arc length on circle http://gowers.wordpress.com/2014/03/02/how-do-the-power-series-definitions-of-sin-and-cos-relate-to-their-geometrical-interpretations/#more-5401
I need an explanation of this bit on the blog:
One non-rigorous but informative way of thinking about this is that for each $t$ between $x$ and $1$, we should take an interval $[x,x+dt]$, work out the length of the bit of the circle vertically above this interval, and sum up all those lengths. The bit of the circle in question is a straight line (since $dt$ is infinitesimally small) and by similar triangles its length is $\frac{dt}{\sqrt{1-t^2}}$.
How did I write that down? Well, the big triangle I was thinking of was one with vertices $(0,0), (t,0)$ and the point on the circle directly above $(t,0)$, which is $(t,\sqrt{1-t^2})$, by Pythagoras’s theorem. The little triangle has one side of length $dt$, which corresponds to the side in the big triangle of length $\sqrt{1-t^2}$. So the hypotenuse of the little triangle is$ \frac{dt}{\sqrt{1-t^2}}$, as I claimed.


why does $dt$ correspond to $\sqrt{1-t^2}$ ?

The big (right) triangle $\triangle OPQ$ is similar to the little (right) triangle $\triangle SPR$, because by construction the side $PQ$ is  perpendicular to the side $PR$ and the side $OP$ is perpendicular to the side $PS$ as $S$ tends to $P$. These triangles have two identical angles, and thus three identical ones. $\angle OPQ=\angle RPS$ and $\angle OQP=\angle SRP$, as shown in the picture, where the $x$-coordinates of $P$ and $S$ are, respectively,  $t$ and $t+dt$.  As such corresponding sides have lengths in the same ratio: $$\dfrac{PR}{PQ}=\dfrac{PS}{OP}.$$ 
Since $OP=1$, $PQ=\sqrt{1-t^2}$ and $PR=dt$, the infinitesimally small length $ds$ of the arc of circle between $P$ and $S$ is given by  $$ds=PS=\dfrac{dt}{\sqrt{1-t^2}}.$$

why for each $t$ between $x$ and $1$?

Because one is evaluating the length of the arc from the generic point $P$ whose $x,y$-coordinates are $(x,\sqrt{1-x^2})$ to the point whose coordinates are $(1,0)$. This length is (see the original post)
$$\int_x^1\dfrac{dt}{\sqrt{1-t^2}}.$$

ADDED: If you apply the arc length formula
\begin{equation*}
s=\int_{x}^{1}\sqrt{1+\left( \frac{dy}{dt}\right) ^{2}}dt
\end{equation*}
to the function $y=\sqrt{1-t^{2}}$, since $\dfrac{dy}{dt}=-\dfrac{t}{\sqrt{1-t^{2}}}$ you get the same result as above
\begin{equation*}
s=\int_{x}^{1}\sqrt{1+\left( -\frac{t}{\sqrt{1-t^{2}}}\right) ^{2}}
dt=\int_{x}^{1}\sqrt{\frac{1}{1-t^{2}}}dt=\int_{x}^{1}\frac{dt}{\sqrt{1-t^{2}
}}.
\end{equation*}
ADDED 2. For further information on the derivation and application of the arc length formula see Paul's Online Math Notes. 
