<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 way to show that $\int_0^1 \frac{dx}{\sqrt{1-x^2}} = {\pi \over 2}$ without the use of trigonometric functions. i will use the fact the area of unit circle is $\pi.$ that is $\int_0^1 \sqrt{1-x^2}dx = {\pi \over 4}$ twice and integration by parts.
\begin{eqnarray}
{\pi \over 4}  & = & \int_0^1 \sqrt{1-x^2}\ dx \\
              & = &
\int_0^1{1-x^2 \over \sqrt{1-x^2}} \ dx  \\
& = &\int_0^1 \frac{dx}{\sqrt{1-x^2}}- \int_0^1{x^2 \over \sqrt{1-x^2}} \ dx \\
& = &\int_0^1 \frac{dx}{\sqrt{1-x^2}} + \int_0^1 x\  d \sqrt{1-x^2} \\
& = &\int_0^1 \frac{dx}{\sqrt{1-x^2}} +{x \sqrt{1-x^2}} |_0^1 -
\int_0^1\sqrt{1-x^2}\ dx \\
& = &\int_0^1 \frac{dx}{\sqrt{1-x^2}} - {\pi \over 4}.
\end{eqnarray}
and that proves the claim $$ \int_0^1 \frac{dx}{\sqrt{1-x^2}} = {\pi \over 2}. $$
</s>