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: 
This is an improper integral and the only problem to treat is at $1$. We have

$$\frac{1}{\sqrt{1-s^2}}\sim\_1\frac1{\sqrt2\sqrt{1-s}}$$
and since the integral
$$\int\_0^1\frac{ds}{\sqrt{1-s}}$$
is convergent then the given integral is also convergent.
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.