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)
 A: 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.
A: Here's an alternative approach:
Using the Euler substitution $t=\sqrt{\frac{1-s}{1+s}}\implies s=\frac{1-t^2}{1+t^2}$, the integral is transformed to
$$\begin{align}
\int_{0}^{1}\frac{\mathrm{d}s}{\sqrt{1-s^2}}
&=\int_{1}^{0}\frac{t+\frac{1}{t}}{2}\,\frac{(-4t)\,\mathrm{d}t}{(1+t^2)^2}\\
&=2\int_{0}^{1}\frac{\mathrm{d}t}{1+t^2},\\
\end{align}$$
which is not an improper integral.
A: By replacing $t$ with $1-t$ we get:
$$\int_{0}^{1}\frac{dt}{\sqrt{1-t^2}}=\int_{0}^{1}\frac{dt}{\sqrt{t(2-t)}}\leq\int_{0}^{1}\frac{dt}{\sqrt{t}}=2.$$
A: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
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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}
A: 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.
A: 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 $$ \int_0^1 \frac{dx}{\sqrt{1-x^2}} = {\pi \over 2}. $$
