How can a doubly improper integral become a singly improper integral after substitution?
If we take $x=\sin^2{t}$ then:
$$\int_0^1\frac{1}{x\sqrt{1-x}}dx=2\int_0^{\pi/2}\csc{t}dt$$
On the lefthand side the integral is doubly improper on both the lower ($0$) and upper ($1$) bound. But after substitution the new integral is only improper at the lower bound ($0$). How come? What happened to the improper upper bound?
On
Rewrite your integral as $$\int^{t=\pi/2}_{t=0} \frac1{x(t)}\cdot\frac{dx(t)}{\sqrt{1-x(t)}}$$
Note that $dx(t)=(\sin 2t)dt$.
At $t=\frac{\pi}2$, there is a zero of order one in the $\sqrt{1-x(t)}$, however $dx(t)$ also has a zero of order one at $t=\frac{\pi}2$! The two zeroes exactly cancels out, leading to that $\csc t$ is not singular near $t=\frac{\pi}{2}$, as you observed.
You might feel uncomfortable to say ‘there is a zero in the differential’. Alternatively, rewriting your integral as $$\int^{t=\pi/2}_{t=0} \frac1{x(t)}\cdot\frac{\frac{dx(t)}{dt}}{\sqrt{1-x(t)}}dt$$, you may comfortably say that, at $t=\frac{\pi}2$, the first order zero in $\sqrt{1-x(t)}$ in the denominator is cancelled by the first order zero in $\frac{dx(t)}{dt}$.
Actually, we can always treat differentials as functions.
If you perform a contour integral of $f(z)=\frac{z}{1+z^2}$ along a small circle around $z=\infty$, i.e. $$-\oint_{|z|=\infty}\frac{z}{1+z^2}dz$$ one may expect the integral to be zero, since $\frac z{1+z^2}$ is analytic at $z=\infty$.
However $dz$ has a pole at $z=\infty$, thus indeed the integral is non-zero. It turns out that the integral equals $2\pi i$.
The same happens if you consider $$ \int_0^1\frac{1}{\sqrt{x}}\,dx $$ with the substitution $\sqrt{x}=t$ or $x=t^2$ that transforms the integral into $$ \int_0^1\frac{2t}{t}\,dt=\int_0^1 2\,dt=2 $$ no longer an improper integral.
Note that an antiderivative of $1/\sqrt{x}$ is $2\sqrt{x}$, which is not differentiable at $x$.