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 is a proof in single-variable calculus language. On the interval $0\leq s \leq 1$, we have
$$
\sqrt{1-s^2} = \sqrt{(1+s)(1-s)} \leq \sqrt{2} \sqrt{1-s}
$$
So
$$
\frac{1}{\sqrt{1-s^2}} \geq \frac{1}{\sqrt{2}} \frac{1}{\sqrt{1-s}}
$$
We *would like to* use the Comparison Test, comparing $\int\_0^1 \frac{ds}{\sqrt{1-s}}$ to $\int\_0^1 \frac{ds}{\sqrt{1-s}}=\int\_0^1 \frac{du}{\sqrt{u}}$. The latter converges by the $p$-test ($p=1/2 < 1$).

Unfortunately, the comparison goes in the wrong direction to apply the Comparison Test. We need to instead bound $\sqrt{1-s^2}$ from *below* in order to bound $\frac{1}{\sqrt{1-s^2}}$ from above. So assume instead that $\frac{1}{2} \leq s \leq 1$. Then
$$
\sqrt{1-s^2} = \sqrt{(1+s)(1-s)} \geq \sqrt{\frac{3}{2}} \sqrt{1-s}
$$
So
$$
\frac{1}{\sqrt{1-s^2}} \leq \sqrt{\frac{2}{3}} \frac{1}{\sqrt{1-s}}
$$
on this interval.

To put it together, you have
$$
\begin{aligned}
\int\_0^1 \frac{ds}{\sqrt{1-s^2}} &= \int\_0^{1/2}\frac{ds}{\sqrt{1-s^2}} + \int\_{1/2}^1 \frac{ds}{\sqrt{1-s^2}}\\
&\leq \int\_0^{1/2}\frac{ds}{\sqrt{1-s^2}} + \sqrt{\frac{2}{3}} \int\_{1/2}^1 \frac{ds}{\sqrt{1-s}}
\end{aligned}
$$
The first integral is not improper and the second converges.