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.
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\begin{align}&\overbrace{\color{#66f}{\large%
\int\_{0}^{1}{\dd s \over \root{1 - s^{2}}}}}^{\ds{\dsc{s}\ =\ \dsc{1 - t^{2}}}}\ =\
\int\_{1}^{0}{-2t\,\dd t \over {\root{1 - \pars{1 - t^{2}}^{2}}}}
=2\int\_{0}^{1}{\dd t \over \root{2 - t^{2}}}
\\[5mm]&< 2\int\_{0}^{1}{\dd t \over \root{2 - 1^2}} = \color{#66f}{\large 2}
\end{align}