Question: Let 

$$\int\_0 ^1 \frac{\text{d}s}{\sqrt{1-s^2}}$$

How to show that it converges with no use of trigonometric functions? 

(trivially, it is the anti-derivative of $\sin ^{-1}$ and therfore can be computed directly)

Answer: 
Here's an alternative approach:

Using the Euler substitution $t=\sqrt{\frac{1-s}{1+s}}\implies s=\frac{1-t^2}{1+t^2}$, the integral is transformed to

$$\begin{align}
\int\_{0}^{1}\frac{\mathrm{d}s}{\sqrt{1-s^2}}
&=\int\_{1}^{0}\frac{t+\frac{1}{t}}{2}\,\frac{(-4t)\,\mathrm{d}t}{(1+t^2)^2}\\
&=2\int\_{0}^{1}\frac{\mathrm{d}t}{1+t^2},\\
\end{align}$$

which is not an improper integral.
here is a way to show that $\int\_0^1 \frac{dx}{\sqrt{1-x^2}} = {\pi \over 2}$ without the use of trigonometric functions. i will use the fact the area of unit circle is $\pi.$ that is $\int\_0^1 \sqrt{1-x^2}dx = {\pi \over 4}$ twice and integration by parts.

\begin{eqnarray}
{\pi \over 4} & = & \int\_0^1 \sqrt{1-x^2}\ dx \\
 & = &
\int\_0^1{1-x^2 \over \sqrt{1-x^2}} \ dx \\
& = &\int\_0^1 \frac{dx}{\sqrt{1-x^2}}- \int\_0^1{x^2 \over \sqrt{1-x^2}} \ dx \\
& = &\int\_0^1 \frac{dx}{\sqrt{1-x^2}} + \int\_0^1 x\ d \sqrt{1-x^2} \\
& = &\int\_0^1 \frac{dx}{\sqrt{1-x^2}} +{x \sqrt{1-x^2}} |\_0^1 -
\int\_0^1\sqrt{1-x^2}\ dx \\
& = &\int\_0^1 \frac{dx}{\sqrt{1-x^2}} - {\pi \over 4}.
\end{eqnarray}

and that proves the claim $$ \int\_0^1 \frac{dx}{\sqrt{1-x^2}} = {\pi \over 2}. $$