aligned}
& \int_a^b \frac{d x}{\sqrt{-x^2+(a+b) x-a b}} \\
& =\int_a^b \frac{d x}{\sqrt{-\left(x^2-(a+b) x+a b\right.}} \\
& =\int_a^b \frac{d x}{\sqrt{-\left[x^2+\left(\frac{a+b}{2}\right)^2+a b-\left(\frac{a+b}{2}\right)^2\right]-(a+b) x}}
\end{aligned}
$$
$\begin{aligned} & =\int_a^b \frac{d x}{\sqrt{\left(\frac{a-b}{2}\right)^2-\left\{x-\left(\frac{a+b}{2}\right)\right\}^2}} \quad[\because b>a] \\ & =\int_a^b \frac{d x}{\sqrt{\left(\frac{b-a}{2}\right)^2-\left\{x-\left(\frac{a+b}{2}\right)\right\}^2}} \\ & \therefore \int \frac{d x}{\sqrt{a^2-x^2}}=\sin ^{-1} \frac{x}{a}+C\end{aligned}$
$\begin{aligned} & =\left\{\sin ^{-1}\left[\frac{x-\left(\frac{a+b}{2}\right)}{\frac{b-a}{2}}\right]\right]_a^b \\ & =\sin ^{-1}\left[\frac{b-\left(\frac{a+b}{2}\right)}{\frac{b-a}{2}}\right]-\sin ^{-1}\left(\frac{a-\frac{a+b}{2}}{\frac{b-a}{2}}\right) \\ & =\sin ^{-1}(+1)-\sin ^{-1}(-1)=+\frac{\pi}{2}+\frac{\pi}{2}=+\pi\end{aligned}$