How to evaluate this improper integral? I got stuck when evaluating these two improper integrals:$$
\int_a^b\frac{dx}{\sqrt{(b-x)(x-a)}}
$$
and$$
\int_0^1\frac{dx}{\sqrt{x-x^3}}
$$
How to evaluate them? Thank you!

\begin{align*}
\int_a^b{\frac{\mathrm dx}{\sqrt{(b-x)(x-a)}}}&=\int_a^b\frac{\mathrm d x}{\sqrt{-ab+(a+b)x-x^2}}\\
&=\int_a^b\frac{\mathrm dx}{\sqrt{-ab+\left(\frac{a+b}{2}\right)^2-\left(x-\frac{a+b}{2}\right)^2}}\\
&=\int_a^b\frac{\mathrm dx }{\sqrt{\left(\frac{b-a}{2}\right)^2-\left(x-\frac{a+b}{2}\right)^2}}
\end{align*}
Let's put $$\frac{b-a}{2}\sin t=x-\frac{a+b}{2},\quad\mathrm dx=\frac{b-a}{2}\cos t\mathrm dt$$
where $$x=a\implies \frac{b-a}{2}\sin t_1=-\frac{b-a}{2}\implies t_1=-\frac{\pi}{2},$$
$$x=b\implies \frac{b-a}{2}\sin t_2=\frac{b-a}{2}\implies t_2=\frac{\pi}{2}$$
It follows
\begin{align*}
\int_a^b\frac{\mathrm d x}{\sqrt{(b-x)(x-a)}}&=\int_{-\pi/2}^{\pi/2}\frac{\frac{b-a}{2}\cos t \,\mathrm dt}{\frac{b-a}{2}\cos t}\\
&=\pi
\end{align*}
