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.
By replacing $t$ with $1-t$ we get:
$$\int\_{0}^{1}\frac{dt}{\sqrt{1-t^2}}=\int\_{0}^{1}\frac{dt}{\sqrt{t(2-t)}}\leq\int\_{0}^{1}\frac{dt}{\sqrt{t}}=2.$$