<s>[INST] Proving that $\int_0 ^1 \frac{\text{d}s}{\sqrt{1-s^2}}$ converges with no trig functions 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)
 [/INST] 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.
</s>